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A bicycle rider accelerates from rest up to full speed on a flat, straight road. The frictional force between the road and the tires pushing her forward is f1. The air drag (and other frictional forces) pushing back is f2. What relationship would f1 and f2 have in the first few seconds of the ride then After she has reached full speed?

My answer: For the first one, I thought F2 would have been greater than f1, but then I realized that if it was, it would have been going backwards, so I decided that f1 is greater than f2 because it is moving forwards. Then, for the second part, I think that after she has reached full speed, it would be f1=f2, but I don't really have any reasoning. I just assumed that at that point, they'd be equal.

Any help or suggestions for figuring it out (please don't just give me an answer; I really need help understanding this stuff).

2007-03-15 15:34:10 · 3 answers · asked by Nikita R 2 in Science & Mathematics Physics

3 answers

Okay, I like this.

I believe you are correct on the explanation of the first few seconds. If the only force accelerating the bike forward is F1, then it must be greater than F2

(F1 > F2)

When the bike is at full speed and is no longer accelerating or slowing down, the forces must be equal like you said.

(F1 = F2)

This is because if (F1 > F2), then the bike would still be accelerating, and thus speeding up and not reaching the full speed. If (F1 < F2), then the bike would be slowing down and losing speed. Thus, if the full speed has been reached and maintained, the forces must be equal.

Hope this helps.

2007-03-15 15:54:38 · answer #1 · answered by dc6984 2 · 0 1

The frictional forces never changes. The wind drag increases as the velocity of the bicycle increases.

So at start the wind f2 would be very small nearly negible. And f1 would be the greater.

And at full speed, the f2 no doubt greater than friction f1

One cannot say that they will or will not ever be equal, as those are entirely independent variables. Not related in any way whatsoever. Both forces are pushing backward trying to stall the ride. They are not ever opposite in directional force.

Now if the wind were blowing at the back side and pushing the bike fprward ( it is indeed NOT) , sure then then at a 'steady coast' the friction would be equal to the frictional force. This, however, is not the case. So you are therefore incorrect on your analyis.

2007-03-16 01:07:14 · answer #2 · answered by James M 6 · 0 0

At the beginning of the ride, f1 is much greater than f2, and the extra force over distance (work, or energy) goes towards accelerating the bike forwards.

As the rider gets going faster, wind forces become significant, and the rate of acceleration slows. (she's still speeding up, but more slowly).

When she's at full speed (terminal velocity), she is not accelerating (not speeding up or slowing down) so we know that f1=f2. If they were not the same, then her speed would not be constant.

If at any time f2 become greater than f1 (she doesn't push as hard, or her tire goes flat) then she will slow down until she reaches a new equilibrium of the new f1=the new f2.

2007-03-15 22:38:57 · answer #3 · answered by violentquaker 4 · 0 0

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