Question

A speeding motorist zooms through a 50 km/h zone at 75 km/h (21 m/s) without noticing a stationary police car. The police officer immediatley heads after the speeder, accelerating at 2.5m/s^2. When the officer catches up to the speeder, how far down the raod are they, and how fast is the police car going?

I know you use the equations x= x_0+ v_0t+ .5at^2 and v= V_0 + at... But I want to know why we use these equations and also how does my book derive that when you set the 1st equation for the police officer and the speeding motorcycle equal it comes out to be

x_s= 2v_s0^2/a_p

THank you

Answer 1

75 km/hr * 1000/1 m/km * 1/60 hr/min * 1/60 min/sec = 75000/3600 m/s = 750/36 = 375/18 = 125/6 m/s

P(t) = 0.5 * a * t^2 + v0 * t + p0
We'll let p0 = 0 where the patrol car is sitting. We'll also make the time when the speeder passes the patrol car be t = 0.

Pspeeder(t) = 0.5 * 0 * t^2 + 125/6 * t + 0 = 125/6 * t
Ppatrolcar(t) = 0.5 * 2.5 * t^2 + 0 * t + 0 = 1.25 * t^2

Step 1 - Find the time from when the patrol car starts until it overtakes the speeder.
At that point, Pspeeder(t) = Ppatrolcar(t)
125/6 * t = 1.25 * t^2
0 = 1.25 * t^2 - 125/6 * t
0 = 0.01 * t^2 - 1/6 * t
0 = t^2 - 100/6 * t
0 = 6 * t^2 - 100 * t
0 = 3 * t^2 - 50 * t
0 = t (3t - 50)

0 = t { this is when the speeder passes the patrol car on the side of the road }
-- OR --
0 = 3t - 50
50/3 = t

Step 2 - Use either position formula to find where they are when the speeder is overtaken
Pspeeder(50/3) = 125/6 * 50/3 = 6250 / 18 = 3125 / 9 = 347.22222222222222222222222222222

Step 3 - How fast is the patrol car going?
V(t) = a * t + v0
V(50/3) = 2.5 * 50/3 + 0
V(5/3) = 125 / 3 m/s

125 / 3 m/s * 1/1000 km/m * 60/1 s/min * 60/1 min/hr = 450000 / 3000 km/hr = 150 km/hr