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velocity, acceleration


amy

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I'm having a little trouble with this problem for my physics homework, could anyone give me a step in the right direction here, thanks.

 

''The rear of a bicycle passes a point P on a road and travels at a steady speed of 12 m/s down the road. At the same instant the front of a car starts from rest at P and moves in the same direction as the bicycle with an acceleration of 2 m/s^2. When and how far from P does the front of the car catch up with the rear of the bicycle.''

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The big step in the right direction is realizing that what is asked for is the time when the positions of the car front and the bicycle back are the same. That said, the idea of the homework is that people asking for help first present their ideas and what they already tried to solve the question (the tradition is that new members ignore or don't know of this rule: Welcome to SFN).

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well yes I did realise that thats what they were asking, but i cant seem to visualise the situation, i drew it out but it doesnt seem right to me, if the rear of the bike and the front of the car are at P at the same instant how could it catch up if they're already at the same point P?

sorry note taken , but there are my ideas so far, still need clarification, thanks .

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They indeed are at the same position at time zero. And they will be at the same position at one later time, too - the one that is asked for. The physical reason is that the bike initially moves aways from P faster than the car (admittedly not what one would expect if one only heard the terms "bike" and "car"), but the car keeps on increasing its speed (accelerate) and will catch up and overtake.

Edited by timo
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i think i get where your coming from, the bike has already just passed the point when the car just takes off so for a short space of time, the bike is in front of the car, until the car accelerates enough to catch up, thanks for your help.

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  • 2 weeks later...

I'm having a little trouble with this problem for my physics homework, could anyone give me a step in the right direction here, thanks.

 

''The rear of a bicycle passes a point P on a road and travels at a steady speed of 12 m/s down the road. At the same instant the front of a car starts from rest at P and moves in the same direction as the bicycle with an acceleration of 2 m/s^2. When and how far from P does the front of the car catch up with the rear of the bicycle.''

Suppose the car catches up with the bicycle after [latex]t[/latex] seconds. The bicycle will have travelled [latex]12t[/latex] metres from P and the car will have travelled [latex]\frac12(2)t^2[/latex] metres from P. The two distances are the same. Equate them and solve for [latex]t[/latex].

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