Q83 / 90 · JEE Main 2010 · Physics

A point $\mathrm{P}$ moves in counter-clockwise direction on a circular path as shown in the figure. The movement of ' $\mathrm{P}$ ' is such that it sweeps out a length $s=t^3+5$, where $s$ is in metres and $t$ is in seconds. The radius of the path is $20 \mathrm{~m}$. The acceleration of ' $\mathrm{P}$ ' when $t=2 \mathrm{~s}$ is nearly Figure

  • A
    $13 \mathrm{~m} / \mathrm{s}^2$
  • B
    $12 \mathrm{~m} / \mathrm{s}^2$
  • C
    $7.2 \mathrm{~m} / \mathrm{s}^2$
  • Correct answer
    D
    $14 \mathrm{~m} / \mathrm{s}^2$

Solution

$$ S=t^3+5 $$ $\therefore \quad$ speed, $\mathrm{v}=\frac{\mathrm{ds}}{\mathrm{dt}}=3 \mathrm{t}^2$ and rate of change of speed $=\frac{\mathrm{dv}}{\mathrm{dt}}=6 \mathrm{t}$ $\therefore \quad$ tangential acceleration at $\mathrm{t}=2 \mathrm{~s}, \mathrm{a}_{\mathrm{t}}=6 \times 2=12 \mathrm{~m} / \mathrm{s}^2$ at $t=2 \mathrm{~s}, \mathrm{v}=3(2)^2=12 \mathrm{~m} / \mathrm{s}$ $\therefore \quad$ centripetal acceleration, $\quad a_c=\frac{v^2}{R}=\frac{144}{20} \mathrm{~m} / \mathrm{s}^2$ $\therefore \quad$ net acceleration $=\sqrt{\mathrm{a}_t^2+\mathrm{a}_{\mathrm{i}}^2}$ $\approx 14 \mathrm{~m} / \mathrm{s}^2$

Hence, the correct answer is option: 'D'.

Relevant Questions