Police officer at rest at the side of the highway

notices a speeder moving at 62 km/h along a

straight level road near an elementary school. When

the speeder passes, the officer accelerates at 3.0 m/s2

in pursuit. The speeder does not notice until the

police officer catches up.

Now assume that the police officer accelerates

until the police car is moving 10.0 km/h faster

than the speeder and then moves at a constant

velocity until the police officer catches up. How

long will it take to catch the speeder?

Let's take the police officers starting position as our origin, that is Sp (t0) = Ss (t0) = 0, and, since we know the police officers starts from a standstill, Vp (t0) = 0 and Vs (t0) = 62km/h

For the first part of the pursuit, during the officers acceleration:

Sp (t) = Sp (t0) + Vp (t0) * t + (3/2) * t^2

Sp (t) = 1.5*t^2

Vp (t) = 3*t

We need to know how much time transpires during this first bit, that is until the officers speed is 72 km/h or 20m/s, and we can find that out by using Vp (t):

Vp (t1) = 20 = 3*t1

t1 = 3.33 ... s

We now look at how much space each of them covers during this time:

Sp (t1) = 1.5 * (t1) ^2 = 16.66 ... m

Ss (t1) = Vs*t1 = (62/3.6) * (3.33 ...) = 57.41 m

For the second part of the pursuit we will take the officers position as our origin:

Sp (t) = 20*t

Ss (t) = (57.41 - 16.66) + (17.22 ...) * t

Now we just need to equate Sp (t) to Ss (t):

20*t = (40.75) + (17.22 ...) * t

t = 40.75 / (2.77 ...) = 14.67 s

But we need to add the time in the first part of the pursuit to the time we just found:

14.67 + 3.33 = 18s

The pursuit lasts 18 seconds.