A particle of mass 1 g moves on a straight line. The variation of its velocity with time is shown in fig. Find the force acting on the particle att = 1 s,4s and 7 s.

Explanation

(i) At t = 1 s, acceleration of the particle is equal to the slope of the straight line from 0 to 3s as 1 s is between 0 and 3 s.
Acceleration of the particle at 1 = 1s is
`a=(v_(f)-v_(i))/(t_(f)-t_(i))=(15-0)/(3-0)=5ms^(-2)`
`therefore` Force acting on the particle
`= ma = 1xx10^(-3)xx5=5xx10^(-3)N`.
(ii) At t = 4 s, acceleration is given by
a = slope of the straight line from t = 3 to 5 s
`=(v_(f)-v_(i))/(t_(f)-t_(i))=(15-15)/(5-3)=(0)/(2)=0 ms^(-2)`
`therefore` Force acting on the particle
`= ma = 1xx10^(-3)xx0=0N`
(iii) At t = 7 s, acceleration is given by
a = slope of the straight line from t = 5 to 8s
`=(v_(f)-v_(i))/(t_(f)-t_(i))=(0-15)/(8-5)=(15)/(3)=-5 ms^(-2)`
`therefore` Force acting on the particle
`= ma = 1xx10^(-3)xx -5 =-5xx10^(-3)N`.
Negative sign indicates that the force is in the opposite direction of the motion of the particle.