Instantaneous Velocity Formula
Instantaneous Velocity Formula Questions:
1) A cat that is walking toward a house along the top of a fence is moving at a varying velocity. The cat's position on the fence is 
. Position x is in meters, and time t is in seconds. What is the cat's instantaneous velocity at time t = 10.0 s?
Answer: The cat's velocity can be found using the formula:
The cat's position has only one component, since it is moving in a straight line along the fence. The positive x direction is chosen to be toward the house. The magnitude of the velocity in the x direction is:
This derivative can be solved using the scalar multiple rule and the power rule for derivatives:
The magnitude of the instantaneous velocity is:
The magnitude of the cat's instantaneous velocity at t = 10.0 s is:
The cat's instantaneous velocity at t=10.0 s is 0.05 m/s in the -x direction (away from the house).
2) A child kicks a ball horizontally, off the edge of a cliff. The horizontal position of the ball is given by the function x(t) = bt, where b = 6.0 m/s. The vertical position of the ball is given by the function y(t) = ct2, where c = -4.90 m/s2. At t = 4.0 s, what are the horizontal and vertical components of the instantaneous velocity?
Answer: The components of the instantaneous velocity can be found using the formula:
If the horizontal direction of the ball is defined as the positive x direction, and vertically upward is defined as the positive y direction, then the magnitudes of the x and y components of the instantaneous velocity are:
and
These can be solved using the scalar multiple rule and the power rule for derivatives:
The horizontal instantaneous velocity is:
The horizontal velocity of the ball is a constant value of 6.0 m/s in the +x direction.
The vertical instantaneous velocity is:
vy = c(2t)
vy = 2ct
At t = 4.0 s, the vertical instantaneous velocity is:
vy = 2ct
vy = 2(-4.90 m/s2)(4.0 s)
The vertical instantaneous velocity at t = 4.0 s is 39.2 m/s in the -y direction.
Sours: https://www.softschools.com/formulas/physics/instantaneous_velocity_formula/156/
Learning Objectives
By the end of this section, you will be able to:
- Explain the difference between average velocity and instantaneous velocity.
- Describe the difference between velocity and speed.
- Calculate the instantaneous velocity given the mathematical equation for the velocity.
- Calculate the speed given the instantaneous velocity.
We have now seen how to calculate the average velocity between two positions. However, since objects in the real world move continuously through space and time, we would like to find the velocity of an object at any single point. We can find the velocity of the object anywhere along its path by using some fundamental principles of calculus. This section gives us better insight into the physics of motion and will be useful in later chapters.
Instantaneous Velocity
The quantity that tells us how fast an object is moving anywhere along its path is the instantaneous velocity, usually called simply velocity. It is the average velocity between two points on the path in the limit that the time (and therefore the displacement) between the two points approaches zero. To illustrate this idea mathematically, we need to express position x as a continuous function of t denoted by x(t). The expression for the average velocity between two points using this notation is [latex] \overset{\text{–}}{v}=\frac{x({t}_{2})-x({t}_{1})}{{t}_{2}-{t}_{1}} [/latex]. To find the instantaneous velocity at any position, we let [latex] {t}_{1}=t [/latex] and [latex] {t}_{2}=t+\text{Δ}t [/latex]. After inserting these expressions into the equation for the average velocity and taking the limit as [latex] \text{Δ}t\to 0 [/latex], we find the expression for the instantaneous velocity:
[latex] v(t)=\underset{\text{Δ}t\to 0}{\text{lim}}\frac{x(t+\text{Δ}t)-x(t)}{\text{Δ}t}=\frac{dx(t)}{dt}. [/latex]
Instantaneous Velocity
The instantaneous velocity of an object is the limit of the average velocity as the elapsed time approaches zero, or the derivative of x with respect to t:
[latex] v(t)=\frac{d}{dt}x(t). [/latex]
Like average velocity, instantaneous velocity is a vector with dimension of length per time. The instantaneous velocity at a specific time point [latex] {t}_{0} [/latex] is the rate of change of the position function, which is the slope of the position function [latex] x(t) [/latex] at [latex] {t}_{0} [/latex]. (Figure) shows how the average velocity [latex] \overset{\text{–}}{v}=\frac{\text{Δ}x}{\text{Δ}t} [/latex] between two times approaches the instantaneous velocity at [latex] {t}_{0}. [/latex] The instantaneous velocity is shown at time [latex] {t}_{0} [/latex], which happens to be at the maximum of the position function. The slope of the position graph is zero at this point, and thus the instantaneous velocity is zero. At other times, [latex] {t}_{1},{t}_{2} [/latex], and so on, the instantaneous velocity is not zero because the slope of the position graph would be positive or negative. If the position function had a minimum, the slope of the position graph would also be zero, giving an instantaneous velocity of zero there as well. Thus, the zeros of the velocity function give the minimum and maximum of the position function.
Figure 3.6 In a graph of position versus time, the instantaneous velocity is the slope of the tangent line at a given point. The average velocities [latex] \overset{\text{–}}{v}=\frac{\text{Δ}x}{\text{Δ}t}=\frac{{x}_{\text{f}}-{x}_{\text{i}}}{{t}_{\text{f}}-{t}_{\text{i}}} [/latex] between times [latex] \text{Δ}t={t}_{6}-{t}_{1},\text{Δ}t={t}_{5}-{t}_{2},\text{and}\,\text{Δ}t={t}_{4}-{t}_{3} [/latex] are shown. When [latex] \text{Δ}t\to 0 [/latex], the average velocity approaches the instantaneous velocity at [latex] t={t}_{0} [/latex].
Example
Finding Velocity from a Position-Versus-Time Graph
Given the position-versus-time graph of (Figure), find the velocity-versus-time graph.
Figure 3.7 The object starts out in the positive direction, stops for a short time, and then reverses direction, heading back toward the origin. Notice that the object comes to rest instantaneously, which would require an infinite force. Thus, the graph is an approximation of motion in the real world. (The concept of force is discussed in Newton’s Laws of Motion.)
Strategy
The graph contains three straight lines during three time intervals. We find the velocity during each time interval by taking the slope of the line using the grid.
Solution
Time interval 0.5 s to 1.0 s: [latex] \overset{\text{–}}{v}=\frac{\text{Δ}x}{\text{Δ}t}=\frac{0.0\,\text{m}-0.0\,\text{m}}{1.0\,\text{s}-0.5\,\text{s}}=0.0\,\text{m/s} [/latex]
Time interval 1.0 s to 2.0 s: [latex] \overset{\text{–}}{v}=\frac{\text{Δ}x}{\text{Δ}t}=\frac{0.0\,\text{m}-0.5\,\text{m}}{2.0\,\text{s}-1.0\,\text{s}}=-0.5\,\text{m/s} [/latex]
The graph of these values of velocity versus time is shown in (Figure).
Figure 3.8 The velocity is positive for the first part of the trip, zero when the object is stopped, and negative when the object reverses direction.
