Work Energy and Power - Result Question 37
40. A car of mass $m$ starts from rest and accelerates so that the instantaneous power delivered to the car has a constant magnitude $P_0$. The instantaneous velocity of this car is proportional to :
[2012M]
(a) $t^{2} P_0$
(b) $t^{1 / 2}$
(c) $t^{-1 / 2}$
(d) $\frac{t}{\sqrt{m}}$
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Answer:
Correct Answer: 40. (b)
Solution:
- (b) Constant power of car $P_0=F \cdot V=$ ma.v
$ P_0=m \frac{d v}{d t} \cdot v $
$P_0 d t=m v d v$ Integrating
$P_0 \cdot t=\frac{m v^{2}}{2}$
$v=\sqrt{\frac{2 P_0 t}{m}}$
$\because P_0, m$ and 2 are constant
$\therefore \quad v \propto \sqrt{t}$
At constant power, $P=$ constant
$P=F v=\frac{m d v}{d t} v=P \quad(\because F=\frac{m d v}{d t})$
$\Rightarrow v d v=\frac{P}{m} d t$
Integrating both sides, we get
$ \begin{align*} & \int v d v=\int \frac{P}{m} t \\ & \Rightarrow \frac{v^{2}}{2}=\frac{P}{m} t+C_1 \tag{i} \end{align*} $
At $t=0, v=0$
$\therefore C_1=0$
From equation (i) we get
$V^{2}=\frac{2 P t}{m} \Rightarrow V=(\frac{2 P t}{m})^{1 / 2}$
