# Solving straight-line motion question for time [closed]

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.

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• If it helps you to think about the problem geometrically, you can look at my answer to a similar question – Justin L. Jan 2 '11 at 9:42
• I don't believe this solution helps as the police-car does not follow a consistent linear relationship. This would involve creating a linear regression which would be inaccurate and impractical for me. – RodgerB Jan 2 '11 at 15:37

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.

• Thanks for the in-depth reply, the relative speeds has really helped me understand the logical process involved. Although this solves the problem, I'm still determined to find the algebraic way to solve it, to get the end answer of 32.5 (the relative speeds answer is 33.75 in which I assume precision has been lost). – RodgerB Jan 2 '11 at 12:50
• So I've tried to find a logical method of accounting for the velocity of the full chase. I have solved the equation for 32.5m and the correct velocity is supposedly 14.9551. LHS = 22.22 * 32.5 (722.15 = v*32.5 + 236.11) v = 14.9551 We need to find the average velocity. At first I tried finding the mean, this left 16.62 which is wrong, because 2/3 of the time it was increasing speed 0-22.22, and 1/3 of the time it was increasing speed 22.22-27.78, and then constantly travelling at 27.78. So my working is: ((2/3)*22.22 + (1/3)*(27.78-22.22) + 27.78) / 3 Is this thinking logically correct? – RodgerB Jan 2 '11 at 14:36
• @RodgerB The relative speeds answer is 32.5. I'm not sure where the 33.75 came from, but there is no loss of precision. My post even explicitly identified the length of the last segment as 17.5 seconds. If you add the 15 seconds before that you get 32.5. I can't understand your second comment. There are too many numbers popping up haphazardly. The correct version of the equation I earlier identified as your problem is $22.22t = 27.78(t-15) + 236.1$ – Mark Eichenlaub Jan 2 '11 at 15:23
• Thank you very much, sorry for misinterpreting your answer (tis' close to 3am). Not even my TI-89 has the capacity to remember what it was that made it 33.75, but I was adding the numbers as your example progressed and didn't quite understand. I'll try reading over it once again after getting some sleep, it's been a long day. – RodgerB Jan 2 '11 at 15:53
• @RodgerB Okay, hope it works out for you. – Mark Eichenlaub Jan 2 '11 at 16:05