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Body is moved along a straight line by a machine delivering a constant power. The distance moved by the body in time $t$ is proportional to

Option 1)
$t^{3/4}$
Option 2)
$t^{3/2}$

Option 3)
$t^{1/4}$
Option 4)
$t^{1/2}$

Answers (2)

best_answer

As we learnt in

Power if the force is constant -

$P=\frac{dw}{dt}= P= \vec{F}\cdot \vec{v}$

- wherein

$\vec{F}\rightarrow force$

$\vec{v}\rightarrow velocity$

As we discussed in

Power if the force is variable -

$P_{av}= \frac{\Delta w}{\Delta t}= \frac{\int_{0}^{t}p\cdot dt}{\int_{0}^{t}dt}$

- wherein

$P\rightarrow power$

$dt\rightarrow short\: interval \: of\: time$

Power is constant P = cosntant (K)

Since $P=F.v(force\times velocity)$

$\therefore\ (m\frac{dv}{dt}).v=K$

$\Rightarrow\ v\frac{dv}{dt} = \frac{K}{m}$

$\Rightarrow\ \int v\ dv=\frac{K}{m}\ \int dt$

$\Rightarrow\ \frac{v^{2}}{2}=\frac{K}{m} t$

$\Rightarrow\ v=\sqrt{}\frac{2K}{m}.t^{1/2}$

$S=\frac{dS}{dt}=\sqrt{}\frac{2K}{m}.t^{2}$

$S=\sqrt{}\frac{2K}{m}\int_{0}^{T} {t}^{1/2}dt=\sqrt{}\frac{2K}{m}.(\frac{t^{3/2}}{3/2})|_{0}^{T}$

$S=\frac{2}{3}\sqrt{}\frac{2K}{m}\ T^{3/2}$

$\therefore\ S\ \alpha\ T^{3/2}$

Correct answer is 2

Option 1)

$t^{3/4}$

This is an incorrect option.

Option 2)

$t^{3/2}$

This is the correct option.

Option 3)

$t^{1/4}$

This is an incorrect option.

Option 4)

$t^{1/2}$

This is an incorrect option.

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