# A simple problem that will take one minute to answer (:

1. Feb 3, 2012

### lNVlNClBLE

Just need someone to tell me if I did this right. xD Thanks!

1. The problem statement, all variables and given/known data
The car passes the police officer traveling at 20m/s. It takes the officer 4 seconds to start his motorcycle and start pursuit after which it takes the officer 17 seconds to catch the offending vehicle. a) Compute the constant acceleration necessary for the officer to catch the car. b) Compute the total displacement the officer traveled to catch the car.

2. Relevant equations
v = Vo + at

3. The attempt at a solution
a) 420/17 = 24.71 m/s

24.71 = 0 + a(21)
21a = 24.71
a = 1.18m/s^2

b)
20(21) = 420m

2. Feb 3, 2012

### BruceW

you've done part b) right, but I don't think you've got part a) correct. On the first line, you have: 420/17=24.71m/s And this is the total distance divided by the time over which the police motorcycle was moving. So this would give the average speed of the motorcycle over that period. Then you've done 24.71 = 0 + a(21). I guess you were trying to use the equation v = v0 + at, but this equation requires v the final speed, but you have used the average speed over that time period.

As a hint, try using the equation for the distance travelled by the motorcycle (as a function of time and acceleration), and remember that the time period it was moving was 17s.

3. Feb 3, 2012

### lNVlNClBLE

Would using the displacement formula be better?

Δx = Vo(t) + (1/2)at^2
420 = 0(17) + (1/2)(a)(17)^2
420 = 144.5a
a = 2.91 m/s^2

Last edited: Feb 3, 2012
4. Feb 4, 2012

### BruceW

Yep, that's the one. Nice work, I got the same answer as that.