Kinematic Equations for a Police Car Giving Chase in 12 Seconds

[demonslayer42]

Homework Statement

1. A car driving at a constant speed of 75 ft/s passes a police car that is initially at rest. If the police car decides to give chase
A) What rate would the police have to accelerate to catch up with the other car in 12 seconds?
B) What is the police car's speed at the end of the 12 seconds?

Homework Equations

Use Kinematic Equations

The Attempt at a Solution

My first step was to assign the variables:

Vo = 0
V = 75 ft/s
a = unknown
x = unknown
t = 12

I used V = Vo + at so 75 ft/s = 0 + a12 which comes out to 6.25 ft/s Is this part right, or am I way off? I'm new to Physics, so please be nice. I want to do well :) If part A is correct, I have no clue how to solve part B. Help would be greatly appreciated.

 

[physlaw]

For part a I believe you have to consider the motion of both cars... just because the police car gets up to the same speed doesn't mean it will catch it, since the other car continues to move at constant speed. Try writing down both of their 'position' equations being sure to note that car one is moving with constant speed (so no acceleration). Those two positions have to be the same for the police car to have caught up, so set them equal and solve 'a.'

Try the equation you tried to use for part b.

 

[demonslayer42]

Thank you for your reply :) I'm not sure if I understand you correctly, but I tried what you said.

The first car's displacement would be 75 ft/s * 12 seconds = 900 feet? So if I use X = Vot + 1/2(at^2)that means that acceleration of the second car would have to be 12.50 ft/s Is this the correct? Or am I still way off? So if the acceleration = 12.50 ft/s I use V = Vo + at and end up with Final V = 150 ft/s ? That just doesn't sound right to me, could you please try to explain what I'm doing wrong? Thank you for your help :)