Police officer catching speeding car with constant acceleration

[beginner]

Hi there,

I can't seem to understand what I'm doing wrong in the following problem:

"a police officer begins pursuing a speeding car when it passes. If the car moves with a constant velocity of 60.0m/s, and the officer moves with a constant acceleration of 6.0m/s^2. How long will it take the officer to catch up to the car if he begins pursuing from rest?"

I used V(final) = V(initial) + at. Substituting 60.0 for v(final), 0 for v(initial), and 6.0 for a. I solved for t and got 10.0s, but the answer is supposed to be 20.0s. What am I doing wrong?

thanks in advance

 

[mani]

a police officer begins pursuing a speeding car when it passes. If the car moves with a constant velocity of 60.0m/s, and the officer moves with a constant acceleration of 6.0m/s^2. How long will it take the officer to catch up to the car if he begins pursuing from rest?"

I used V(final) = V(initial) + at. Substituting 60.0 for v(final), 0 for v(initial), and 6.0 for a. I solved for t and got 10.0s, but the answer is supposed to be 20.0s. What am I doing wrong?


You have found the instant of time when the velocities of the two cars will become the same. At that instant the two may not be in the same position.

You have to find the instant when they are in the same POsition, i.e. the instant when their displacements or distance traveled (from the point where the crimnal's car passes the stationary police car), are the same.

 

[Shyam]

I would suggest drawing a vector diagram. Like u solve other questions in relatve velocity.
I agree with Ambitwistor

 

[mani]

"a police officer begins pursuing a speeding car when it passes. If the car moves with a constant velocity of 60.0m/s, and the officer moves with a constant acceleration of 6.0m/s^2. How long will it take the officer to catch up to the car if he begins pursuing from rest?"


Initial coinciding point is our space origin, the coinciding instant, our time origin.
The path of motion is our X Axis.
X coordinate of criminal's car at any instant t = 60 t meters

X coordinate of police car at any instant t = (1/2)(6)t^2 meters

If the catch up takes place at instant t1;
60(t1) = (1/2)(6)(t1)^2 which has the solutions;
t1 = 0 and t1 = 20.
The first is trivial. The catch up takes place at 20 secs

 

[PrudensOptimus]

Δx = 0.5at^2

a = 6 m/s^2

Δx = 3t^2

Δx = vt

vt = 3t^2

t = v/3 = 60/3 = 20s.