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Physics - Kinematics Question with Solution | TestHub

PhysicsKinematics1-DMedium2 minQB

PhysicsMediumsingle choice

A body is in straight line motion with an acceleration given by $a = 32 - 4 v$ . The initial conditions are $t = 0$ , $v = 4$ . The velocity when $t = ln 2$ is:

Options:

Answer:

Solution:

$\frac{d v}{32 - 4 v} = d t$ integrates over $v = 4$ to $v$ and $t = 0$ to $ln 2$ as $- \frac{1}{4} ln \frac{32 - 4 v}{16} = ln 2$ , so $\frac{32 - 4 v}{16} = \frac{1}{16}$ , giving $32 - 4 v = 1$ and $v = \frac{31}{4}$ . Option D.

Stream:JEESubject:PhysicsTopic:KinematicsSubtopic:1-D

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