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

[lNVlNClBLE]

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

Homework Statement

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.

Homework Equations

v = Vo + at

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

 

[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 traveled by the motorcycle (as a function of time and acceleration), and remember that the time period it was moving was 17s.

 

[lNVlNClBLE]

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 traveled by the motorcycle (as a function of time and acceleration), and remember that the time period it was moving was 17s.

Would using the displacement formula be better?

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

 

[BruceW]

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