August 23, 2019, 05:38:28 pm

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#### Eddict

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##### Re: Physics Help
« Reply #15 on: January 21, 2008, 09:03:33 pm »
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But the bike started moving 5 m's before the 1000m thing a magigie and asks for the instantaneous speed 5 m's into the race, so wouldn't that mean x=10 and not 5, eventhough 5 is right, I don't see how.

#### dcc

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##### Re: Physics Help
« Reply #16 on: January 21, 2008, 10:39:55 pm »
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i did that question yesterday, and the 5 metre isn't BEFORE the start, its after

Quote
bicycle 5.0m from the start of a 1000m sprint.

The wording is ambiguous, but the information provided later on in the question clears up how you interpret the question.

#### Eddict

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##### Re: Physics Help
« Reply #17 on: January 22, 2008, 12:17:56 am »
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i did that question yesterday, and the 5 metre isn't BEFORE the start, its after

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Yeah i know, but the question says that he is 5m's behind the start of the 1000m sprint.. What I dont get is why x=10 to find the acceleration, but 5 is used to find the instantaneous speed.. I understand the acceleration is constant, though how is that to change anything?

#### dcc

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##### Re: Physics Help
« Reply #18 on: January 22, 2008, 09:07:34 am »
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Quote
Q: Two physic students are trying to determine the instantanious speed of a bicycle 5.0m from the start of a 1000m sprint. They use a stopwatch to measure the time taken for the bicycle to cover the first 10m. if the acceleration is constant and the measured time is 4.0s what was the instantanious speen of the bicycle at the 5.0m mark..

no, they say they want to measure his speed 5.0m from the start of a 1000metre sprint.  5 metres from the start could be in either direction (5 metres before or 5 metres after) but it's clear they want to know 5 metres after the start (i.e. 995 metres left go in in the 1000 metre sprint).

They say that the time taken for the bike to cover 10 metres is 4 seconds, so you can subsitute that in to get
$a=\dfrac{5}{4}ms^-2$

Now since the accelleration is constant, you can continue to use those equations of motion to solve at the 5 metre mark, using
$v^2 = u^2 + 2ax$

And then you get 3.5ish m/s

These equations only work with constant accelleration because of the nature of how accelleration works, if its constant, you can easily derive how things will be effected.