Question

A point $P$ moves in counter-clockwise direction on a circular path as shown in the figure. The movement of $'P'$ 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 $20\, m$. The acceleration of $‘P’$ when $t\, =\, 2\, s$ is nearly

• A. $13 \,m/s^2$
• B. $12 \,m/s^2$
• C. $7.2 \,m/s^2$
• D. $14 \,m/s^2$

Answer 1

Answer: (D)

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