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

The Correct Option is A

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:

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