Chapter 2, Problem 26P

PROBLEM A race car starting from rest accelerates at a constant rate of 5.00 m/s², (a) What is the velocity of the car after it has traveled 1.00 × 10² ft? (b) How much time has elapsed? (c) Calculate the average velocity two different ways.

STRATEGY We’ve read the problem, drawn the diagram in Figure 2.16, and chosen a coordinate system (steps 1 and 2). We'd like to find the velocity v after a certain known displacement Δx. The acceleration a is also known, as is the initial velocity v₀ (step 3, labeling, is complete), so the third equation in Table 2.4 looks most useful for solving part (a). Given the velocity, the first equation in Table 2.4 can then be used to find the time in part (b). Part (c) requires substitution into Equations 2.2 and 2.7, respectively.

Figure 2.16 (Example 2.4)

SOLUTION

(a) Convert units of Δx to SI, using the information in the inside front cover.

Write the kinematics equation for v² (step 4):

Solve for v, taking the positive square root because the car moves to the right (step 5):

Substitute v₀ = 0, a = 5.00 m/s², and Δx = 30.5 m:

1.00 × 10²ft = (1.00 × 10² ft)

v² = v₀² + 2a Δx

v = $\sqrt{{\text{v}}_0^2\text{ + 2}\text{a}\Delta \text{x}}$

v = $\sqrt{{\text{v}}_0^2\text{ + 2}\text{a}\Delta \text{x}}$ = $\sqrt{{\left(0\right)}^{\text{2}}\text{ + 2}\left(5.00{\text{ m/s}}^{\text{2}}\right)\left(30.5m\right)}$= 17.5 m/s

(b) Find the trooper's speed at that time. Substitute the time into the trooper’s velocity equation:

v_trooper = v₀ + a_trooper t = 0 + (3.00m/s²)(16.9s)

= 50.7 m/s

Solve Example 2.5, “Car Chase,” by a graphical method. On the same graph, plot position versus time for the car and the trooper. From the intersection of the two curves, read the time at which the trooper overtakes the car.