Question

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:

• A. $a = -2 \alpha v^3$
• B. $a = -5 \alpha v^5$
• C. $a = -3 \alpha v^2$
• D. $a = -4 \alpha v^4$

Answer 1

Answer: (A)

$a = -2 \alpha v^3$

Answer 2

Given:

$t = \alpha x^2 + \beta x$

Differentiating with respect to $x$:

$\frac{dt}{dx} = 2\alpha x + \beta$

Using:

$\frac{1}{v} = 2\alpha x + \beta \quad \implies \quad v = \frac{1}{2\alpha x + \beta}$

Differentiating with respect to time:

$-\frac{1}{v^2} \frac{dv}{dt} = 2\alpha \quad \implies \quad \frac{dv}{dt} = -2\alpha v^3$