JEE MAIN - Physics (2021 - 27th August Morning Shift - No. 23)
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
12
Explanation
$$V = \sqrt {5000 + 24x} $$
$${{dV} \over {dx}} = {1 \over {2\sqrt {5000 + 24x} }} \times 24 = {{12} \over {\sqrt {5000 + 24x} }}$$
Now, $$a = V{{dV} \over {dx}}$$
$$ = \sqrt {5000 + 24x} \times {{12} \over {\sqrt {5000 + 24x} }}$$
a = 12 m/s2
$${{dV} \over {dx}} = {1 \over {2\sqrt {5000 + 24x} }} \times 24 = {{12} \over {\sqrt {5000 + 24x} }}$$
Now, $$a = V{{dV} \over {dx}}$$
$$ = \sqrt {5000 + 24x} \times {{12} \over {\sqrt {5000 + 24x} }}$$
a = 12 m/s2
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