Significance
During the time interval between 0 s and 0.5 s, the object’s position is moving away from the origin and the position-versus-time curve has a positive slope. At any point along the curve during this time interval, we can find the instantaneous velocity by taking its slope, which is +1 m/s, as shown in (Figure). In the subsequent time interval, between 0.5 s and 1.0 s, the position doesn’t change and we see the slope is zero. From 1.0 s to 2.0 s, the object is moving back toward the origin and the slope is −0.5 m/s. The object has reversed direction and has a negative velocity.
Speed
In everyday language, most people use the terms speed and velocity interchangeably. In physics, however, they do not have the same meaning and are distinct concepts. One major difference is that speed has no direction; that is, speed is a scalar.
We can calculate the average speed by finding the total distance traveled divided by the elapsed time:
[latex] \text{Average speed}=\overset{\text{–}}{s}=\frac{\text{Total distance}}{\text{Elapsed time}}. [/latex]
Average speed is not necessarily the same as the magnitude of the average velocity, which is found by dividing the magnitude of the total displacement by the elapsed time. For example, if a trip starts and ends at the same location, the total displacement is zero, and therefore the average velocity is zero. The average speed, however, is not zero, because the total distance traveled is greater than zero. If we take a road trip of 300 km and need to be at our destination at a certain time, then we would be interested in our average speed.
However, we can calculate the instantaneous speed from the magnitude of the instantaneous velocity:
[latex] \text{Instantaneous speed}=|v(t)|. [/latex]
If a particle is moving along the x-axis at +7.0 m/s and another particle is moving along the same axis at −7.0 m/s, they have different velocities, but both have the same speed of 7.0 m/s. Some typical speeds are shown in the following table.
| Speed | m/s | mi/h |
|---|---|---|
| Continental drift | [latex] {10}^{-7} [/latex] | [latex] 2\,×\,{10}^{-7} [/latex] |
| Brisk walk | 1.7 | 3.9 |
| Cyclist | 4.4 | 10 |
| Sprint runner | 12.2 | 27 |
| Rural speed limit | 24.6 | 56 |
| Official land speed record | 341.1 | 763 |
| Speed of sound at sea level | 343 | 768 |
| Space shuttle on reentry | 7800 | 17,500 |
| Escape velocity of Earth* | 11,200 | 25,000 |
| Orbital speed of Earth around the Sun | 29,783 | 66,623 |
| Speed of light in a vacuum | 299,792,458 | 670,616,629 |
Calculating Instantaneous Velocity
When calculating instantaneous velocity, we need to specify the explicit form of the position function x(t). For the moment, let’s use polynomials [latex] x(t)=A{t}^{n} [/latex], because they are easily differentiated using the power rule of calculus:
[latex] \frac{dx(t)}{dt}=nA{t}^{n-1}. [/latex]
The following example illustrates the use of (Figure).
Example
Instantaneous Velocity Versus Average Velocity
The position of a particle is given by [latex] x(t)=3.0t+0.5{t}^{3}\,\text{m} [/latex].
- Using (Figure) and (Figure), find the instantaneous velocity at [latex] t=2.0 [/latex] s.
- Calculate the average velocity between 1.0 s and 3.0 s.
Strategy(Figure) gives the instantaneous velocity of the particle as the derivative of the position function. Looking at the form of the position function given, we see that it is a polynomial in t. Therefore, we can use (Figure), the power rule from calculus, to find the solution. We use (Figure) to calculate the average velocity of the particle.
Solution
- [latex] v(t)=\frac{dx(t)}{dt}=3.0+1.5{t}^{2}\,\text{m/s} [/latex].Substituting t = 2.0 s into this equation gives [latex] v(2.0\,\text{s})=[3.0+1.5{(2.0)}^{2}]\,\text{m/s}=9.0\,\text{m/s} [/latex].
- To determine the average velocity of the particle between 1.0 s and 3.0 s, we calculate the values of x(1.0 s) and x(3.0 s):
[latex] x(1.0\,\text{s})=[(3.0)(1.0)+0.5{(1.0)}^{3}]\,\text{m}=3.5\,\text{m} [/latex]
[latex] x(3.0\,\text{s})=[(3.0)(3.0)+0.5{(3.0)}^{3}]\,\text{m}=22.5\,\text{m.} [/latex]
Then the average velocity is
[latex] \overset{\text{–}}{v}=\frac{x(3.0\,\text{s})-x(1.0\,\text{s})}{t(3.0\,\text{s})-t(1.0\,\text{s})}=\frac{22.5-3.5\,\text{m}}{3.0-1.0\,\text{s}}=9.5\,\text{m/s}\text{.} [/latex]
Significance
In the limit that the time interval used to calculate [latex] \overset{\text{−}}{v} [/latex] goes to zero, the value obtained for [latex] \overset{\text{−}}{v} [/latex] converges to the value of v.
Example
Instantaneous Velocity Versus Speed
Consider the motion of a particle in which the position is [latex] x(t)=3.0t-3{t}^{2}\,\text{m} [/latex].
- What is the instantaneous velocity at t = 0.25 s, t = 0.50 s, and t = 1.0 s?
- What is the speed of the particle at these times?
Strategy
The instantaneous velocity is the derivative of the position function and the speed is the magnitude of the instantaneous velocity. We use (Figure) and (Figure) to solve for instantaneous velocity.
Solution
- Show Answer
[latex] v(t)=\frac{dx(t)}{dt}=3.0-6.0t\,\text{m/s} [/latex]
- Show Answer
[latex] v(0.25\,\text{s})=1.50\,\text{m/s,}v(0.5\,\text{s})=0\,\text{m/s,}v(1.0\,\text{s})=-3.0\,\text{m/s} [/latex]
- Show Answer
[latex] \text{Speed}=|v(t)|=1.50\,\text{m/s},0.0\,\text{m/s,}\,\text{and}\,3.0\,\text{m/s} [/latex]
Significance
The velocity of the particle gives us direction information, indicating the particle is moving to the left (west) or right (east). The speed gives the magnitude of the velocity. By graphing the position, velocity, and speed as functions of time, we can understand these concepts visually (Figure). In (a), the graph shows the particle moving in the positive direction until t = 0.5 s, when it reverses direction. The reversal of direction can also be seen in (b) at 0.5 s where the velocity is zero and then turns negative. At 1.0 s it is back at the origin where it started. The particle’s velocity at 1.0 s in (b) is negative, because it is traveling in the negative direction. But in (c), however, its speed is positive and remains positive throughout the travel time. We can also interpret velocity as the slope of the position-versus-time graph. The slope of x(t) is decreasing toward zero, becoming zero at 0.5 s and increasingly negative thereafter. This analysis of comparing the graphs of position, velocity, and speed helps catch errors in calculations. The graphs must be consistent with each other and help interpret the calculations.
Figure 3.9 (a) Position: x(t) versus time. (b) Velocity: v(t) versus time. The slope of the position graph is the velocity. A rough comparison of the slopes of the tangent lines in (a) at 0.25 s, 0.5 s, and 1.0 s with the values for velocity at the corresponding times indicates they are the same values. (c) Speed: [latex] |v(t)| [/latex] versus time. Speed is always a positive number.
