A particle of mass 2KG starts moving on a straight line with an initial velocity of 2 m/s at a constant acceleration of 2 m/s squared .Then rate of change of kinetic energy is a.four times the velocity at any moment b.two times the displacement at any moment c.four times the rate of change of velocity at any momen d.constant throughout

Dear Student ,
Here in this question the mass of the body is 2 Kg and its initial velocity is 2 m/s at a constant acceleration 2 m/s2 .
Now from the equation of motion ,the velocity of the the particle at any time on the straight line or in its path is ,
v = 2+ 2t ......(1)
Therefore the kinetic energy of the body at any time instant 't' is ,
Ek$\frac{1}{2}m{v}^{2}=\frac{1}{2}×2×{\left(2+2t\right)}^{2}=4+4{t}^{2}+8t$
Now the rate of change of kinetic energy is ,

Therefore , the rate of change of momentum is four times the velocity at any moment .
So the correct option is a .
Regards

• 20
using v=u +at

velocity at any time will be 2 + 2t
KE at any time will be given by 1/2 * 2 * (2+2t)2= (2+2t)​= 4 +4t2 + 8t

rate of change of KE = d(KE)/dt = 8t + 8

So clearly the first option is correct.

displacement at any time will be equal to 2t+ 1/2* 2 * t2  = 2t+t2  so the second option is incorrect.
rate of change of velocity is fixed as it is possessing constant acceleration so the third option is also incorrect.
and as the rate of change of KE has 't' in its expression, it is dependent on time therefore the last option is aso incorrect
• 2
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