Question

The relation between time \( t \) and displacement \( x \) is \( t = \alpha x^2 + \beta x \), where \(\alpha\) and \(\beta\) are constants. If \( \nu \) is the velocity, the retardation is:

• A. \( 2\alpha \nu^3 \beta^2 \)
• B. \( 2\alpha \beta \nu^3 \)
• C. \( -2\beta \nu^3 \)
• D. \( 2\alpha \nu^3 \)

Hint:

In kinematics, carefully differentiate each term and make sure to apply the chain rule when variables depend on each other.

Answer 1

The Correct Option is D

Given the equation of motion: \[ t = \alpha x^2 + \beta x \] To find the velocity \( \nu \), differentiate \( x \) with respect to \( t \): \[ \frac{dx}{dt} = \frac{d}{dt} (\alpha x^2 + \beta x) \] Using the chain rule for \( x \) as a function of \( t \), we get: \[ \frac{dx}{dt} = 2\alpha x \frac{dx}{dt} + \beta \frac{dx}{dt} \] This simplifies to: \[ \nu = 2\alpha x + \beta \] Now differentiate \( \nu \) with respect to \( t \) to find the acceleration \( a \) (which is the derivative of velocity): \[ a = \frac{d\nu}{dt} = 2\alpha \nu \frac{d\nu}{dt} + \beta \frac{d\nu}{dt} \] Thus, the retardation, which is the negative acceleration, is: \[ \text{Retardation} = -a = - 2 \alpha \nu^3 \]