Motion In One Dimension Question 11
Question 11 - 2024 (31 Jan Shift 1)
The relation between time ’ $t$ ’ and distance ’ $x$ ’ is $t=\alpha x^{2}+\beta x$, where $\alpha$ and $\beta$ are constants. The relation between acceleration (a) and velocity (v) is:
(1) $a=-2 \alpha v^{3}$
(2) $a=-5 \alpha v^{5}$
(3) $a=-3 \alpha v^{2}$
(4) $a=-4 \alpha v^{4}$
Show Answer
Answer: (1)
Solution:
$\mathrm{t}=\alpha \mathrm{x}^{2}+\beta \mathrm{x} \quad$ (differentiating wrt time )
$\frac{\mathrm{dt}}{\mathrm{dx}}=2 \alpha x+\beta$
$\frac{1}{\mathrm{v}}=2 \alpha x+\beta$
(differentiating wrt time)
$-\frac{1}{v^{2}} \frac{d v}{d t}=2 \alpha \frac{d x}{d t}$
$\frac{d v}{d t}=-2 \alpha v^{3}$
