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

Show Answer

Time interval 0 s to 0.5 s: $$ \overset{\text{–}}{v}=\frac{\text{Δ}x}{\text{Δ}t}=\frac{0.5\,\text{m}-0.0\,\text{m}}{0.5\,\text{s}-0.0\,\text{s}}=1.0\,\text{m/s}$$

Time interval 0.5 s to 1.0 s: $$ \overset{\text{–}}{v}=\frac{\text{Δ}x}{\text{Δ}t}=\frac{0.0\,\text{m}-0.0\,\text{m}}{1.0\,\text{s}-0.5\,\text{s}}=0.0\,\text{m/s}$$

Time interval 1.0 s to 2.0 s: $$ \overset{\text{–}}{v}=\frac{\text{Δ}x}{\text{Δ}t}=\frac{0.0\,\text{m}-0.5\,\text{m}}{2.0\,\text{s}-1.0\,\text{s}}=-0.5\,\text{m/s}$$

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.

Example

Instantaneous Velocity Versus Average Velocity

The position of a particle is given by $$ x(t)=3.0t+0.5{t}^{3}\,\text{m}$$.

1. Using (Figure) and (Figure), find the instantaneous velocity at $$ t=2.0 $$ 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. $$v(t)=\frac{dx(t)}{dt}=3.0+1.5{t}^{2}\,\text{m/s}$$.Substituting t = 2.0 s into this equation gives $$ v(2.0\,\text{s})=[3.0+1.5{(2.0)}^{2}]\,\text{m/s}=9.0\,\text{m/s}$$.
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):
   $$x(1.0\,\text{s})=[(3.0)(1.0)+0.5{(1.0)}^{3}]\,\text{m}=3.5\,\text{m}$$
   $$x(3.0\,\text{s})=[(3.0)(3.0)+0.5{(3.0)}^{3}]\,\text{m}=22.5\,\text{m.}$$

   Then the average velocity is

   $$\overset{\text{–}}{v}=\frac{x(3.0\,\text{s})-x(1.0\,\text{s})}{t(3.0\,\text{s})-t(1.0\,\text{s})}=\frac{22.5-3.5\,\text{m}}{3.0-1.0\,\text{s}}=9.5\,\text{m/s}\text{.}$$

Significance

In the limit that the time interval used to calculate $$ \overset{\text{−}}{v} $$ goes to zero, the value obtained for $$ \overset{\text{−}}{v} $$ converges to the value of v.

Example

Instantaneous Velocity Versus Speed

Consider the motion of a particle in which the position is $$ x(t)=3.0t-3{t}^{2}\,\text{m}$$.

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. Show Answer

   $$v(t)=\frac{dx(t)}{dt}=3.0-6.0t\,\text{m/s}$$
2. Show Answer

   $$v(0.25\,\text{s})=1.50\,\text{m/s,}v(0.5\,\text{s})=0\,\text{m/s,}v(1.0\,\text{s})=-3.0\,\text{m/s}$$
3. Show Answer

   $$\text{Speed}=|v(t)|=1.50\,\text{m/s},0.0\,\text{m/s,}\,\text{and}\,3.0\,\text{m/s}$$

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: $$ |v(t)| $$ versus time. Speed is always a positive number.