Question

A particle of mass 2 kg starts moving in a straight line with an initial velocity of 2m per second at a constant acceleration of 2 metre per second square the rate of change of kinetic energy is 4 times the velocity at any point

Answer 1

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 = 12mv2=12×2×(2+2t)2=4+4t2+8t

Now the rate of change of kinetic energy is ,

dEkdt=ddt(4+4t2+8t)=8+8t

Therefore , the rate of change of momentum is four times the velocity at any moment .

So the correct option is a .

Regards

Answer 2

Concept Introduction:-

Velocity is the directional velocity of a item in movement as a demonstration of its fee of extrade in role as located from a specific frame.

Given Information:-

We have been given that A particle of mass $2$ kg starts moving in a straight line with an initial velocity of $2$m per second at a constant acceleration of $2$ metre per second square the rate of change of kinetic energy is $4$ times the velocity at any point.

To Find:-

We have to prove that the kinetic value.

Solution:-

According to the problem

We know that kinetic energy is given by $K.E. =\frac{1}{2}\times {mv}^{2}$

$d(\mathrm{KE}) / \mathrm{dt}=\mathrm{mv}\left(\frac{d v}{d t}\right)=m v a$

where $m$ is the mass

$a$ is the acceleration and

$v$ is the velocity

Given that $\mathrm{m}=2 \mathrm{~kg}, \mathrm{a}=2 \mathrm{~m} / \mathrm{s}^{2}$ and $\mathrm{v}=2 \mathrm{~m} / \mathrm{s}$

Therefore, $\frac{\mathrm{d}(\mathrm{K} \cdot \mathrm{E} .)}{\mathrm{dt}}=(2)(2)(\mathrm{v})=4 \mathrm{v}$

Therefore, the rate of the change of kinetic energy is four times the numerical value of velocity.

Final Answer:-

The rate of the change of kinetic energy is four times the numerical value of velocity.

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