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WBJEE · Maths · Application of Derivatives

A particle is moving in a straight line. At time \(t\), the distance between the particle from its starting point is given by \(x=t-6 t+z\). Its acceleration will be zero at

  1. A \(t=1\) unit time
  2. B \(t=2\) unit time
  3. C \(t=3\) unit time
  4. D \(t=4\) unit time
Verified Solution

Answer & Solution

Correct Answer

(B) \(t=2\) unit time

Step-by-step Solution

Detailed explanation

Hints : \(x=t-6 t^2+t^3\) \[ \begin{aligned} & \frac{d x}{d t}=1-12 t+3 t^2 \\ & \frac{d^2 x}{d t^2}=-12+6 t \end{aligned} \] Acceleration \(=\frac{d^2 x}{d t^2}\) \(\therefore\) Acceleration \(=0 \Rightarrow 6 t-12=0 \Rightarrow t=2\)