Answer:
A.) Time = 17.13 s
B.) Distance = 32 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
