A particle of mass 2 kg starts moving in a straight line with an initial velocity of 2 ms -1 at a constant acceleration of 2 ms -2. Then rate of change of kinetic energy

Question

A particle of mass 2 kg starts moving in a straight line with an initial velocity of 2 ms -1 at a constant acceleration of 2 ms -2. Then rate of change of kinetic energy (A) is four times the velocity at any moment (B) is two times the displacement at any moment (C) is four times the rate of change of velocity at any moment (D) is constant through out

Tardigrade Admin · Accepted Answer

As, K=(1/2) m v2 (d K/d t)=m v ⋅ (d v/d t)=(m (d v/d t)) v= textmav=4 v ∵ m=2 kg and a = ms -2

Q. A particle of mass $2 k g$ starts moving in a straight line with an initial velocity of $2 m s^{- 1}$ at a constant acceleration of $2 m s^{- 2}$ . Then rate of change of kinetic energy

is four times the velocity at any moment

is two times the displacement at any moment

is four times the rate of change of velocity at any moment

is constant through out

As, $K = \frac{1}{2} m v^{2}$
$\frac{d K}{d t} = m v \cdot \frac{d v}{d t} = (m \frac{d v}{d t}) v = mav = 4 v$
$∵ m = 2 k g$ and $a = m s^{- 2}$

Q. A particle of mass 2kg starts moving in a straight line with an initial velocity of 2ms−1 at a constant acceleration of 2ms−2. Then rate of change of kinetic energy