Question

A particle of mass 0.5 kilogram is kept at rest. A force of 2 Newton acts on it for 5 seconds. Find the distance covered by the particle in
(a) This 5 Seconds
(b) The next 5 Seconds

Answer 1

Answer:

(a) The acceleration of the particle is

a=

m

F

=

0.5kg

2.0N

=4.0m/s

2

.

The distance moved by the particle in 5.0s is

s=ut+

2

1

at

2

=0+

2

1

×(4.0m/s

2

)×(5s)

2

=50m.

(b) The velocity at the end of the first 5.0s is

v=u+at=0+(4.0m/s

2

)×(5.0s)=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After this, the force is withdrawn, and hence, the particle moves with uniform velocity.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=(20m/s)×(5.0s)=100m.