A motorist travelling with a constant speed of 15 m/s

Question

A motorist travelling with a constant speed of 15 m/s (about 34 mi/h) passes a school-crossing corner, where the speed limit is 10 m/s (about 22 mi/h). Just as the motorists pass, a police officer on a motorcycle stopped at the corner and starts off in pursuit with a constant acceleration of 3.0 m/s².

a. How much time elapses before the officer catches up with the motorist?

b. What is the officer’s speed at that point?

c. What is the total distance each vehicle has travelled at that point?

A motorist travelling with a constant speed of 15 ms (about 34 mi/h) passes — A motorist travelling with a constant speed of 15 m/s (about 34 mi/h) passes

Answer

The final answer for:

a. is t = 0 or t = 10 seconds

b. is 30 meters per second

c. is 150 meters for both police and the motorist.

Explanation

Data: Information from the question

The police officer’s position, x_p = ?

Motorist position, x_m = ?

Origin at the corner, x₀ = 0

Initial velocity of the police, u_p = 0

The initial velocity of the motorist, u_m = 15 m/s

Constant acceleration of the police, a_p = 3 m/s²

Acceleration of the motorist, a_m = 0

a.

Formula: The equations that will help us solve the problem for a

For the motorist, x_m = x₀ + u_mt + (1/2)a_mt²

While for the police officer, x_p = x₀ + u_pt + (1/2)a_pt²

Solution

using x₀ = 0, a_m = 0, and u_p = 0 and equating both equations for the motorist and the police officer, we will get:

u_mt = (1/2)a_pt²

Therefore, we have:

t = 0 or t = 2(u_m/a_p) = 2 (15/3) = 10 seconds

Therefore, the motorist passes the parked motorcycle at the corner at a time t = 0. Additionally, the police officer catches up with the motorist at the time t = 10 seconds.

b.

Formula: The equations that will help us solve the problem for b

By using v_p = u_p + a_pt

and u_p = 0, while a_p = 3 m/s²

Thus, by substituting our equation with our data, we will obtain

v_p = u_p + a_pt = 0 + 3t

v_p = 3t

At t = 10 s, the above equation will become v_p = 3 x 10 = 30 m/s

Therefore, v_p = 30 m/s shows that when the police officer overtakes the motorist, she is now using a speed of as twice as the motorist is.

c.

To find the distance the motorist cover in 10 seconds, we will say

x_m = u_mt = 15 x 10 = 150 m

While the distance for the police officer is

x_p = (1/2)a_pt² = (1/2) x 3 x 10² = 150 m

Therefore, the police officer catches up with the motorist when they cover equal distances.

You may also like to read:

A motorcyclist heading east through a small Iowa city