A bicyclist and his bicycle have a mass of 105 kg. What force is necessary to bring him to rest when he is traveling at 25 m/s and must stop to avoid a tree 4 m in front of him?

Question

A bicyclist and his bicycle have a mass of 105 kg. What force is necessary to bring him to rest when he is traveling at 25 m/s and must stop to avoid a tree 4 m in front of him?

Answer 1

Let

m = 105 kg be the given mass,

v = 25 m/s be his velocity,

s = 4m be the distance over which he must decelerate to 0

a be his deceleration (yet unknown)

t be the time it takes him to stop (yet unknown)

F be the force needed to make him stop.

From Newton's second law

F = m.a

From definition of acceleration (or deceleration)

v = a.t

s = a.t²/2

Combining the two above equations

s = a.(v/a)²/2 = v²/2a

Solving the above for a

a = v²/2s

Plugging into the equation for F

F = m.v²/2s

= 105 kg . (25 m/s)^2 / (2 . 4m)

= 8203N

The moral of the story, I suppose, is that if you travel at highway speed

then stopping would subject you to about 8g deceleration -- hardly survivable.