Physics - Kinematics Question with Solution | TestHub

PhysicsKinematics1-DEasy2 minPYQ_2010

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The position $x$ of a particle varies with time $(t)$ as $x = A t^{2} - B t^{3}$ . The acceleration at time $t$ of the particle will be equal to zero. What is the value of $t ?$

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Answer:

Solution:

Given that $x = A t^{2} - B t^{3}$ $∴ velocity = \frac{dx}{dt} = 2 A t - 3 B t^{2}$ and acceleration $= \frac{d}{dt} (\frac{dx}{dt}) = 2 A - 6 Bt$ For acceleration to be zero $2 A - 6 Bt = 0$ . $∴ t = \frac{2 A}{6 B} = \frac{A}{3 B}$

Stream:BITSATSubject:PhysicsTopic:KinematicsSubtopic:1-D

⏱ 2mℹ️ Source: PYQ_2010

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