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February 23, 2020, 06:51:47 pm

### AuthorTopic: Speeding - Two solutions  (Read 829 times) Tweet Share

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#### /0

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##### Speeding - Two solutions
« on: February 05, 2008, 06:31:52 pm »
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This could also go in maths section but whatever.

A speeding motorbike travels past a stationary police car. The police car starts accelerating immediately, and keeps accelerating until it has passed the bike.

DATA: motorbike speed: 35ms-1; police car acceleration: 4ms-2

How far does the police car travel before it overtakes the motorbike?

Ok so there are two ways which I think should work in solving this, but they give different answers.

Solution 1 - Motion Formulae

Motorbike:
$x=ut+\frac{1}{2}at^2 = 35t$

Police car:
$x=ut+\frac{1}{2}at^2=2t^2$

So $2t^2=35t \Rightarrow t= 0$ or $t=17.5$

$\Rightarrow x = 612.5m$

Solution 2 - Summation

Motorbike:
$x=35t$

Police car:
$x=4+8+...+t=\sum_{a=1}^{t} 4a=\frac{4t(t+1)}{2}$

So $35t=\frac{4t(t+1)}{2} \Rightarrow t=0$ or $t=16.5$

$\Rightarrow x=577.5m$

How is it that this can be wrong? Thanks
« Last Edit: February 05, 2008, 06:34:05 pm by DivideBy0 »

#### Collin Li

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##### Re: Speeding - Two solutions
« Reply #1 on: February 05, 2008, 09:42:40 pm »
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Summation? I don't know why you think it is correct. It may be flawed because summation is based on $t$ being an integer only.

A fast way to do this question, however, would be to draw a velocity-time graph, and then realise that the time taken to reach 35 m/s is $\frac{35}{4} = 8.75$ seconds. Then, you can now use symmetry and say that in $8.75 \times 2 = 17.5$ seconds, the total displacement between the car and motorbike will be zero, so that the distance covered will be $17.5 \times 35 = 612.5\mbox{ m}$
« Last Edit: February 05, 2008, 09:44:34 pm by coblin »

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##### Re: Speeding - Two solutions
« Reply #2 on: February 05, 2008, 10:22:56 pm »
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Thanks coblin, that method is quite good too

But are you sure t must be an integer? I mean, for the formula for summation it never says that the upper limit must be an integer, as far as I know.

#### Collin Li

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##### Re: Speeding - Two solutions
« Reply #3 on: February 05, 2008, 11:02:32 pm »
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The formula for summation is based on the addition of how many metres is moved per second. The formula $\sum_{a=1}^{t} a= \frac{t(t+1)}{2}$ is only true for integer values of $t$. It is based on the fact that $t$ moves up by increments of 1. What you could do is calculate the first 0.75 seconds, then do the rest from 0.75 to 15.75, or something similar to that - although both of these attempts would be particularly long-winded and difficult to understand.
« Last Edit: February 05, 2008, 11:26:57 pm by coblin »

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##### Re: Speeding - Two solutions
« Reply #4 on: February 06, 2008, 06:51:19 pm »
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I talked to my physics teacher today and he said that the police car actually travels 2 metres in its first second, and 6 in the next, so the sum should be

$\sum_{a=1}^t 4a-2=\frac{4t(t+1)}{2}-2t=2t^2$