Easy MCQ +4 / -1 PYQ · JEE Mains 2022

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

  1. A $-30 \hat{y}$
  2. B $30 \hat{y}$ Correct answer
  3. C $3 \hat{x}+15 \hat{y}$
  4. D $3 \hat{x}+15 \hat{y}+7 \hat{z}$

Solution

<p>$x = 3t \Rightarrow {a_x} = 0$</p> <p>$y = 5{t^3} \Rightarrow {a_y} = 30t$</p> <p>$\overrightarrow a (t = 1) = 30\widehat y$</p>

About this question

Subject: Physics · Chapter: Kinematics · Topic: Motion in a Straight Line

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