Physics Problem

Question

A cop car going 2 m/s is passed by a truck going 42 m/s which is above the speed limit. The cop accelerates at 4m/s² to catch the truck. The trucks velocity remains constant.

a. How much time does it take for the cop to catch up.

b. What’s the total distance the cop traveled.

c. What is the cops speed when catching the truck

Answer 1

We set up two equations for displacement after the instant the truck passes the police car

The truck has a constant velocity

x=42t

The police car has an initial velocity of 2 and an acceleration of 4 m/s2

x=2t+0.5(4)(t²) =2t+2t²

a) They meet when displacements are the same 2t+2t²=42t

You can now solve for t (Note two answers but we already kow they are the same at t=0)

b)Now knowing t we can find displacement by plugging into either of the displacement equations (truck one is easier)

c) Knowing t we can find velocity of cop car using v=v0+at

v=2 + (4)t

Answer 2

The cop:

Initial speed v_ic = 2 m/sec

Acceleration a = 4 m/sec²

The truck:

Constant speed v_iT = 42 m/sec

Since the cop is accelerating, his velocity is a function of time, t:

v_c(t) = v_ic + at = 2 + 4t m/sec²

When the cop catches up with the truck

v_c(t) = v_{iT = 42}

so 2 + 4t = 42 and t = 40/4 = 10 sec.

The distance traveled is the integral of the velocity evaluated at t = 10 sec:

d_c(t) = 2t + 2t² + d_c(0)

We can set d_c(0) = 0 so

d_c(10) = 2(10) + 2(10)² = 20 + 200 = 220 m

The cop's speed when catching the truck is the same as the truck's speed: 42 m/sec

Answer 3

Let's start by writing down everything we know. Let's assume the instant the truck passes the car is time t = 0 s and take this to be the origin for the position of the cop and truck. This will make things easier.

x_cop(0) = 0 m

x_truck(0) = 0 m

v_cop(0) = 2 m/s

v_truck(0) = 42 m/s

v_truck(t) = 42 m/s

a_truck = 0 m/s²

a_cop = 4 m/s²

First, we want to know a what time, t, the cop car reaches the truck. In other words, we know that at some time, t, x_cop(t) = x_truck(t). Let's start with the kinematic equation for position.

x_truck(t) = x_truck(0) + v_truck(0)t + (1/2)a_truckt²

= v_truck(0)t

x_cop(t) = x_cop(0) + v_cop(0)t + (1/2)a_copt²

= v_cop(0)t + (1/2)a_copt²

Now set these two equations equal to each other and solve for t.

v_truck(0)t = v_cop(0)t + (1/2)a_copt²

[v_truck(0) - v_cop(0)]t = (1/2)a_copt²

t = 2[v_truck(0) - v_cop(0)] / a_cop

= 2[42 m/s - 2 m/s] / 4 m/s²

= 2[40 m/s] / 4 m/s²

= 80 m/s / 4 m/s²

= 20 s.

So it takes 20 s for the cop to catch up to the truck. Plug this time back into the kinematic equation for the cop's position and we get:

x_cop(t) = v_cop(0)t + (1/2)a_copt²

x_cop(20 s) = (2 m/s)(20 s) + (1/2)(4 m/s²)(20 s)²

= (40 m) + (800 m)

= 840 m.

The cop has traveled 840 m to catch up to the truck. Now we can use the kinematic equation for the cops speed:

v_cop(t) = v_cop(0) + a_copt

v_cop(20 s) = (2 m/s) + (4 m/s²)(20 s)

= (2 m/s) + (80 m/s)

= 82 m/s.

And there you have it. The cop is going 82 m/s when he reaches the truck.