Question

A particle moves along the x-axis and has its displacement x varying with time t according to the equation \( x = c_0 (t^2 - 2) + c (t - 2)^2 \) where \( c_0 \) and \( c \) are constants of appropriate dimensions. Then, which of the following statements is correct?

• A. the acceleration of the particle is \( 2c_0 \)
• B. the acceleration of the particle is \( 2c \)
• C. the initial velocity of the particle is \( 4c \)
• D. the acceleration of the particle is \( 2(c + c_0) \)

Hint:

To find the velocity and acceleration of a particle given its displacement as a function of time, differentiate the displacement with respect to time once for velocity and twice for acceleration. The initial velocity is the velocity at \( t = 0 \).

Answer 1

The Correct Option is D

Given the displacement equation:

\( x = c_0 (t^2 - 2) + c (t - 2)^2 \)

Expanding the equation yields:

\( x = c_0 t^2 - 2c_0 + c (t^2 - 4t + 4) \)

\( x = c_0 t^2 - 2c_0 + c t^2 - 4ct + 4c \)

\( x = (c_0 + c) t^2 - 4ct + (4c - 2c_0) \)

Velocity is found by differentiating the displacement with respect to time:

\( v = \frac{dx}{dt} = \frac{d}{dt} [(c_0 + c) t^2 - 4ct + (4c - 2c_0)] \)

\( v = 2(c_0 + c) t - 4c \)

Acceleration is found by differentiating the velocity with respect to time:

\( a = \frac{dv}{dt} = \frac{d}{dt} [2(c_0 + c) t - 4c] \)

\( a = 2(c_0 + c) \)

The particle's acceleration is \( 2(c_0 + c) \).

Therefore, the correct statement is:

the acceleration of the particle is \( 2(c + c_0) \)

Final Answer: The final answer is the acceleration of the particle is \( 2(c + c_0) \)