The velocity of an object with a mass of #2 kg# is given by #v(t)= 3 t^2+ 2 t +8#. What is the impulse applied to the object at #t= 4 #?

1 Answer
Jan 15, 2016

Answer:

The impulse at #t=4# is #52 kg ms^-1#

Explanation:

Impulse is equal to the rate of change of momentum:

#I = Delta p = Delta (mv)#.

In this instance the mass is constant so #I = mDeltav#.

The instantanteous rate of change of the velocity is simply the slope (gradient) of the velocity-time graph, and can be calculated by differentiating the expression for the velocity:

#v(t)=3t^2+2t+8#
#(dv)/dt = 6t +2 #

Evaluated at t = 4, this gives #Delta v=26 ms^-1#

To find the impulse, then, #I = mDeltav = 2*26 = 52 kgms^-1#