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**Answer:**

a) The time the police officer required to reach the motorist was 15 s.

b) The speed of the officer at the moment she overtakes the motorist is 30 m/s

c) The total distance traveled by the officer was 225 m.

**Explanation:**

The equations for the position and velocity of an object moving in a straight line are as follows:

x = x0 + v0 · t + 1/2 · a · t²

v = v0 + a · t

Where:

x = position at time t

x0 = initial position

v0 = initial velocity

t = time

a = acceleration

v = velocity at time t

a)When the officer reaches the motorist, the position of the motorist is the same as the position of the officer:

x motorist = x officer

Using the equation for the position:

x motirist = x0 + v · t (since a = 0).

x officer = x0 + v0 · t + 1/2 · a · t²

Let´s place our frame of reference at the point where the officer starts following the motorist so that x0 = 0 for both:

x motorist = x officer

x0 + v · t = x0 + v0 · t + 1/2 · a · t² (the officer starts form rest, then, v0 = 0)

v · t = 1/2 · a · t²

Solving for t:

2 v/a = t

t = 2 · 15.0 m/s/ 2.00 m/s² = 15 s

The time the police officer required to reach the motorist was 15 s.

b) Now, we can calculate the speed of the officer using the time calculated in a) and the equation for velocity:

v = v0 + a · t

v = 0 m/s + 2.00 m/s² · 15 s

v = 30 m/s

The speed of the officer at the moment she overtakes the motorist is 30 m/s

c) Using the equation for the position, we can find the traveled distance in 15 s:

x = x0 + v0 · t + 1/2 · a · t²

x = 1/2 · 2.00 m/s² · (15s)² = 225 m

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