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:

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

We can expand this:

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

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

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

To find the velocity, we differentiate the displacement equation with respect to time:

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

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

To find the acceleration, we differentiate the velocity equation with respect to time:

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

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

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

Thus, 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 \( 2(c + c_0) \)

Answer 2

To determine the correct statement about the particle's motion, we need to analyze the displacement equation given:

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

Let's proceed step-by-step to compute the velocity and acceleration of the particle:

1. First, differentiate the displacement equation with respect to time \(t\) to find the expression for velocity \(v\):

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

\(= c_0 \cdot 2t + c \cdot 2(t - 2)\)

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

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

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

1. Next, differentiate the velocity expression with respect to time \(t\) to find the acceleration \(a\):

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

\(= 2c_0 + 2c\)

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

1. Verify the initial velocity by substituting \(t=0\) in the velocity expression:

\(v_0 = (2c_0 + 2c)\cdot 0 - 4c = -4c\)

This confirms that the initial velocity is not \(4c\).

Thus, the correct statement is that the acceleration of the particle is \(2(c + c_0)\).