JEE Main · 2010 · 11 Apr 2010 · Offline · Kinematics

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

Answer: (D)

Solution

The path length is

s = t^{3} + 5

, so the speed is

v = \frac{d s}{d t} = 3 t^{2}

and the tangential acceleration is

a_{t} = \frac{d v}{d t} = 6 t

. At

t = 2

s the speed is

v = 3 \times 2^{2} = 12

m/s and

a_{t} = 12

m/s

^{2}

. Moving on a circle of radius

20

m there is also a centripetal acceleration

a_{c} = \frac{v ^{2}}{R} = \frac{144}{20} = 7.2

m/s

^{2}

towards the centre. These two are perpendicular, so the net acceleration is

a = a_{t}^{2} + a_{c}^{2} = 144 + 51.84 \approx 14

m/s

^{2}

JEE Main · 2010 · 11 Apr 2010 · Offline · Single correct option