If the velocity of a body related to displacement x is given by $\upsilon = \sqrt {5000 + 24x}$ m/s, then the acceleration of the body is .................... m/s2.
Answer (integer)
12
Solution
$V = \sqrt {5000 + 24x}$<br><br>$${{dV} \over {dx}} = {1 \over {2\sqrt {5000 + 24x} }} \times 24 = {{12} \over {\sqrt {5000 + 24x} }}$$<br><br>Now, $a = V{{dV} \over {dx}}$<br><br>$= \sqrt {5000 + 24x} \times {{12} \over {\sqrt {5000 + 24x} }}$<br><br>a = 12 m/s<sup>2</sup>
About this question
Subject: Physics · Chapter: Kinematics · Topic: Motion in a Straight Line
This question is part of PrepWiser's free JEE Main question bank. 112 more solved questions on Kinematics are available — start with the harder ones if your accuracy is >70%.