An unmarked police car traveling a constant 80km/h is passed by a speeder traveling 135km/h.
Precisely 3.00s after the speeder passes, the police officer steps on the accelerator; if the police car's acceleration is 1.50m/s^2 , how much time passes before the police car overtakes the speeder after the speeder passes (assumed moving at constant speed)?
80km/hr = 22.22 m/s
135km/hr = 37.5 m/s
so, at time t when the police overtake the scoff-law,
37.5t = 22.22t + 0.75 (t-3)^2
To solve this problem, we need to understand the concept of relative motion and calculate the distance covered by both the speeder and the police car. Let's break it down step by step:
Step 1: Determine the distance covered by the speeder in 3.00 seconds.
We know that the speeder is traveling at 135 km/h. To convert this to meters per second, we need to multiply by 1000/3600:
Speeder's speed = 135 km/h * (1000 m/3600 s) = 37.5 m/s
The distance covered by the speeder in 3.00 seconds is:
Distance = Speed * Time = 37.5 m/s * 3.00 s = 112.5 m
Step 2: Determine the distance covered by the police car in the time it takes to catch up to the speeder.
Since the police car starts moving after 3.00 seconds, we need to find the time it takes for the police car to catch up to the speeder. Let's call this time "t".
The equation for the distance covered by the police car is given by the kinematic equation:
Distance = Initial Velocity * Time + 0.5 * Acceleration * Time^2
The initial velocity of the police car is 80 km/h. Converting this to meters per second:
Initial Velocity = 80 km/h * (1000 m/3600 s) = 22.22 m/s
Using the given acceleration of the police car as 1.50 m/s^2, the equation becomes:
112.5 m + 22.22 m/s * t + 0.5 * 1.50 m/s^2 * t^2
Step 3: Solve the equation for time.
Setting the distance covered by the police car equal to the distance covered by the speeder:
112.5 m + 22.22 m/s * t + 0.5 * 1.50 m/s^2 * t^2 = 112.5 m
Rearranging the equation and simplifying:
0.5 * 1.50 m/s^2 * t^2 + 22.22 m/s * t = 0
This is a quadratic equation of the form: at^2 + bt = 0, where a = 0.5 * 1.50 m/s^2 and b = 22.22 m/s.
Using the quadratic formula to solve for t:
t = [-b ± sqrt(b^2 - 4ac)] / 2a
t = [-22.22 ± sqrt((22.22)^2 - 4 * 0.5 * 1.50 * 0)] / 2 * (0.5 * 1.50)
t = [-22.22 ± sqrt(493.13)] / 1.5
t ≈ [-22.22 ± 22.20] / 1.5
Simplifying:
t ≈ -0.02 s (ignoring negative value)
t ≈ 14.80 s
Therefore, it takes approximately 14.80 seconds for the police car to overtake the speeder after the speeder passes.