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. 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:

For motion along a straight line: \[ v = \frac{dx}{dt}, \quad a = \frac{d^2x}{dt^2} \] Always simplify the displacement equation before differentiating.

Answer 1

The Correct Option is D

Concept:

Velocity is the first derivative of displacement with respect to time.
Acceleration is the second derivative of displacement with respect to time.
Step 1: Simplify the displacement equation. \[ x = c_0(t^2 - 2) + c(t - 2)^2 \] \[ x = c_0 t^2 - 2c_0 + c(t^2 - 4t + 4) \] \[ x = (c_0 + c)t^2 - 4ct + (4c - 2c_0) \]
Step 2: Find velocity by differentiating with respect to time. \[ v = \frac{dx}{dt} = 2(c_0 + c)t - 4c \]
Step 3: Find acceleration by differentiating velocity. \[ a = \frac{dv}{dt} = 2(c_0 + c) \]
Step 4: Conclusion. Acceleration is constant and equal to: \[ a = 2(c + c_0) \]