A particle is moving in a straight line. The variation of position '$x$' as a function of time '$t$' is given as $x=\left(t^3-6 t^2+20 t+15\right) m$. The velocity of the body when its acceleration becomes zero is :
Solution
<p>The position equation is given by:</p>
<p>$x = t^3 - 6t^2 + 20t + 15$</p>
<p>First, compute the velocity $v$ by differentiating the position function with respect to time:</p>
<p>$\frac{d x}{d t} = v = 3t^2 - 12t + 20$</p>
<p>Next, compute the acceleration $a$ by differentiating the velocity function with respect to time:</p>
<p>$\frac{d v}{d t} = a = 6t - 12$</p>
<p>We need to find the time $t$ when the acceleration is zero:</p>
<p>$6t - 12 = 0$<br>
<p>$t = 2 \, \mathrm{sec}$</p></p>
<p>Now, find the velocity at $t = 2 \, \mathrm{sec}$:</p>
<p>$v = 3(2)^2 - 12(2) + 20$</p>
<p>$v = 8 \, \mathrm{m/s}$</p>
About this question
Subject: Physics · Chapter: Kinematics · Topic: Motion in a Straight Line
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