Check Your Understanding
The position of an object as a function of time is [latex] x(t)=-3{t}^{2}\,\text{m} [/latex]. (a) What is the velocity of the object as a function of time? (b) Is the velocity ever positive? (c) What are the velocity and speed at t = 1.0 s?
(a) Taking the derivative of x(t) gives v(t) = −6t m/s. (b) No, because time can never be negative. (c) The velocity is v(1.0 s) = −6 m/s and the speed is [latex] |v(1.0\,\text{s})|=6\,\text{m/s} [/latex].
Summary
- Instantaneous velocity is a continuous function of time and gives the velocity at any point in time during a particle’s motion. We can calculate the instantaneous velocity at a specific time by taking the derivative of the position function, which gives us the functional form of instantaneous velocity v(t).
- Instantaneous velocity is a vector and can be negative.
- Instantaneous speed is found by taking the absolute value of instantaneous velocity, and it is always positive.
- Average speed is total distance traveled divided by elapsed time.
- The slope of a position-versus-time graph at a specific time gives instantaneous velocity at that time.
Conceptual Questions
There is a distinction between average speed and the magnitude of average velocity. Give an example that illustrates the difference between these two quantities.
Average speed is the total distance traveled divided by the elapsed time. If you go for a walk, leaving and returning to your home, your average speed is a positive number. Since Average velocity = Displacement/Elapsed time, your average velocity is zero.
Does the speedometer of a car measure speed or velocity?
If you divide the total distance traveled on a car trip (as determined by the odometer) by the elapsed time of the trip, are you calculating average speed or magnitude of average velocity? Under what circumstances are these two quantities the same?
Average speed. They are the same if the car doesn’t reverse direction.
How are instantaneous velocity and instantaneous speed related to one another? How do they differ?
Problems
A woodchuck runs 20 m to the right in 5 s, then turns and runs 10 m to the left in 3 s. (a) What is the average velocity of the woodchuck? (b) What is its average speed?
Sketch the velocity-versus-time graph from the following position-versus-time graph.
Sketch the velocity-versus-time graph from the following position-versus-time graph.
Given the following velocity-versus-time graph, sketch the position-versus-time graph.
An object has a position function x(t) = 5t m. (a) What is the velocity as a function of time? (b) Graph the position function and the velocity function.
A particle moves along the x-axis according to [latex] x(t)=10t-2{t}^{2}\,\text{m} [/latex]. (a) What is the instantaneous velocity at t = 2 s and t = 3 s? (b) What is the instantaneous speed at these times? (c) What is the average velocity between t = 2 s and t = 3 s?
a. [latex] v(t)=(10-4t)\text{m/s} [/latex]; v(2 s) = 2 m/s, v(3 s) = −2 m/s; b. [latex] |v(2\,\text{s})|=2\,\text{m/s},|v(3\,\text{s})|=2\,\text{m/s} [/latex]; (c) [latex] \overset{\text{–}}{v}=0\,\text{m/s} [/latex]
Unreasonable results. A particle moves along the x-axis according to [latex] x(t)=3{t}^{3}+5t\text{} [/latex]. At what time is the velocity of the particle equal to zero? Is this reasonable?
Glossary
- instantaneous velocity
- the velocity at a specific instant or time point
- instantaneous speed
- the absolute value of the instantaneous velocity
- average speed
- the total distance traveled divided by elapsed time
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Displacement, velocity, and time
Video transcript
Instantaneous velocity tells us about the motion of a particle at a specific instant of time anywhere along its path.
Instantaneous velocityis taken as the limit of average velocity as the time tends towards zero.To Calculate Vinst we can use the displacement-time graph/ Instantaneous Velocity Formula. i.e.,the derivative of displacement (s) with respect to time(t) taken.
![\[\vec{V_{inst}}=\lim_{\Delta t\rightarrow 0}\frac{\Delta\vec {s} }{\Delta t}=\frac{d \vec{s}}{dt}\]](https://lambdageeks.com/wp-content/ql-cache/quicklatex.com-1e3992051897fa96339202d3a94461ba_l3.png)
To know how to calculate instantaneous velocity of an object, we have steps to follow. Let us see it with an example.
Consider an equation for velocity in terms of position/displacement.
To calculate instantaneous velocity, we must consider an equation that tells us its position‘s’ at a certain time ‘t’. It means the equation must contain the variable ‘s‘ on one side and ‘t‘ on the other side,
s = -2t2 + 10t +5 at t = 2 second.
In this equation, the variables are:
Displacement = s, measured in meters.
Time = t, measured in seconds.
Consider the derivative of the given equation.
To find the derivative of a given displacement equation, differentiate the function with respect to time,
ds/dt = -(2) 2t (2-1) + (1)10t1 – 1 + (0)5t0
ds/dt= -4t1 + 10t0
ds/dt= -4t + 10
Substitute the given value of “t” in the derivative equation to find instantaneous velocity.
Find the instantaneous velocity at t = 2, substitute “2” for t in the derivative ds/dt = -4t + 10. Then, we can solve the equation,
ds/dt = -4t + 10
ds/dt = -4(2) + 10
ds/dt = -8 + 10
ds/dt = -2 meters/second
Here, “meters/second” is the SI Unit of instantaneous velocity.
How to calculate instantaneous velocity from a graph
Instantaneous velocity at any specific point of time is given by the slope of tangent drawn to the position-time graph at that point.
- Plot a graph of distance vs. time.
- Mark a point at which you have to find instantaneous velocity, say A.
- Determine the point on the graph corresponding to time t1and t2.
- Calculate the vavg and draw a tangent at point A.
- In the graph, vinst at point A is found by tangent, drawn at that point
- Longer the tangent, the more accurate will be the values.
- In the image shown, blue lineis the position vs. time graph, and the red line is an approximated slope for the line at t = 2.5 seconds.
- If we keep choosing points that are closer and closer to one another, the line will begin approaching the slope of the line tangent to a single point.
- If we take the limit of the function at that point, we will get the value of the slope of tangent at that point.
- The distance is approximately 140 m, and the time interval is 4.3s. Therefore, the approximated slope is 32.55 m/s.
How to calculate instantaneous velocity from a position-time graph.
To calculate the instantaneous velocity from a position-time graph.
Plot the displacement function with respect to time.
- Use the x-axis and y-axis to represent time and displacement.
- Then plot the values of time and displacement on the graph.
Choose any two points on the s-t graph.
- The displacement line contains the points (3,6) and (5,8).
- In this example, if we want to find slope at (3,6), we can set A = (3,6) and B=(5,8)
Find the line slope connecting the two points, i.e., between A and B.
Find the average velocity between those two-time intervals, i.e.,
![\[slope=\textbf{K}=\frac{Y_{B}- Y_{A}}{X_{B}-X_{A}}\]](https://lambdageeks.com/wp-content/ql-cache/quicklatex.com-71b02ea4e37b709b45813b44323d967e_l3.png)
where K is the slope between the two points.
