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
- A \(t=1\) unit time
- B \(t=2\) unit time
- C \(t=3\) unit time
- D \(t=4\) unit time
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 \(=0 \Rightarrow 6 t-12=0 \Rightarrow t=2\)
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