Question

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?

Answer 1

Answer:

$f_1 > f_2$

Explanation:

Frictional force due to road is pushing the cycle in forward direction

so it is given as f1

also the air drag is always in opposite direction of motion so it is given as f2

so here net force on the bicycle is given as

$F_{net} = f_1 - f_2$

now initially the speed of cycle is increasing with time

so here we can say that net force on the cycle must be positive in direction

so we have

$f_1 - f_2 > 0$

so we have

$f_1 > f_2$

Answer 2

Answer:

$f_1>f_2$

Explanation:

The bicycle is able to move forward due to the frictional force between the Tyre of the bicycle and the road. The air drag and other frictional forces act in the opposite direction of motion. The net force is:  

$F_{net}=f_1-f_2$ is non-zero due to which the bicycle is able to accelerate forward. Therefore, the frictional force between the Tyre and the road must be greater than the air drag.

Later on, as the bicylce attains full constant speed, the two forces would be equal.