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 at t = 1 s, 4 s and 7 s.

A particle of mass 1 g moves on a straight line. The variation of its

1.	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 at t = 1 s, 4 s and 7 s.
Answer» Solution : (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 3s. Acceleration of the particle at t = 1 s 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)xx10=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.