An unmarked police car traveling a constant 90 km/h is passed by a speeder. Precisely 1.00 s after the speeder passes, the police officer steps on the accelerator. If the police car accelerates uniformly at 3.00 m/s2 and overtakes the speeder after accelerating for 8.00 s , what was the speeder's speed?

The speeder's speed was 129 km/h.

Explanation:

Hi there!

The equation of position of the police car is the following:

x = x0 + v0 · t + 1/2 · a · t²

Where

x = position of the police car at time t.

x0 = initial position.

v0 = initial velocity.

t = time.

a = acceleration

Let's place the origin of the frame of reference at the point where the police car is passed by the speeder so that the initial position, x0, is zero.

The distance traveled by the police car during 1.00 s before accelerating can be calculated with this equation:

x = x0 + v · t

converting km/h in m/s:

90 km/h · (1000 m/ 1 km) · (1 h / 3600 s) = 25 m/s

x = 0 m + 25 m/s · 1.00 s

x = 25 m

The distance traveled by the police car in 8.00 s can be calculated using the equation of position:

x = x0 + v0 · t + 1/2 · a · t²     (x0 = the position at which the car is 1.00 s after the speeder passes = 25 m).

x = 25 m + 25 m/s · 8.00 s + 1/2 · 3.00 m/s² · (8.00 s)²

x = 321 m

If the police car catches the speeder after traveling 321 m, then, the speeder traveled 321 m in 9.00 s. Using the equation of the position of an object moving at constant speed, we can calculate the velocity of the speeder:

x = x0 + v · t

321 m = 0 m + v · 9.00 s

321 m / 9.00s = v

v = 35.7 m/s = 129 km/h

The speeder's speed was 129 km/h.

d is the correct answer