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 = t³ + 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/s² (b) 12 m/s² (c) 7.2 m/s² (d) 14 m/s²

Answer: (d). v = ds/dt = 3t² = 12 m/s at t = 2 s Tangential: at = dv/dt = 6t = 12 m/s ² Centripetal: ac = v²/R = 144/20 = 7.2 m/s ² a = √(at² + ac²) = √(144 + 51.84) ≈ 14 m/s ² So the answer is (d) 14 m/s².

A point P moves in counter-clockwise direction on a circular path as… | JEE Main 2010 (AIEEE)

Q21JEE Main 2010 · AIEEEMedium

Non-Uniform Circular Motion

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 $20\text{ m}.$ The acceleration of $'P'$ when $t = 2\text{ s}$ is nearly.

Show answer and solution

Answer: (d)

$v = \frac{ds}{dt} = 3t^2 = 12$ m/s at $t = 2$ s
Tangential: $a_t = \frac{dv}{dt} = 6t = 12$ m/s $^2$
Centripetal: $a_c = \frac{v^2}{R} = \frac{144}{20} = 7.2$ m/s $^2$
$a = \sqrt{a_t^2 + a_c^2} = \sqrt{144 + 51.84} \approx 14$ m/s $^2$
So the answer is (d) $14\text{ m/s}^2$ .

Report a mistake

Non-Uniform Circular Motion question from JEE Main 2010 (AIEEE)

More questions on Non-Uniform Circular Motion