A speeder traveling at 36 m/s passes a motorcycle policeman at rest at the side of the road. The policeman accelerates at 2.54 m/s^2. To the nearest tenth of a second how long does it take the policeman to catch the speeder?

Answer 1

we want the distances to be equal, so

36t = 1/2 * 2.54 t^2
t = 28.3 s

Answer 2

distance speeder goes = 36 t
distance cop goes = .5(2.54) t^2
so
36 = 1.27 t

Answer 3

How are you factoring out the t and t^2?

Answer 4

d=v*t and d=1/2at^2

For speeder, d=v*t=(36m/s)*t

For officer, d=1/2at^2=1/2(2.54m/s^2)t^2

set equations equal to each other:

1/2(2.54m/s^2)t^2=(36m/s)*t

t's are on both sides of the equation, so you can cancel out one t on each side, and the equation becomes the following:

1/2(2.54m/s^2)=(36m/s)t

solve for t

1.27t=36

t=36/1.27

t=28.3s

Answer 5

t's are on both sides of the equation, so you can cancel out one t on each side, and the equation becomes the following:

1/2(2.54m/s^2)t=(36m/s)

solve for t

1.27t=36

t=36/1.27

t=28.3s