Amotorcycle officer hidden at an intersection observes a car driven by an oblivious driver who ignores a stop sign and continues through the intersection at constant speed. the police officer takes off in pursuit 1.79 s after the car has passed the stop sign. she accelerates at 5.0 m/s2 until her speed is 117 km/h, and then continues at this speed until she catches the car. at that instant, the car is 1.4 km from the intesection.

(a) how long did it take for the officer to catch up to the car?
(b) how fast was the car traveling?

a) It takes the officer 48.12 s to catch up to the car.

b) The velocity of the car was 29.09 m/s

Explanation:

The equations for the position and velocity of objects 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

First let´s find how much time it takes the officer to reach max-speed using the equation of the velocity:

v = v0 + a · t (v0 = 0 because the officer starts from rest)

v/a = t

Let´s convert km/h into m/s

117 km/h · 1000 m / km · 1 h / 3600 s = 32.5 m/s

Then:

32.5 m/s / 5.0 m/s² = 6.50 s

We know that when the officer catches the car, her position is 1.4 km from the intersection. So let´s find her position after the acceleration period (6.5 s) so we can then calculate how much time it takes her to reach the 1.4 km traveling at constant velocity:

x = x0 + v0 · t + 1/2 · a · t² (x0 = 0 and v0 = 0 because the officer starts from rest and the origin of the frame of reference is located where the officer starts the persecution)

x = 1/2 · a · t²

x = 1/2 · 5.0 m/s² · (6.5 s)²

x = 105.6 m

Now let´s calculate the time it takes her to reach 1.4 km:

In this case, a = 0 and x0 = 105.6 m. Then:

x = x0 + v · t

(x-x0)/v = t

(1400 m - 105.6 m)/ 32.5 m/s = 39.83 s

Then, the total time it takes the officer to reach the 1.4 km from the intersection and catch the car will be:

total time = 1.79 s + 6.50 s + 39.83 s = 48.12 s

It takes the officer 48.12 s to catch up to the car.

b) Using the equation of the position with a = 0 and knowing that after 48.12 s the position of the car is 1.4 km:

x = x0 + v · t (x0 = 0 becaue the origin of the frame of reference is located at the intersection and the final distance traveled (1.4 km) is relative to that intersection).

x = v · t

1400 m / 48.12 s = v

v = 29.09 m/s

The velocity of the car was 29.09 m/s