Question

A particle of mass $0.5kg$ is kept at rest. A force of $2.0N$ acts on it for $5.0s.$ Find the distance moved by the particle in (a) this $5.0s,$ and (b) the next $5.0s.$

Answer 1

(a) The acceleration of the particle is
$a=\frac{F}{m}=\frac{2.0N}{0.5kg}=4.0m/s^2.$
The distance moved by the particle in $5.0s$ is
$s=ut+\frac{1}{2}a{t}^2=0+\frac{1}{2}\times \left(4.0m/s^2\right)\times {\left(5s\right)}^2=50m.$
(b) The velocity at the end of the first $5.0s$ is
$v=u+at=0+\left(4.0m/s^2\right)\times \left(5.0s\right)=20m/s.$
After this, the force is withdrawn, and hence, the particle moves with uniform velocity. The distance moved in the next $5.0s$ is,
$s=ut=\left(20m/s\right)\times \left(5.0s\right)=100m.$