Circular MotionHardBloom L3

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 = t2 + 5, where s is in metres and t is in seconds. The radius of the path is 20 m. The acceleration of 'P' when $t = 5\sqrt{\frac{3}{10}}$seconds is nearly :

Options

A.2 m/s2
B.1.5 m/s2
C.2.5 m/s2
D.3 m/s2

Solution

Sol. v = 2t

at = 2 m/s2

$a_{c} = \frac{v^{2}}{r} = \frac{4t^{2}}{20} = \frac{t^{2}}{5} = \frac{25 \times 3}{50} = \frac{3}{2}$ m/s2

So atotal = $\sqrt{4 + \frac{9}{4}} = \sqrt{\frac{25}{4}}$ = 2.5

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