Here, the slope between A and B is:
![\[slope=\textbf{K}=\frac{(8-6)}{(5-3)}=2\]](https://lambdageeks.com/wp-content/ql-cache/quicklatex.com-7da94feb02fc9056bcfea0933687275f_l3.png)
Repeat to find slope several times, moving B nearer to A.
- Keep choosing points closer to one another; then, it will begin to approach the slope of the tangent line.
- If we consider the limit of the function at that point, we will get the value of the slope at that point.
- Here we can use the points (4,7.7), (3.5, 6.90), and (3.25, 6.49) for B and the original point of (3,6) for A.
![\[slope=\textbf{K}=\frac{(7.7-6)}{(4-3)}=1.7\]](https://lambdageeks.com/wp-content/ql-cache/quicklatex.com-5117ceaf150946301fa3a88d458af477_l3.png)
![\[slope=\textbf{K}=\frac{(6.90-6)}{(3.5-3)}=1.8\]](https://lambdageeks.com/wp-content/ql-cache/quicklatex.com-645d3749a4eac1c3fd735561489022e2_l3.png)
![\[slope=\textbf{K}=\frac{(6.49-6)}{(3.25-3)}=1.96\]](https://lambdageeks.com/wp-content/ql-cache/quicklatex.com-676c56cda3ac555aee906a9627251aa0_l3.png)
Calculate the slope for an infinitely small interval on the tangent line.
In the example, as we move B closer to A, we get values of 1.7, 1.8, and 1.96 for K. Since these numbers are approximately equal to 2, we can say that 2 is A’s slope.
Here, instantaneous velocity is 2m/s.
Instantaneous velocity formula
In mathematical terms, we can write the instantaneous velocity formula as,
![\[Instantaneous{\enskip} velocity=\frac{Change{\enskip} in {\enskip} position}{Time {\enskip} interval}\]](https://lambdageeks.com/wp-content/ql-cache/quicklatex.com-cc7593bb0e4a199e871cdd100d14b23e_l3.png)
Here, ds/dt is the derivative of displacement (s) with respect to time (t).
The above derivative holds a finite value when both the denominator and the numerator tend to zero.
Instantaneous velocity formula calculus
By using calculus, it is always possible to calculate the velocity of an object at any moment along its path. It is called instantaneous velocity and is given by the equation v = ds/dt.
Instantaneous velocity = limit as change in time approaches zero (change in position/change in time) = derivative of displacement with respect to time
![\[V_{inst}=\lim_{\Delta t\rightarrow 0}\frac{ds}{dt}\]](https://lambdageeks.com/wp-content/ql-cache/quicklatex.com-4f6213a22fabfbcdc7f684f31be98d8b_l3.png)
![\[\vec{V}= instantaneous{\enskip} velocity\]](https://lambdageeks.com/wp-content/ql-cache/quicklatex.com-1ff9e40d221b7d7cd29e817587f31af2_l3.png)
![\[\Delta {\vec{S}} = vector{\enskip} change{\enskip} in {\enskip}position(m)\]](https://lambdageeks.com/wp-content/ql-cache/quicklatex.com-e3f9debff3790caeda37e680508d92ab_l3.png)
![\[\Delta {t} =change{\enskip} in {\enskip}time(s)\]](https://lambdageeks.com/wp-content/ql-cache/quicklatex.com-60124b458b46c3899765c6b7b711441a_l3.png)
![\[\frac{ds} {dt}=derivative{\enskip} of{\enskip} position{\enskip} vector{\enskip} with {\enskip}respect {\enskip}to {\enskip}time (m/s)\]](https://lambdageeks.com/wp-content/ql-cache/quicklatex.com-d8a9ae6c0917eff8b60d69a221df9e14_l3.png)
![\[s = displacement\]](https://lambdageeks.com/wp-content/ql-cache/quicklatex.com-b39aae1c1579547eca7327e9676bfbc7_l3.png)
![\[t = time\]](https://lambdageeks.com/wp-content/ql-cache/quicklatex.com-171cdd7cdd7a0756a7fbdc090787ed02_l3.png)
Average velocity and instantaneous velocity formula
| Formula | Symbol | Definition | |
| Average velocity |  | sf= Final displacement si = Initial displacement tf = Final time ti = Initial time | Average velocity is total distance divided by the total time taken. |
| Instantaneous velocity |  |  | Velocity at any instant of time. |
Instantaneous angular velocity formula
The instantaneous angular velocity is the rate at which a particle moves in a circular path at a particular moment of time.
The instantaneous angular velocity of a rotating object is given by
![\[\omega_{av}=\lim_{\Delta t\rightarrow 0}\frac{\Delta\theta}{\Delta t}=\frac{d \theta}{dt}\]](https://lambdageeks.com/wp-content/ql-cache/quicklatex.com-f5675885d71bcc9fbcfc2295d48ca01f_l3.png)
= derivative of angular positionθ with respect to time, found by taking the limit Δ t → 0 in the average angular velocity.
![\[\omega_{av}=\frac{\theta_{2}- \theta_{1}}{t_{2}-t_{1}}= \frac{\Delta{\theta}}{\Delta t}\]](https://lambdageeks.com/wp-content/ql-cache/quicklatex.com-4ee7b44c0fc17a1294718a76190949ed_l3.png)
The direction of the angular velocity in a circular path is along the axis of rotation and points away from you for a body rotating clockwise and toward you for a body rotating anticlockwise. In mathematics, this is generally described by the right-hand rule.
Instantaneous velocity and speed formula
The formula of instantaneous velocity
![\[\vec{ V_{inst}}=\lim_{\Delta t\rightarrow 0}\frac{\Delta\vec {s} }{\Delta t}=\frac{d \vec{s}}{dt}\]](https://lambdageeks.com/wp-content/ql-cache/quicklatex.com-58cb366dad1378d55cd74605e9b3f86f_l3.png)
The formula for instantaneous speed
![\[speed_{inst}=\frac{ds}{dt}\]](https://lambdageeks.com/wp-content/ql-cache/quicklatex.com-8968ed3012cab5b7e34f5e8d3dddb8aa_l3.png)
Difference between Instantaneous Speed and Instantaneous Velocity.
| Instantaneous velocity | Instantaneous speed |
| It is the velocity of a particle in motion at a particular moment of t. | It is the measure of speed of a particle at a specific moment of t. |
| Instantaneous velocity measures how fast and in which direction an object is moving. | Instantaneous speed measures how quickly a particle is moving in motion. |
| Vector quantity | Scalar quantity |
Instantaneous velocity definition and formula
Instantaneous velocity definition
Instantaneous velocity is described as the velocity of an object in motion. We can find it by using average velocity, but we must narrow the time to approach zero.
In total, we can say that instantaneous velocity is the velocity of a particle in motion at a particular instant of time.
