A speeder traveling at 40 m/s passes a motorcycle policeman at rest at the side of the road. The policeman accelerates at 1.92 m/s2. To the nearest tenth of a second how long does it take the policeman to catch the speeder?

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To find the time it takes for the policeman to catch the speeder, we can use the concept of relative motion.

Let's assume that at time t, the policeman catches up to the speeder. At this time, the positions of both the speeder and the policeman are the same.

We can first find out how far the speeder has traveled when the policeman catches up. This can be done by using the formula:

distance = initial velocity * time + (1/2) * acceleration * time^2

For the speeder:
Initial velocity (u) = 40 m/s
Time (t) = ?
Acceleration (a) = 0 (since the speeder maintains a constant speed)

For the policeman:
Initial velocity (u) = 0 (the policeman is at rest)
Time (t) = ?
Acceleration (a) = 1.92 m/s^2

Now, let's set up the equation for distance traveled. Since their positions are the same when the policeman catches up, we have:

40t = (1/2) * 1.92t^2

Simplifying the equation:

20t = 0.96t^2

Dividing both sides by t:

20 = 0.96t

Now, we can solve for t:

t = 20 / 0.96

Calculating this, we find:

t ≈ 20.83 seconds (to two decimal places)

Therefore, it takes the policeman approximately 20.8 seconds (to the nearest tenth) to catch the speeder.