Question

A 30 kg bicycle is at rest. A 60 N force to the right acts on the bicycle during a time interval of 1.5 seconds. What is the velocity of the bicycle at the end of this interval?

Answer 1

According to Newton's second law,

`F=ma`

Where F is the force acting on a body, m is the mass of the body, and a is the acceleration of the body due to the applied force.

The acceleration is given by,

`a=(v)/(t)`

Where v is the velocity and t is the time interval.

Substituting this equation in the first equation,

`F=(mv)/(t)`

From the question,

F=60 N

m=30 kg

t=1.5 s

On substituting these values in the equations,

`60=30*(v)/(1.5)`

Which gives,

`v=(60*1.5)/(30)=3\text{ m/s}`

Therefore the velocity of the bicycle is 3 m/s