A speeder traveling at 41 m/s passes a motorcycle policeman at rest at the side of the road. The policeman accelerates at 3.3 m/s2. To the nearest tenth of a second how long does it take the policeman to catch the speeder?

Write equations for the distance travelled by speeder and policemen after the speeder passes.

Set them equal and solve for t.

Im sorry I am new to physics this is my first year and I still have no idea what to do for this problem. The other two that you answered I had them correct before you answered them and was using it as a check. When I reworked them your way, I got the same answer. However on this problem, I am at a loss because I feel there is not enough data, and I am obviously wrong. I worked it differently than u said and got 12.4 seconds, which sounds wrong. Please help!

To find out how long it takes for the policeman to catch the speeder, we can use the concept of relative velocity. We need to consider the initial separation between the two and how their velocities change over time.

To get started, let's find out how long it takes for the speeder to come to a stop. We'll use the following equation of motion:

v² = u² + 2as

where:
v = final velocity (0 m/s, since the speeder comes to a stop)
u = initial velocity of the speeder (41 m/s)
a = acceleration of the speeder (0 m/s², since the speeder comes to a stop)
s = displacement (unknown, what we need to find)

Rearranging the equation, we have:

0 = (41 m/s)² + 2(0 m/s²)s
0 = 1681 m²/s² + 0 m/s² * s
0 = 1681 m²/s²

Since the resulting equation is always zero, we can conclude that the displacement s is zero. This means that the speeder comes to a stop at the location of the motorcycle policeman.

Now, let's determine how much time it takes for the policeman to catch up to the speeder. We'll use the equation:

v = u + at

where:
v = final velocity of the policeman (unknown, the same as the speeder's velocity when caught)
u = initial velocity of the policeman (0 m/s since he is at rest)
a = acceleration of the policeman (3.3 m/s²)
t = time taken (unknown, what we need to find)

Rearranging the equation, we get:

v = (0 m/s) + (3.3 m/s²)t
v = 3.3 m/s² * t

Since the final velocity of the policeman is the same as the speeder's velocity when caught (41 m/s), we have:

41 m/s = 3.3 m/s² * t

Solving for t:

t = 41 m/s / 3.3 m/s²
t ≈ 12.42 seconds (rounded to the nearest tenth of a second)

Therefore, it takes the policeman approximately 12.4 seconds to catch the speeder.