A particle moves along the x-axis and has its displacement x varying with time t according to the equation:
x=c₀(t²-2)+c(t-2)²
where c₀ and c are constants of appropriate dimensions.
Then, which of the following statements is correct?

Solution

Physics sir · 4 steps

Differentiate twice: both brackets carry a t², so both constants show up in the acceleration.

1. Velocity. Differentiate x once with respect to t. The constant -2c₀ drops out, and the chain rule on (t-2)² gives 2(t-2).

   v = dx/dt = 2c₀ t + 2c(t-2)

2. Acceleration. Differentiate once more. Both terms are linear in t, so each gives a constant: a = dv/dt = 2c₀ + 2c = 2(c + c₀).
3. Check option (d). Put t = 0 in v: v₀ = 0 + 2c(0 - 2) = -4c. The magnitude is 4c, but the sign is negative, so (d) is not correct.
4. So. The acceleration is constant and equal to 2(c + c₀), which is option (a).

The idea to keep. Before differentiating, look for every term that hides a t². Each one adds to the acceleration. Signs matter too: always put t = 0 in carefully for the initial velocity.

Answer from the NTA final answer key. Solution by PrepWiser.