Question

a particle is performing circular motion of radius 1m.its speed is v=(2t^2)m/s. what will be the magnitude of its acceleration at t=1s?​

Answer 1

$\Large\bold{\underline{\underline{Answer:-}}}$

$\Large\bold{\boxed{\boxed{4\sqrt{2} \:m/s^2 }}}$

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$\bf\small\bold{\underline{\underline{Step-By-Step\:Explanation:-}}}$

Radial/Centripetal acceleration,$\bf\bold{a_c= \frac{v^2}{r}}$

at t = 1s, v= 2 m/s  $\bf\bold{a_c = \frac{2^2}{1} = 4 m/s^2}$

Tangential acceleration, $\bf\bold{a_t= \frac{dv}{dt}=4t}$

at t = 1 s, $\bf\bold{a_t= 4 m/s^2}$

So, $\bf\bold{a_{res} =\sqrt{a_c^2 + a_t^2}}$

$\bf\bold{\Rightarrow a_{res} =\sqrt{4^2 + 4^2}}$

$\bf\bold{\Rightarrow a_{res} = 4\sqrt{2} m/s^2}$

Hence the acceleration of the particle is $\bf\bold{ 4\sqrt{2} \:m/s^2}$

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Centripetal acceleration — It is always directed towards the centre of the circle. Since v and R are constants for a given uniform circular motion, therefore the magnitude of centripetal acceleration is also constant. However, the direction of centripetal acceleration changes continuously. Therefore, a centripetal acceleration is not a constant vector.

Tengential acceleration— Tangential acceleration is a measure of how quickly the tangential velocity changes. The tangential velocity will be a tangent to the path of the particle that is moving in a curved path