## Chapter 2

## Q. 2.2

## Q. 2.2

**Average and instantaneous velocities**

Here we will look at the difference between average and instantaneous velocities via a specific example of a cheetah chasing its prey. The cheetah is crouched in ambush 20.0 m to the east of an observer’s vehicle, as shown in Figure 2.10a. At time t = 0, the cheetah charges an antelope in a clearing 50.0 m east of the observer. The cheetah runs along a straight line; the observer estimates that, during the first 2.00 s of the attack, the cheetah’s coordinate x varies with time t according to the equation

x = 20.0 m + (5.00 m/s²)t².

(a) Find the displacement of the cheetah during the interval between t_1 = 1.00 s and t_2 = 2.00 s. (b) Find the average velocity during this time interval. (c) Estimate the instantaneous velocity at time t_1 = 1.00 s by taking ∆t = 0.10 s.

## Step-by-Step

## Verified Solution

**SET UP** Figure 2.10b shows the diagram we sketch. First we define a coordinate system, orienting it so that the cheetah runs in the +x direction. We add the points we are interested in, the values we know, and the values we will need to find for parts (a) and (b).

**SOLVE** Part (a): To find the displacement ∆x, we first find the cheetah’s positions (the values of x) at time t_1 = 1.00 s and at time t_2 = 2.00 s by substituting the values of t into the given equation. At time t_1 = 1.00 s, the cheetah’s position x_1 is

x_1 = 20.0 m + (5.00 m/s^2) (1.00 s)^2 = 25.0 m

At time t_2 = 2.00 s, the cheetah’s position x_2 is

x_2 = 20.0 m + 15.00 m/s^)(2.00 s)^2 = 40.0 m.

The displacement during this interval is

\Delta x = x_2 – x_1 = 40.0 m – 25.0 m = 15.0 m.

Part (b): Knowing the displacement from 1.00 s to 2.00 s, we can now find the average velocity for that interval:

\upsilon _{\mathrm{av},x}=\frac{\Delta x}{\Delta t}=\frac{40.0 m-25.0 m}{2.00 s -1.00 s} =\frac{15.0 m}{1.00 s}=15.0 m/s.

Part (c): The instantaneous velocity at 1.00 s is approximately (but not exactly) equal to the average velocity in the interval from t_1 = 1.00 s to t_2 = 1.10 s (i.e., \Delta t = 0.10 s). At t_2 = 1.10 s,

x_2 = 20.0 m + (5.00 m/s)(1.10 s)^2 = 26.05 m,

so that

\upsilon _{\mathrm{av},x}=\frac{\Delta x}{\Delta t}=\frac{26.05 m-25.0 m}{1.10 s- 1.00 s}=10.5 m/s

If you use the same procedure to find the average velocities for time intervals of 0.01 s and 0.001 s, you get 10.05 m/s and 10.005 m/s, respectively. As ∆t gets smaller and smaller, the average velocity gets closer and closer to 10.0 m/s. We conclude that the instantaneous velocity at time t = 1.0 s is 10.0 m/s.

**REFLECT** As the time interval ∆t approaches zero, the average velocity in the interval is closer and closer to the limiting value 10.0 m/s, which we call the instantaneous velocity at time t = 1.00 s. Note that when we calculate an average velocity, we need to specify two times— the beginning and end times of the interval—but for instantaneous velocity at a particular time, we specify only that one time.

**Practice Problem:** What is the cheetah’s average speed over (a) the first second of the attack and (b) the first two seconds of the attack?

*Answers:* (a) 5.00 m/s, (b) 10.0 m/s.