Instantaneous velocity formula
For any equation of motion s(t), for instantaneous velocity as t approaches zero, we can write the formula as,
Instantaneous velocitylimit formula
The instantaneous velocity of any object is the limit of the average velocity as the time approaches zero.
![\[Instantaneous {\enskip}velocity = v = \frac{\Delta s(t)}{\Delta t}\]](https://lambdageeks.com/wp-content/ql-cache/quicklatex.com-6d6b5c182b53473794d5eac5a8ff1e88_l3.png)
![\[Instantaneous{\enskip} velocity=\vec{ V_{inst}}=\lim_{\Delta t\rightarrow 0}\frac{\Delta\vec {s} }{\Delta t}\]](https://lambdageeks.com/wp-content/ql-cache/quicklatex.com-481dc050cd1b04458f7c91958eee7022_l3.png)
![\[\vec{ V_{inst}}=\lim_{\Delta t\rightarrow 0}\frac{\Delta\vec {s} }{\Delta t}=\frac{s(t_{2})- s(t_{1})}{t_{2}-t_{1}}\]](https://lambdageeks.com/wp-content/ql-cache/quicklatex.com-8931b1184d9c1b1c11644d891b8c3c0b_l3.png)
![\[\vec{ V_{inst}}=\lim_{\Delta t\rightarrow 0}\frac{\Delta\vec {s} }{\Delta t}=\frac{s(t + \Delta t)- s(t)}{(t+{\Delta t})-t}\]](https://lambdageeks.com/wp-content/ql-cache/quicklatex.com-39b0217d0a82bc6a41d2c15256234a14_l3.png)
![\[\vec{ V_{inst}}=\lim_{\Delta t\rightarrow 0}\frac{\Delta\vec {s} }{\Delta t}=\frac{s(t + \Delta t)- s(t)}{\Delta t}\]](https://lambdageeks.com/wp-content/ql-cache/quicklatex.com-49540a402f2ddbb949c55e721574775e_l3.png)
Insert the values of t1= t and t2 = t + Δt into the equation for the average velocity and take the limit as Δt→0, we find the instantaneous velocity limit formula
How do you find instantaneous velocity on a graph
Instantaneous velocity equals the slope of tangent line of the position-time graph.
Instantaneous Velocity interpretation from s-t graph
- Instantaneous velocity equals the slope of tangent line of the position-time graph.
- Instantaneous Velocity interpretation from s-t graph
- The slope of the purple line (tangent) in the displacement v/s time graph gives instantaneous velocity.
- If the purple line makes an angle
with the positive x-axis.
= slope of purple line =
How do you find instantaneous velocity from average velocity
To find the instantaneous velocity at a point, we have to first find the average velocity at that point.
You can find the instantaneous velocity at t=a by calculating the average velocity of the position vs. time graph by taking the smaller and larger increments of a point at which you want to determine V
Instantaneous velocity example
While riding his bicycle, a cyclist changes his velocity depending on the distance and time he travels.
If we want to find the velocity at one particular point, we must use instantaneous velocity.
Let us see an example,
a). Find out the Instantaneous Velocity of a particle traveling along a straight path for t=2 seconds, with a position function “s” defined as 4t² + 2t + 3?
Solution:
Given s = 4t² + 2t + 3
Differentiate the given function with respect to time, we calculate the Instantaneous Velocity as follows:
Substitute value of t = 2, we get the instantaneous velocity as,
![\[v_{inst} = \frac{ds}{dt}\]](https://lambdageeks.com/wp-content/ql-cache/quicklatex.com-cfe546f5c81c56dc72b865541507a832_l3.png)
Substituting function s,
![\[v_{inst} = \frac{d(4t^2 +2t +3)}{dt}} \]](https://lambdageeks.com/wp-content/ql-cache/quicklatex.com-f98520c8985f132187c616a05a7fe554_l3.png)
![\[v_{inst} =8t+2\]](https://lambdageeks.com/wp-content/ql-cache/quicklatex.com-450fd97ab9c07728ae829e466116258d_l3.png)
![\[v_{inst} = (8 * 2)+2\]](https://lambdageeks.com/wp-content/ql-cache/quicklatex.com-0bfb66d6538246498eb668cfe09d4461_l3.png)
![\[v_{inst} =18 ms ^{-1}\]](https://lambdageeks.com/wp-content/ql-cache/quicklatex.com-c23f9290a5e0f4f39e3a6ea139485b78_l3.png)
Thus, the instantaneous velocity for the above function is 18 m/s.
Instantaneous velocity problem
Some Instantaneous velocity problems,
Problem 1:
The motion of the truck is given by the function s = 3t2 + 10t + 5. Calculate its Instantaneous Velocity at time t = 4s.
Solution:
Given function is s = 3t2 + 10t + 5.
Differentiate the above function with respect to time, we get
![\[{v_{inst} = \frac{ds}{dt}=\frac{d(3t^2 +10t +5)}{dt}}\]](https://lambdageeks.com/wp-content/ql-cache/quicklatex.com-5608095c05c72ace9a4d11618805802f_l3.png)
Substituting function s,
[v_{inst} = v(t)=6t+10]
Substitute value of t = 4s, we get the instantaneous velocity as,
![\[v(4)= 6(4)+10\]](https://lambdageeks.com/wp-content/ql-cache/quicklatex.com-9391fb95ae5b19490104d1b7017a65b1_l3.png)
![\[v(4) =34 ms ^{-1}\]](https://lambdageeks.com/wp-content/ql-cache/quicklatex.com-99fd11836ad3d0a47ae80b2cce44c43d_l3.png)
For the given function, Instantaneous Velocity is 34m/s
Problem 2:
A bullet fired travels along a straight path, and its equation of motion is S(t) = 3t + 5t2. So, for example, if it travels for 12 seconds before impact, find the instantaneous velocity at t = 7s.
Solution: We know the equation of motion:
![\[s(t) = 3t + 5t^2\]](https://lambdageeks.com/wp-content/ql-cache/quicklatex.com-30698bcd54653987ba6f608cfab5d401_l3.png)
![\[v_{inst} = \frac{ds}{dt} = \frac{d(3t +5t^2)}{dt}=3+10t}\]](https://lambdageeks.com/wp-content/ql-cache/quicklatex.com-bfe4214783fcef422a8e6a7e6cd175d8_l3.png)
![\[v_{inst} {\enskip} at (t = 7) = 3 + (10 * 7)\]](https://lambdageeks.com/wp-content/ql-cache/quicklatex.com-fcba84c67b4bb44c0ad03022c9aefd05_l3.png)
![\[v_{inst} = 73 m/s.\]](https://lambdageeks.com/wp-content/ql-cache/quicklatex.com-d1f4344c523d82e616ec977e617dbce5_l3.png)
Problem 3:
An object is released from a certain height to fall freely under the influence of gravitation. The equation of motion for displacement is s(t) = 5.1 t2. What will be the instantaneous velocity of an object at t=6s after release?
