1. An automobile travels along a straight road at 15.65 m/s through a 11.18 m/s

speed zone. A police car observed the automobile. At the instant that the two

vehicles are abreast of each other, the police car starts to pursue the automobile at

a constant acceleration of 1.96 m/s². The motorist noticed the police car in his rear

view mirror 12 s after the police car started the pursuit and applied his brakes and decelerates at 3.05 m/s². (Hint: The police will not go against the law.)

a) Find the total time required for the police car to overtake the automobile.

b) Find the total distance travelled by the police car while overtaking the

automobile.

c) Find the speed of the police car at the time it overtakes the automobile.

d) Find the speed of the automobile at the time it was overtaken by the police car.

a.) Time = 17.13 seconds

b.) 31.88 m

c.) V = 11.18 m/s

d.) V = 7.1 m/s

Explanation:

The initial velocity U of the automobile is 15.65 m/s.

At the instant that the two vehicles are abreast of each other, the police car starts to pursue the automobile with initial velocity U = 0 at a constant acceleration of 1.96 m/s². Because the police is starting from rest.

For the automobile, let us use first equation of motion

V = U - at.

Acceleration a is negative since it is decelerating with a = 3.05 m/s². And

V = 0.

Substitute U and a into the formula

0 = 15.65 - 3.05t

15.65 = 3.05t

t = 15.65/3.05

t = 5.13 seconds

But the motorist noticed the police car in his rear view mirror 12 s after the police car started the pursuit and applied his brakes and decelerates at 3.05 m/s².

The total time required for the police car to overtake the automobile will be

12 + 5.13 = 17.13 seconds.

b.) Using the third equation of motion formula for the police car at V = 11.18 m/s and a = 1.96 m/s²

V^2 = U^2 + 2aS

Where S = distance travelled.

Substitute V and a into the formula

11.18^2 = 0 + 2 * 1.96 * S

124.99 = 3.92S

S = 124.99/3.92

S = 31.88 m

c.) The speed of the police car at the time it overtakes the automobile will be in line with the speed zone which is 11.18 m/s

d.) That will be the final velocity V of the automobile car.

We will use third equation of motion to solve that.

V^2 = U^2 + 2as

V^2 = 15.65^2 - 2 * 3.05 * 31.88

V^2 = 244.9225 - 194.468

V = sqrt (50.4545)

V = 7.1 m/s