Question

A truck is traveling at a constant speed of 20 m/s through a school zone. At time t = 0 seconds, he passes a hidden police car that is at rest. Five seconds after the truck passes, the police car begins accelerating at a constant rate of 2 m/s in order to catch the truck.

Determine the time at which the speed of the truck is equal to the speed of the police car. Mark this time as t↓1. How will the positions of the police car and the truck compare when they have the same speed and why?​

Answer 1

Let's define:

The position zero will be in the first point where the truck and police car meet. (So the initial position of both vehicles is zero)

Then, the position equation for the truck is:

T(t) = 20m/s*t

Where t is our variable, time in seconds.

Now, at t = 5s, the car starts accelerating.

Ac(t) = 2m/s^2 for( t ≥ 5s)

For the velocity of the car, we must integrate that.

V(t) = (2m/s^2)*(t - 5) for (t ≥ 5s)

Where i introduced a little change in the variable because the velocity of the car starts to increase for t larger than 5 seconds.

For the position of the car we integrate again.

C(t) = (1m/s)*(t - 5)^2 for (t ≥ 5s)

Now, let's answer the questions:

Determine the time at which the speed of the truck is equal to the speed of the police car.

Then we must have:

V(t) = (2m/s^2)*(t - 5) = 20m/s. (Remember that we only can use times requal or larger than 5 seconds).

2m/s^2*t - 10m/s = 20m/s

2m/s^2*t = 30m/s

t = (30/2) s = 15s

The velocities of both vehicles will be the same after 15 seconds.

t1 = 15s

How the positions will compare at this time?

The easier thing will be to evaluate the position equation of each vehicle in this time:

T(15s) = 20m/s*15s = 300m

C(15s) = (1m/s^2)*(15s - 5s)^2 = (1m/s^2)*(10s)^2 = 100m.

Then we can see that the truck is 200m ahead of the car.

But remember that the police car is accelerating, so the velocity will keep increasing meaning that eventually, the car will catch the truck.