Solution:
The equation of motion is
s(t) = 5.1 t2
Instantaneous velocity at t = 6s
![\[v_{inst} = [\frac{ds}{dt}]_{t=6} = [\frac{d(3t +5t^2)}{dt}]_{t=6}=3+10t}\]](https://lambdageeks.com/wp-content/ql-cache/quicklatex.com-3d576a0285bb9454c0381f153f61fbf3_l3.png)
![\[v_{inst} = [5.1 * 2 * t]_{t=6}\]](https://lambdageeks.com/wp-content/ql-cache/quicklatex.com-7e3f276920fa5411ee4613db8a0acf74_l3.png)
![\[v_{inst} = [5.1 * 2 * 6]\]](https://lambdageeks.com/wp-content/ql-cache/quicklatex.com-0a55321b11c3a3f77214f0f98d60fd6a_l3.png)
![\[v_{inst} = 61.2 ms ^{-1}\]](https://lambdageeks.com/wp-content/ql-cache/quicklatex.com-ef450e1ddd00e795798caae321d105b1_l3.png)
Problem 4:
Find the velocity at t = 2, given thedisplacement equation is s = 3t3 – 3t2 + 2t + 7.
Solution:
It is just like previous problems, except they have given a cubic equation instead of a quadratic equation to solve it in the same way.
The equation of motion is
s(t) = 3t3 – 3t2 + 2t + 7.
![\[v_{inst} = \frac{ds}{dt} = \frac{d(3t^3+3t^2 +2t+7)}{dt}=(3*3t^2) - (2 * 3t) + 2}\]](https://lambdageeks.com/wp-content/ql-cache/quicklatex.com-fd5b812d38970c7ab8553dfb6949f378_l3.png)
![\[v_{inst} = [9t^2-6t+2]\]](https://lambdageeks.com/wp-content/ql-cache/quicklatex.com-04551e923ea44f243bc122ece3b5ed44_l3.png)
Instantaneous velocity at t = 7s
![\[v_{inst} = 9(7)^{2} - 6(7) +2\]](https://lambdageeks.com/wp-content/ql-cache/quicklatex.com-9944447f97ce8c762a7a996e0a677751_l3.png)
![\[v_{inst} = 441 - 42 +2\]](https://lambdageeks.com/wp-content/ql-cache/quicklatex.com-366b37971a06a8de817cb562a360f749_l3.png)
![\[v_{inst} = 401{\enskip} meters/second\]](https://lambdageeks.com/wp-content/ql-cache/quicklatex.com-8eb5a15149315a945fe70a1164a3e62b_l3.png)
Problem 5:
The position of a person moving along a straight line is given by s(t)= 7t2+ 3t + 19, where t is time(seconds). Find the equation for instantaneous velocity v(t) of the particle at time t.
Solution:
Given: s(t)= 7t2+ 3t + 19
![\[v_{inst} = \frac{ds}{dt} = \frac{d(7t^2 +3t+19)}{dt}\]](https://lambdageeks.com/wp-content/ql-cache/quicklatex.com-03827695fdd6bd86d1861d286086e764_l3.png)
![\[v_{inst} = 14t+ 3\]](https://lambdageeks.com/wp-content/ql-cache/quicklatex.com-2e281a3e7aee31e99c4ea84fc966cea8_l3.png)
vinst = v(t) = (14t + 3) m/s is equation for instantaneous velocity.
Suppose if we assume t = 3s, then
![\[v_{inst}=v(t) = [14(3) + 3)] = 45 m/s\]](https://lambdageeks.com/wp-content/ql-cache/quicklatex.com-13d8a1f5530b8f6ea2946bc3bf3ec3d8_l3.png)
Problem 6:
The motion of an auto is described by the equation of motion s = gt2 + b, where b=20 m and g = 12 m. Therefore, find the instantaneous velocity at t=4s.
Solution:
s(t) = gt2 + b
v (t) = 2gt + 0
v (t) = 2gt
Here, g = 12 and t = 4s,
v (4) = [2 x 12 x 4] = 96 m/s.
v (t) = 96 m/s.
Problem 7:
A table dropped off a 1145 ft building, has a height (in feet) above the ground is given by s(t) = 1145 -12 t2. Then, calculate the instantaneous velocity of the table at 3s?
Solution:
![\[ V_{inst}=\lim_{\Delta t\rightarrow 0}\frac{\Delta {s} }{\Delta t}=\frac{s(t_{2})- s(t_{1})}{t_{2}-t_{1}}\]](https://lambdageeks.com/wp-content/ql-cache/quicklatex.com-52c92a0654107516381c489dbd2cd8ec_l3.png)
![\[ V_{inst}=\lim_{\Delta t\rightarrow 0}\frac{\Delta {s} }{\Delta t}=\frac{s(t + \Delta t)- s(t)}{(t+{\Delta t})-t}\]](https://lambdageeks.com/wp-content/ql-cache/quicklatex.com-1b8a811b415a88aaabd78c5a8e0b39aa_l3.png)
![\[ V_{inst}=\lim_{a \rightarrow 0}\frac{[1145 - 12((t + \Delta t)^{2}]-[1145-12(t)^{2}]} {\Delta t}\]](https://lambdageeks.com/wp-content/ql-cache/quicklatex.com-445b46fd66f851b53fc6fd1723eb33d2_l3.png)
![\[consider{\enskip} \Delta {t} = a {\enskip} and {\enskip} t =3s\]](https://lambdageeks.com/wp-content/ql-cache/quicklatex.com-5788bb5e12d108bd088d9cfbdf8829ba_l3.png)
![\[ V_{inst}=\lim_{a \rightarrow 0}\frac{[1145 - 12(3 + a)^{2}]-[1145-12(3)^{2}]} {a}\]](https://lambdageeks.com/wp-content/ql-cache/quicklatex.com-dd97af9d28ce61e933789d709b3ad1d5_l3.png)
![\[ V_{inst}=\lim_{a \rightarrow 0}\frac{[1145 - 12(3^{2} + a ^{2} + 6a]-[1145-12(9)]} {a}\]](https://lambdageeks.com/wp-content/ql-cache/quicklatex.com-d5a957bd2d2eb87a74a5ded4473b4046_l3.png)
![\[ V_{inst}=\lim_{a \rightarrow 0}\frac{[1145 - 108 - 12a ^{2} - 72a]-1145 + 108]} {a}\]](https://lambdageeks.com/wp-content/ql-cache/quicklatex.com-22e9209c6f50d80c490f9985c90b0961_l3.png)
![\[ V_{inst}=\lim_{a \rightarrow 0}\frac{ - 12a ^{2} - 72a} {a}\]](https://lambdageeks.com/wp-content/ql-cache/quicklatex.com-d0950948deceabb9f98b67a05b4b4a5b_l3.png)
![\[ V_{inst}=\lim_{a \rightarrow 0}\frac{ - 12a - 72} {1}\]](https://lambdageeks.com/wp-content/ql-cache/quicklatex.com-10a344f49dc8ec278a92b7c10f2ce06a_l3.png)
![\[ V_{inst}= -72 m/s\]](https://lambdageeks.com/wp-content/ql-cache/quicklatex.com-133365403d5e2cf7b436e27200749d00_l3.png)
Instantaneous velocity at t = 3s is -72m/s.
