Instantaneous Velocity and Speed – University Physics Volume 1 (2024)

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\left(t\right)=\frac{d}{dt}x\left(t\right).[/latex]

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]\stackrel{\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}\phantom{\rule{0.2em}{0ex}}\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].

Finding Velocity from a Position-Versus-Time Graph
Given the position-versus-time graph of (Figure), find the velocity-versus-time graph.

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 s to 0.5 s: [latex]\stackrel{\text{–}}{v}=\frac{\text{Δ}x}{\text{Δ}t}=\frac{0.5\phantom{\rule{0.2em}{0ex}}\text{m}-0.0\phantom{\rule{0.2em}{0ex}}\text{m}}{0.5\phantom{\rule{0.2em}{0ex}}\text{s}-0.0\phantom{\rule{0.2em}{0ex}}\text{s}}=1.0\phantom{\rule{0.2em}{0ex}}\text{m/s}[/latex]

Time interval 0.5 s to 1.0 s: [latex]\stackrel{\text{–}}{v}=\frac{\text{Δ}x}{\text{Δ}t}=\frac{0.0\phantom{\rule{0.2em}{0ex}}\text{m}-0.0\phantom{\rule{0.2em}{0ex}}\text{m}}{1.0\phantom{\rule{0.2em}{0ex}}\text{s}-0.5\phantom{\rule{0.2em}{0ex}}\text{s}}=0.0\phantom{\rule{0.2em}{0ex}}\text{m/s}[/latex]

Time interval 1.0 s to 2.0 s: [latex]\stackrel{\text{–}}{v}=\frac{\text{Δ}x}{\text{Δ}t}=\frac{0.0\phantom{\rule{0.2em}{0ex}}\text{m}-0.5\phantom{\rule{0.2em}{0ex}}\text{m}}{2.0\phantom{\rule{0.2em}{0ex}}\text{s}-1.0\phantom{\rule{0.2em}{0ex}}\text{s}}=-0.5\phantom{\rule{0.2em}{0ex}}\text{m/s}[/latex]

The graph of these values of velocity versus time is shown in (Figure).

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.

[latex]\text{Average speed}=\stackrel{\text{–}}{s}=\frac{\text{Total distance}}{\text{Elapsed time}}.[/latex]

[latex]\text{Instantaneous speed}=|v\left(t\right)|.[/latex]

[latex]\frac{dx\left(t\right)}{dt}=nA{t}^{n-1}.[/latex]

Instantaneous Velocity Versus Average Velocity
The position of a particle is given by [latex]x\left(t\right)=3.0t+0.5{t}^{3}\phantom{\rule{0.2em}{0ex}}\text{m}[/latex].

1. Using (Figure) and (Figure), find the instantaneous velocity at [latex]t=2.0[/latex] s.
2. 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

1. [latex]v\left(t\right)=\frac{dx\left(t\right)}{dt}=3.0+1.5{t}^{2}\phantom{\rule{0.2em}{0ex}}\text{m/s}[/latex].

   Substituting t = 2.0 s into this equation gives [latex]v\left(2.0\phantom{\rule{0.2em}{0ex}}\text{s}\right)=\left[3.0+1.5{\left(2.0\right)}^{2}\right]\phantom{\rule{0.2em}{0ex}}\text{m/s}=9.0\phantom{\rule{0.2em}{0ex}}\text{m/s}[/latex].

2. 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\left(1.0\phantom{\rule{0.2em}{0ex}}\text{s}\right)=\left[\left(3.0\right)\left(1.0\right)+0.5{\left(1.0\right)}^{3}\right]\phantom{\rule{0.2em}{0ex}}\text{m}=3.5\phantom{\rule{0.2em}{0ex}}\text{m}[/latex]

   [latex]x\left(3.0\phantom{\rule{0.2em}{0ex}}\text{s}\right)=\left[\left(3.0\right)\left(3.0\right)+0.5{\left(3.0\right)}^{3}\right]\phantom{\rule{0.2em}{0ex}}\text{m}=22.5\phantom{\rule{0.2em}{0ex}}\text{m.}[/latex]

   Then the average velocity is

   See Also

   7 Negative Velocity Example: Examples And Problems - TechieScience

   [latex]\stackrel{\text{–}}{v}=\frac{x\left(3.0\phantom{\rule{0.2em}{0ex}}\text{s}\right)-x\left(1.0\phantom{\rule{0.2em}{0ex}}\text{s}\right)}{t\left(3.0\phantom{\rule{0.2em}{0ex}}\text{s}\right)-t\left(1.0\phantom{\rule{0.2em}{0ex}}\text{s}\right)}=\frac{22.5-3.5\phantom{\rule{0.2em}{0ex}}\text{m}}{3.0-1.0\phantom{\rule{0.2em}{0ex}}\text{s}}=9.5\phantom{\rule{0.2em}{0ex}}\text{m/s}\text{.}[/latex]

Significance
In the limit that the time interval used to calculate [latex]\stackrel{\text{−}}{v}[/latex] goes to zero, the value obtained for [latex]\stackrel{\text{−}}{v}[/latex] converges to the value of v.

Instantaneous Velocity Versus Speed
Consider the motion of a particle in which the position is [latex]x\left(t\right)=3.0t-3{t}^{2}\phantom{\rule{0.2em}{0ex}}\text{m}[/latex].

1. What is the instantaneous velocity at t = 0.25 s, t = 0.50 s, and t = 1.0 s?
2. 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

1. [latex]v\left(t\right)=\frac{dx\left(t\right)}{dt}=3.0-6.0t\phantom{\rule{0.2em}{0ex}}\text{m/s}[/latex]
2. [latex]v\left(0.25\phantom{\rule{0.2em}{0ex}}\text{s}\right)=1.50\phantom{\rule{0.2em}{0ex}}\text{m/s,}\phantom{\rule{0.5em}{0ex}}v\left(0.5\phantom{\rule{0.2em}{0ex}}\text{s}\right)=0\phantom{\rule{0.2em}{0ex}}\text{m/s,}\phantom{\rule{0.5em}{0ex}}v\left(1.0\phantom{\rule{0.2em}{0ex}}\text{s}\right)=-3.0\phantom{\rule{0.2em}{0ex}}\text{m/s}[/latex]
3. [latex]\text{Speed}=|v\left(t\right)|=1.50\phantom{\rule{0.2em}{0ex}}\text{m/s},0.0\phantom{\rule{0.2em}{0ex}}\text{m/s,}\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}3.0\phantom{\rule{0.2em}{0ex}}\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.

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 circ*mstances 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?

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?

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\left(t\right)=10t-2{t}^{2}\phantom{\rule{0.2em}{0ex}}\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\left(t\right)=\left(10-4t\right)\text{m/s}[/latex]; v(2 s) = 2 m/s, v(3 s) = −2 m/s; b. [latex]|v\left(2\phantom{\rule{0.2em}{0ex}}\text{s}\right)|=2\phantom{\rule{0.2em}{0ex}}\text{m/s},|v\left(3\phantom{\rule{0.2em}{0ex}}\text{s}\right)|=2\phantom{\rule{0.2em}{0ex}}\text{m/s}[/latex]; (c) [latex]\stackrel{\text{–}}{v}=0\phantom{\rule{0.2em}{0ex}}\text{m/s}[/latex]

Unreasonable results. A particle moves along the x-axis according to [latex]x\left(t\right)=3{t}^{3}+5t\text{​}[/latex]. At what time is the velocity of the particle equal to zero? Is this reasonable?