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?

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

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

we want the distances to be equal, so

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

distance speeder goes = 36 t

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

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

To find out how long it takes for the policeman to catch the speeder, we need to find the time it takes for the policeman to reach the same speed as the speeder. We can do this by using the kinematic equation:

v = u + at

Here, v represents the final velocity, u represents the initial velocity, a represents the acceleration, and t represents the time.

For the speeder:
Initial velocity (u1) = 36 m/s
Final velocity (v1) = ?
Acceleration (a1) = 0 m/s² (since the speeder maintains a constant velocity)

For the policeman:
Initial velocity (u2) = 0 m/s (at rest)
Final velocity (v2) = 36 m/s (same as the speeder's velocity)
Acceleration (a2) = 2.54 m/s²

Let's calculate the time it takes for the policeman to reach the speeder's velocity:

For the speeder:
v1 = u1 + a1t
Since the speeder maintains a constant velocity, a1 = 0 m/s², so the equation simplifies to:
v1 = u1
v1 = 36 m/s

For the policeman:
v2 = u2 + a2t
36 = 0 + (2.54)t
2.54t = 36
t = 36 / 2.54
t ≈ 14.2 seconds (rounded to the nearest tenth)

Therefore, it takes approximately 14.2 seconds (to the nearest tenth) for the policeman to catch the speeder.