Question

A body is initially at rest. It undergoes one – dimensional motion with constant acceleration. The power delivered to it at time t is proportional to
A. ${{t}^{\dfrac{1}{2}}}$
B. $t$
C. ${{t}^{\dfrac{3}{2}}}$
D. ${{t}^{2}}$

Answer 1

Hint: At constant acceleration we are free to use three equations of motion. Executing formula $P=Fv$ of Power because it has time as a factor in it. And velocity equals to $v=u+at$ where u=0 because the body is at rest.

Formula Used:
$P=Fv$
Where:
$P=$ Power
$F=$ Force
$v=$ velocity

Complete step by step answer:
Using equation of motion:
$v=u+at$
Where:
$v=$final velocity
$u=$ initial velocity
$a=$ acceleration
$t=$time
In our question a body is initially at rest that means initial velocity is zero (the body is not moving at all).
$u=0$
Therefore,
$~v=at$
And we know that Force is the product of mass and acceleration
$F=ma$
Therefore Power,

& P=Fv \\\ & P=ma\times at \\\ & P=m{{a}^{2}}t \\\ & \\\ \end{aligned}$$ Here, mass and acceleration both are constant. Therefore Power is directly proportional to time. Answer option B. Additional Information: One Dimensional Motion: A body is said to be in one dimensional motion if it travels in a straight line or only one coordinate is associated with that motion. Example of one dimensional motion: You walking on a road to buy something from the market. Meaning of constant acceleration: We know acceleration is known as change in velocity per unit time. If the acceleration is constant that means change in velocity is always constant. Example : acceleration due to gravity i.e. $$9.8m{{s}^{-2}}$$is constant acceleration. Note: Students must remember that acceleration is constant and while in the formula of power, don’t replace acceleration by derivative of velocity i.e. $$a=\dfrac{dv}{dt}$$, because acceleration is constant. There is another alternative to solve this question: Power is also given as rate of work done i.e. $$P=\dfrac{W}{t}$$ We know that Work is force times displacement $$\begin{aligned} & P=\dfrac{Fs}{t} \\\ & P=\dfrac{ma}{t}\left( ut+\dfrac{1}{2}a{{t}^{2}} \right) \\\ \end{aligned}$$ Where u = 0 and mass and acceleration are constant $$\begin{aligned} & P=\dfrac{ma}{t}\left( \dfrac{1}{2}a{{t}^{2}} \right) \\\ & P=\left( \dfrac{1}{2}ma \right)t \\\ & P\propto t \\\ \end{aligned}$$