Solving straight-line motion question for time

Question

I apologise in advance if this question doesn't appeal to the advanced questions being asked in this Physics forum, but I'm a great fan of the Stack Exchange software and would trust the answers provided here to be more correct than that of Yahoo! Answers etc.

A car is travelling with a constant speed of 80km/h and passes a stationary motorcycle policeman. The policeman sets off in pursuit, accelerating to 80km/h in 10 seconds reaching a constant speed of 100 km/h after a further 5 seconds. At what time will the policeman catch up with the car?

The answer in the back of the book is 32.5 seconds.

The steps/logic I completed/used to solve the equation were:
- If you let x equal each other, the displacement will be the same, and the time can be solved algebraically.

Therefore:
$$x=vt$$ As the car is moving at 80km/h, we want to convert to m/s. 80/3.6 = 22.22m/s
$$x=22.22t$$

As for the policeman, he reaches 22.22m/s in 10 seconds.
$$\begin{aligned} x &= \frac12 (u+v) t \\ &= \frac12 \times 22.22 \times 10 \\ &= 111.11 \mathrm m \end{aligned}$$

The policeman progresses to travel a further 5 seconds and increases his speed to 100km/h.
100km/h -> m/s = 100 / 3.6 = 27.78m/s.

$$\begin{aligned} x &= \frac12 (u+v) t \\ x &= \frac12 \times (22.22 + 27.78) \times 5 \\ x &= \frac12 \times 50 \times 5 \\ x &= 250 / 2 \\ x &= 125 \mathrm m \end{aligned}$$

By adding these two distances together we get 236.1m. So the equation I have is:
$$22.22t = 27.78t - 236.1 $$ Which solves to let t = 42.47s which is really wrong.

Answer 1

Your mistake is in the equation

$$22.22t = 27.78t - 236.1$$

Everything up to there made good sense, but if the police officer has already traveled 236 meters, you should add that to his distance traveled, not subtract it. You'll also need to account for the way the police officer only began traveling at full speed 15 seconds into the chase.

Anyway, it is much easier to do the problem by thinking about the relative speeds. During the first ten seconds, the car is going 80kph and the police officer is going 40kph on average. So the police officer loses ground at an average of 40kph for 10 seconds. We can think of this as 10 seconds' worth of loss, and ask how many seconds' worth of loss the police officer gains as he speeds up further.

In the next segment, the police officer gains ground at an average of 10kph for 5 seconds. He's gaining ground 1/4 as fast as he lost it earlier and does it for 5 seconds, so this makes up for 5/4 of a second's worth of loss, leaving 8 3/4 seconds' lost ground remaining.

Finally he gains ground at 20kph until he catches up. He's gaining here at half the rate he was originally losing ground, so it takes him double the remaining seconds' worth time, or 17.5 seconds, to finish the pursuit.

This method is much simpler to calculate, eliminating many opportunities for errors.