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Physics - Circular Motion Question with Solution | TestHub

PhysicsCircular MotionKinematicsMedium2 minQB

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A particle rotates in a circular path of radius 54 m with varying speed $v = 4 t^{2}$ . Here $v$ is in m/s and $t$ in second. Find angle between velocity and acceleration at $t = 3 s$ .

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Answer:

Solution:

$a_{t} = \frac{d v}{d t} = 8 t = 24 m / s^{2} at t = 3 s; a_{r} = \frac{v ^{2}}{R} = \frac{( 4 \times 9 ) ^{2}}{54} = \frac{1296}{54} = 24 m / s^{2} . tan θ = \frac{a _{r}}{a _{t}} = 1, so θ = 4 5^{\circ} .$

Stream:JEESubject:PhysicsTopic:Circular MotionSubtopic:Kinematics

⏱ 2mℹ️ Source: QB

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