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Physics-

General

Easy

Question

A particle located at $x equals 0$ at time $t equals 0$ , starts moving along the positive $x$ -direction with a velocity ‘ $v$ ’ that varies as $v equals a square root of x$ . The displacement of the particle varies with time as

$t$
$t^{1 / 2}$
$t^{3}$
$t^{2}$

The correct answer is: $t to the power of 2 end exponent$

$v equals alpha square root of x rightwards double arrow fraction numerator d x over denominator d t end fraction equals proportional to square root of x rightwards double arrow fraction numerator d x over denominator square root of x end fraction equals alpha d t$
By integrating both sides $not stretchy integral x to the power of negative 1 divided by 2 end exponent d x equals not stretchy integral alpha blank d t$
$rightwards double arrow fraction numerator square root of x over denominator 1 divided by 2 end fraction equals alpha t rightwards double arrow square root of x equals fraction numerator 1 over denominator 2 end fraction alpha t rightwards double arrow x equals fraction numerator 1 over denominator 4 end fraction alpha to the power of 2 end exponent t to the power of 2 end exponent blank therefore x proportional to t to the power of 2 end exponent$

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