An unmarked police car traveling a constant 95km/h is passed by a speeder traveling 14km/h .Precisely 2.50s after the speeder passes, the police officer steps on the accelerator; if the police car's acceleration is 2.30m/s^2 , how much time passes before the police car overtakes the speeder after the speeder passes (assumed moving at constant speed)?

Question

An unmarked police car traveling a constant 95km/h is passed by a speeder traveling 14km/h .Precisely 2.50s after the speeder passes, the police officer steps on the accelerator; if the police car's acceleration is 2.30m/s^2 , how much time passes before the police car overtakes the speeder after the speeder passes (assumed moving at constant speed)?

Answer 1

The time of police car motion is t,

the time of the speeder motion is (t+2.5).
The distance covered by
- the police car is v(p) •t+at^2/2,
- by speeder is v(s)•(t+2.5).
v(p) •t + at^2/2 = v(s)•(t+2.5).
v(p) = 95 km/h = 26.4 m/s
v(s) = 140 km/h = 38.9 m/s
v(p) •t + at^2/2 = v(s)• t + v(s)• 2.5,
at^2/2 + [v(p) – v(s)] •t -2.5•v(s) = 0,
t^2 –(2/a)•[v(s) – v(p)] •t – 2.5•2/a = 0,
t^2 – 10.9t - 84.6 = 0,
t = 5.24 s.
The total time is 5.24+2.5 = 7.74 s.

Answer 2

To find the time it takes for the police car to overtake the speeder, we need to determine the distance the speeder travels in the 2.50 seconds after passing the police car.

First, we need to convert the speed of the speeder from km/h to m/s since the acceleration is given in m/s^2.

Speed of the speeder in m/s = (Speed of the speeder in km/h) * (1 km/3.6 m) = 14 km/h * (1 km/3.6 m) = 3.89 m/s (rounded to two decimal places)

Next, we can calculate the distance traveled by the speeder in 2.50 seconds using the formula:

Distance = Initial velocity * Time + (1/2) * Acceleration * Time^2

Distance = (3.89 m/s) * (2.50 s) + (1/2) * (0 m/s^2) * (2.50 s)^2 (since the speeder is assumed to be moving at a constant speed)

Distance = 9.73 m

Now, we can calculate the distance the police car needs to cover to catch up with the speeder.

Distance = Initial velocity * Time + (1/2) * Acceleration * Time^2

Distance = (0 m/s) * Time + (1/2) * (2.30 m/s^2) * Time^2 (since the police car starts from rest and accelerates)

Using the quadratic formula to solve for time:

(1/2) * (2.30 m/s^2) * Time^2 = 9.73 m

Time^2 = (9.73 m) / (1.15 m/s^2) (simplifying the equation)

Time^2 = 8.46 s^2

Time = √(8.46 s^2) = 2.91 s (rounded to two decimal places)

Therefore, it will take approximately 2.91 seconds for the police car to overtake the speeder after the speeder passes.