Answer:
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
