Question

A particle of mass 0.5 kg is kept at rest. A force of 2.0 N acts on it for 5.0 second.find the distance moved by the particle on (a) these 5.0 second,and (b) the next 5.0 second.​

Answer 1

Answer:

50 m in both

Explanation:

a=F/m

=2/0.5

=4

s=ut+1/2at^2

u=0

s=[4×(5)^2]÷2=50m

now v=u+at

v=0+(4×5)=20

for next 5 sec

u=20 t=5

s=ut+1/2at^2

s=(20×5)+1/2 ×(-4)(5^2)

s=50m