Question

The motion of a particle along a straight line is described by equation $ x = 8 + 12t - t^3 $ where $x$ is in metre and $t$ in second. The retardation of the particle when its velocity becomes zero is

• A. $24\, m\, s^{-2}$
• B. zero
• C. $6\, m\, s^{-2}$
• D. $12\, m\, s^{-2}$

Answer 1

The Correct Option is D

We are given the motion of a particle along a straight line described by the equation:

x = 8 + 12t - t^3

where x is the position in meters and t is the time in seconds. We need to find the retardation of the particle when its velocity becomes zero.

1. First, calculate the velocity v(t) by differentiating x(t) with respect to time t:

\frac{dx}{dt} = v(t) = \frac{d}{dt} (8 + 12t - t^3) = 12 - 3t^2

2. Set the equation for velocity to zero to find the time when velocity is zero:

12 - 3t^2 = 0

Simplifying gives:

3t^2 = 12 \Rightarrow t^2 = 4 \Rightarrow t = 2\, \text{or} \, -2

3. We consider t = 2 s because time cannot be negative in this context. Now, calculate the acceleration at time t using the second derivative of x(t):

\frac{dv}{dt} = a(t) = \frac{d}{dt}(12 - 3t^2) = -6t

4. Substitute t = 2 s into the acceleration equation:

a(2) = -6 \times 2 = -12\, \text{m/s}^2

The negative sign indicates retardation. Thus, the retardation of the particle when its velocity becomes zero is 12\, \text{m/s}^2.

The correct answer is 12\, \text{m/s}^2.