A bicyclist traveling at speed [v] can brake in a distance d if she squeezes on

Question

A bicyclist traveling at speed [v] can brake in a distance d if she squeezes on one of her two front brakes, thereby
applying a constant frictional force of magnitude F in the backward direction. Suppose the bicyclist is out riding one
day and traveling at speed 2 [v]. She sees a squirrel ahead of her and squeezes both front brakes, thereby applying a
constant frictional force of magnitude 2F in the backward direction. How far does the bicyclist travel before she stops?
(a) d/4
(b) d/2
(c) d
(d) d sq. root 2
(e) 2d
(f) 4d

Explanation / Answer

the acceleration will be twice as much and the velocity will also be twice as much as initially vfinalsquared - vinitialsquared = 2acceleration*distance (vfinal is 0 in both cases so we drop it) Initially: -vinitialsquared = 2F/m * d Second time: -(2*vinitial)squared = 4F/m * d(sub2) Solve for F initial: F = -vinitialsquared*m/2d second: F = -vinitialsquared*m/d(sub2) Set the two equations equal to each other -vinitialsquared * m/2d = -vinitialsquared * m /d(sub2) Do some cancellations 1/2d = 1/d(sub2) Thus, d(sub2) = 2d answer is e

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