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JEE Mains · Physics · STD 11 - 3.2 motion in plane

At time \(t =0\) a particle starts travelling from a height \(7\,\hat{z} cm\) in a plane keeping \(z\) coordinate constant. At any instant of time it's position along the \(x\) and \(y\) directions are defined as \(3\,t\) and \(5\,t^{3}\) respectively. At \(t =1\,s\) acceleration of the particle will be.

  1. A \(-30\,y\)
  2. B \(30\,y\)
  3. C \(3 x+15 y\)
  4. D \(3 x+15 y+7 \hat{z}\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(30\,y\)

Step-by-step Solution

Detailed explanation

\(\overrightarrow{ r }=3 t \hat{ i }+5 t ^{3} \hat{ j }+7 k\) \(\frac{ d ^{2} \overrightarrow{ r }}{ dt ^{2}}=30 t \hat{ j }\) At \(t=1 \Rightarrow \frac{d^{2} \overrightarrow{r}}{{d t^{2}}}=30\,\hat{j}\)
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