Problem 8:
A particles position function is given by s = (3t2)i – (4t)k + 2. what is its instantaneous velocity at t=2? What is its instantaneous acceleration as a function of time?
Solution:
s(t) = (3t2)i – (4t)k +2
v (t) = (6t)i – 4k…………..(Eq.1)
v (2) = (6 * 2)i – 4k
v (2) = 12i – 4k m/s
To calculate instantaneous acceleration as a function of time
a (t) = v1(t)
differentiate Eq.1 w.r.to t, we get
a (t) = 6i m/s
Problem 9:
The position of an insect is given by s = 44 + 20t – 3t3, where t is in seconds and s is in meters.
a. Find the average velocity of object between t = 0 and t = 4 s.
b. At what time between 0 and 4 is the instantaneous velocity zero.
solution:
To calculate average velocity
![\[\vec{ V_{avg}}=\frac{d\vec {s}}{dt}=\frac{s_{f}- s_{i}}{t_{f}-t_{i}}=\frac{s(4)- s(0)}{4-0}\]](https://lambdageeks.com/wp-content/ql-cache/quicklatex.com-6b7ece0a24625fa03277b013776955fb_l3.png)
![\[\vec{v_{avg}}= \frac{[44 + 20(4) - 3 (4)^{3} ] - 44]}{4}\]](https://lambdageeks.com/wp-content/ql-cache/quicklatex.com-ff61ae77b01c1f12c767aa85c2717de4_l3.png)
![\[\vec{v_{avg}} = -28 m/s\]](https://lambdageeks.com/wp-content/ql-cache/quicklatex.com-c465dec943511df655f46e096a93e884_l3.png)
To find the time at which instantaneous velocity is zero.
![\[\vec{ V_{inst}}=\frac{d\vec {s}}{dt}= 20-9 t^{2}\]](https://lambdageeks.com/wp-content/ql-cache/quicklatex.com-76909d6d4f569d1d3c1520776ac5823b_l3.png)
![\[20-9 t^{2}=0\]](https://lambdageeks.com/wp-content/ql-cache/quicklatex.com-9a0d4bfee218fe8237a78714893dcc6c_l3.png)
![\[ t = \sqrt {20}{9}\]](https://lambdageeks.com/wp-content/ql-cache/quicklatex.com-7324f4b18482a9015090e1f41a3b1b5f_l3.png)
![\[t = 1.49 s\]](https://lambdageeks.com/wp-content/ql-cache/quicklatex.com-23069848fba109bc3d98d436c2c0470e_l3.png)
Problem 10:
A particle is in motion with displacement function s = t2 + 3.
Find the position at t = 2.
Find average velocity from t = 2 to t = 3.
Find its instantaneous velocity at t = 2.
Solution:
To find position at t = 2
s(t) = t2 + 3
s (2) = (2)2 + 3
s (2) = 7
To find theaverage velocity.
![\[\vec{ V_{avg}}=\frac{\Delta\vec {s}}{\Delta t}\]](https://lambdageeks.com/wp-content/ql-cache/quicklatex.com-1e6937e4f632de3bb0e3f92529ec7b1c_l3.png)
![\[\vec{ V_{avg}}=\frac{s_{f}- s_{i}}{t_{f}-t_{i}}=\frac{s(12)- s(7)}{3-2}= 5 m/s\]](https://lambdageeks.com/wp-content/ql-cache/quicklatex.com-486ab3e6b45382401b252004aa16e38c_l3.png)
To find instantaneous velocity
![\[\vec{ V_{inst}}=\frac{d\vec {s}}{dt}\]](https://lambdageeks.com/wp-content/ql-cache/quicklatex.com-8f7d70b5ed23ac52a30386b04ae89112_l3.png)
![\[\vec{ V_{inst}}= 2t\]](https://lambdageeks.com/wp-content/ql-cache/quicklatex.com-f15ccc2d9866e77a79ea695fd3252cc4_l3.png)
At t = 2s
![\[\vec{ V_{inst}} = 2(2) = 4 m/s \]](https://lambdageeks.com/wp-content/ql-cache/quicklatex.com-d01a63c9d18887cf5c20f5566b8e0a3e_l3.png)
Instantaneous velocity vs. average velocity
| Instantaneous velocity | Average velocity |
| The instantaneous velocity is the average velocity between two points. | Average velocity is the ratio of change of distance with respect to time over a period. |
| Instantaneous velocity tells about the motion between two points on the path taken. | Average velocity does not give information about motion between the points. The path may be straight/curved, and the motion may be steady/variable. |
| Instantaneous velocity is equal to slope of the tangent of displacement(s) vs. the time graph. | It equals to the slope of the secant lineof the s-t graph. |
| vector | vector |
How to findinstantaneous velocity without calculus
We can find instantaneous velocity by approximation on the displacement vs. time graph without calculus at a particular point. We need to draw a tangent at a point along the curved line and estimate the slope where you need to find instantaneous velocity.
How do I calculate instantaneous velocity and instantaneous acceleration
| Instantaneous velocity | Instantaneous acceleration | |
| From formula | To calculate Instantaneous velocity, take the limit of change of distance with respect to time taken as time approaches zero. i.e., by taking the first derivate of the displacement function. | To calculate Instantaneous acceleration, take the limit of change of velocity with respect to time as the change in time approaches zero. i.e., by taking the second derivate of the displacement function. |
| From graph | Equal to slope of the tangent of the s-t graph. | Equal to slope of the tangent of the v-t graph. |
Problem 11:
A bullet fired in space travels along a straight path, and its equation of motion is s(t) = 2t +4t2. If it travels for 12 seconds before impact, find the instantaneous velocity and instantaneous acceleration at the t = 3s.
Solution: We know the equation of motion: s(t) = 2t + 4t2
![\[v_{inst}= \frac{ds}{dt} = \frac{d(2t+4t^{2})}{dt} = 2+ 8t\]](https://lambdageeks.com/wp-content/ql-cache/quicklatex.com-40fe9be54fd5539cf6d23884daff1e25_l3.png)
![\[v_{inst} {\enskip}at{\enskip} v(t=7) = 2 + (8 * 3)\]](https://lambdageeks.com/wp-content/ql-cache/quicklatex.com-f2c1878bd35dbb609baf272fb9bf1c9d_l3.png)
![\[v_{inst}=26m/s\]](https://lambdageeks.com/wp-content/ql-cache/quicklatex.com-31200731c8f738a6d0047f49adf23ae7_l3.png)
![\[a(t)= \frac{dv}{dt} = \frac{d(2+8t)}{dt} = 8\]](https://lambdageeks.com/wp-content/ql-cache/quicklatex.com-622ac2bc85ef8f9763df8f30b73515db_l3.png)
![\[a(t)=8 m/s\]](https://lambdageeks.com/wp-content/ql-cache/quicklatex.com-da0055e1ee9ab34119d6f26e6400c9be_l3.png)
How to find instantaneous speed and velocity
Instantaneous speed is given as the magnitude of the instantaneous velocity.
