The speed of the car when you hit the police car is 26.154 m/s.
The speed of the car's v = 110km/h = 30.56 m/s
The total time travel will also include the reaction time:
t = 2 + 0.4 = 2.4 s
For police,
The acceleration, a = - 5 m/s²
The distance traveled is:
S(p) = vt + (1/2)at²
S(p) = (30.56 × 2.4 + (1/2) (-5) (-2.4)² )
S(p) = 58.93 m
Now, the original gap between the car was 25 m.
Therefore, the final gap = (25 − 73.33 + 58.93) m = 10.6 m
As a result, let's say that at time t₀, the distance between the cars is 10.6 meters.
The speed of the police car at t₀ will be:
v(p) = (30.56 − 5 × 2.4) = 18.56m/s.
Now, the collision occurs at a time t when S = S(p)
We pick your position's coordinates to be S = 0 and the police car's position to be S(p) = 10.6m at time t₀ .
Therefore,
S(p) - 10.6 = 18.56(t - t₀) - (1/2) (5) (t - t₀)²
S(p) = 10.6 + 18.56t − 2.5t²
And,
S = 30.56 (t - t₀) - (1/2)(5)(t - t₀)²
When the distances are equal.
S = S(p)
30.56 (t - t₀) - (1/2)(5)(t - t₀)² = 10.6 + 18.56t − 2.5t²
t = 0.883 sec
So, the speed of your car during the collision will be:
v = 30.56 - 5(0.883) = 26.154 m/s
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The complete is mentioned below:
You are arguing over a cell phone while trailing an unmarked police car by 25m; both your car and the police car are traveling at 110km/h. Your argument diverts your attention from the police car for 2.0s (long enough for you to look at the phone and yell, “I won’t do that!”). At the beginning of that 2.0s, the police officer begins braking suddenly at 5.0m/s². What is the separation between the two cars when your attention finally returns? Suppose that you take another 0.40s to realize your danger and begin braking. If you too brake at 5.0m/s², what is your speed when you hit the police car?
