Question

a particle moves in circular path with uniform speed of 5 m/s. Find the acceleration of the particle when it moves from one end of a diameter to its other end in 3s.​

Answer 1

the answer is 2 the answer is 2

Answer 2

Given:

A particle moves in circular path with uniform speed of 5 m/s.

To find:

Acceleration of the particle when it moves from one end of a diameter to its other end in 3s.

Calculation:

Acceleration can be calculated by change in velocity divided by time taken.

$\rm \: a = \dfrac{change \: in \: v}{t}$

• For a angular displacement of $\theta$ , change in velocity comes as $\Delta v=2v\sin(\theta/2)$

$\rm \implies \: a = \dfrac{2v \sin( \frac{ \theta}{2} ) }{t}$

• Here $\theta$ = 180°.

$\rm \implies \: a = \dfrac{2v \sin( \frac{ {180}^{ \circ} }{2} ) }{t}$

$\rm \implies \: a = \dfrac{2v \sin({90}^{ \circ} ) }{t}$

$\rm \implies \: a = \dfrac{2v }{t}$

$\rm \implies \: a = \dfrac{2 \times 5 }{3}$

$\rm \implies \: a = 3.33 \: m {s}^{ - 2}$

So, acceleration is 3.33 m/s².