If displacement as a function of time is known, we can find out the instantaneous speed at any time.
Let’s understand this by an example.
Problem 12:
Equation of motion is s(t) = 3t3
![\[Instantaneous{\enskip} speed = \frac{ds}{dt} \]](https://lambdageeks.com/wp-content/ql-cache/quicklatex.com-d3008e7c7d0e32e143f4e91a957dbee8_l3.png)
![\[s_{inst} = \frac{d(3t^{3})}{dt} = 9t^{2}\]](https://lambdageeks.com/wp-content/ql-cache/quicklatex.com-c41a66c4a039cc9beb80f0a42b9413dd_l3.png)
Consider t = 2s
![\[s_{inst} = 9(2)^{2}= 36m/s \]](https://lambdageeks.com/wp-content/ql-cache/quicklatex.com-183a0b26e5896687abdac0d4be739e9f_l3.png)
Why is it possible to calculate instantaneous velocity using kinematic formulas only when acceleration is constant
Kinematics equations can be used only when the acceleration of the object is constant.
In the case of variable accelerations, Kinematics equations will be different depending on the function form the acceleration takes; at that time; we should use the integrated approach to calculate instantaneous velocity. Which will be a bit complex.
Why do we take small time intervals while calculating instantaneous velocity. How does it give velocity at that instant if we are calculating it over a certain time interval
The instantaneous velocity is given by,
The smaller the value of “t” , the more closely will be the slope of the tangent line, i.e., instantaneous velocity.
When you want to calculate the velocity at a specific time, you need to first calculate the average velocities by taking small intervals of time. If those average velocities give the same value, then it will be the required instantaneous velocity.
Is velocity and instantaneous velocity different
Instantaneous velocity is different from velocity.
Velocity is generally known as the rate of change of position with time. In contrast, in instantaneous velocity, the time interval is narrowed to approach zero to give velocity at a particular instant of time.
For example,
A particle moving in a circle has zero displacements, and it is required to know the velocity of a particle. In this case, we can calculate instantaneous velocity because it has a tangential velocity at any given point of time.
What is instantaneous velocity with real-life examples
Instantaneous velocity real-life examples
If we consider an example of a squash ball, the ball comes back to its initial point; at that time, the total displacement and average velocity will be zero. In such cases, the motion is calculated by instantaneous velocity.
- The speedometer of a vehicle gives information about the instantaneous velocity/speed of a vehicle. It shows velocity at a particular instant of time.
- In a race, photographers take snapshots of runners, their average velocity doesn’t change, but their instantaneous velocity, as captured in the “snapshots,” changes. So it will be an example of instantaneous velocity.
- If you are near a shop and a vehicle crossed in front of you at “t“second, and you start to think about its velocity at a particular time, here you would be referring to the instantaneous velocity of the vehicle.
Frequently asked questions | FAQs
Is instantaneous velocity a vector
Instantaneous velocity is a vector quantity.
Instantaneous velocity is a vector because it has both magnitude and direction. It shows both speed (refers to magnitude) and direction of a particle. It has a dimension of LT-1.We can determine it by taking the slope of the distance-time graph.
How do you find instantaneous velocity with only a position vs. time graph and without an equation given
We can determine instantaneous velocity by taking the slope of the position-time graph.
- Plot a graph of displacement over time.
- Choose point A and another point B that is near to A on the line.
- Find the slope between A and B, calculate several times, moving A nearer to B.
- Calculate the slope for an infinitely small interval on the line.
- The slope obtained is instantaneous velocity.
Is it possible to instantaneously change velocity
It is not possible to bring an instantaneous change in velocity since it would require infinite acceleration.
In general, acceleration is the result of F = ma
![\[a= \frac{F}{m} = (Force{\enskip} over{\enskip} a {\enskip}mass)\]](https://lambdageeks.com/wp-content/ql-cache/quicklatex.com-4f7a8022e89df4b4e04f949ce10e53a8_l3.png)
and velocity is the outcome of the acceleration (from integration).If a change in velocity is a step function and as the time approaches zero, it would require infinite acceleration and force to change the velocity of mass instantaneously.
How can I calculate displacement when acceleration is a function of instantaneous velocity Initial velocity is given
We can calculate displacement in two ways, when Initial velocity is given
From derivation
Here acceleration is a function of instantaneous velocity,
![\[a =\frac{dv}{dt}\]](https://lambdageeks.com/wp-content/ql-cache/quicklatex.com-44ec6c4a9da8d82e472229e48196214b_l3.png)
Initial velocity
![\[v = \frac{ds}{dt}\]](https://lambdageeks.com/wp-content/ql-cache/quicklatex.com-8720a2771743fb4c460557783a31d0cd_l3.png)
![\[a = \frac{d(ds)}{dt^{2}}\]](https://lambdageeks.com/wp-content/ql-cache/quicklatex.com-e21ec51a8b9653251d2e631f1e53b865_l3.png)
![\[d(ds) = a dt^{2}\]](https://lambdageeks.com/wp-content/ql-cache/quicklatex.com-c3c786f3337998806c0b0e3d1adba9c5_l3.png)
By integrating,
![\[ds =\int {a dt^{2}\]](https://lambdageeks.com/wp-content/ql-cache/quicklatex.com-7ee13cb9951496fc4069fc695dacb5f0_l3.png)
Using this form, you can get displacement ds.
From the formula
By using the below kinematic equation, we can find displacement,
![\[S = ut + \frac{1}{2} at^{2}\]](https://lambdageeks.com/wp-content/ql-cache/quicklatex.com-264ac7cec65789b89fbf473e5026d272_l3.png)
What is average andinstantaneous velocity
The average velocity and instantaneous velocity are expressed as follows,
| Average velocity | Instantaneous velocity |
| The average velocity for a particular time interval is total displacement divided by total time. | Both time interval and displacement approach zero at some point. But the limit of the derivative of displacement to total time interval is non-zero, called instantaneous velocity. |
| Average velocity is the velocity of the whole path in motion | while instantaneous velocity is the velocity of a particle at a specific time |
| v = | v = |
Is instantaneous acceleration perpendicular to instantaneous velocity
Instantaneous acceleration of the body is always perpendicular to the instantaneous velocity.
In a circular motion, the instantaneous acceleration of the body is always perpendicular to the instantaneous velocity, and that acceleration is called centripetal acceleration. The speed remains unchanged; only the direction changes as the perpendicular acceleration changes the body’s trajectory.
About Raghavi Acharya
I am Raghavi Acharya, I have completed my post-graduation in physics with a specialization in the field of condensed matter physics. Having a very good understanding in Latex, gnu-plot and octave. I have always considered Physics to be a captivating area of study and I enjoy exploring the various fields of this subject. In my free time, I engage myself in digital art. My articles are aimed towards delivering the concepts of physics in a very simplified manner to the readers.
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Instantaneous velocity find
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