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)?
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.
i think their is a mistake, the answer is incorrect
The solution is right. The mistake may be in calculations. Calculate it carefully.
To find the time it takes for the police car to overtake the speeder, we need to determine the distance traveled by both the speeder and the police car during the 2.50s after the speeder passes.
Let's calculate the distance traveled by the speeder:
Speed = 14 km/h = 14000 m/3600 s = 3.889 m/s
Time = 2.50 s
Distance = Speed * Time = 3.889 m/s * 2.50 s = 9.7225 m
Now, let's calculate the distance traveled by the police car:
Initial speed = 95 km/h = 95000 m/3600 s = 26.389 m/s
Acceleration = 2.30 m/s^2
Time = ? (What we're trying to find)
To find the time it takes for the police car to overtake the speeder, we can use the kinematic equation:
Distance = Initial Speed * Time + (0.5 * Acceleration * Time^2)
Since the police car starts from rest and accelerates, the initial speed is 0 m/s:
Distance = (0.5 * Acceleration * Time^2)
Substituting the known values:
9.7225 m = (0.5 * 2.30 m/s^2 * Time^2)
Simplifying the equation:
19.445 m = 2.30 m/s^2 * Time^2
Dividing both sides by 2.30 m/s^2:
Time^2 = 19.445 m / 2.30 m/s^2
Time^2 = 8.464 s^2
Taking the square root of both sides:
Time ≈ √(8.464 s^2)
Time ≈ 2.91 s
Therefore, it takes approximately 2.91 seconds for the police car to overtake the speeder after the speeder passes.