Question

If x=2t^3-3t^2-36t +100 find the position when velocity is zero. (Assume SI units)
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Answer 1

Need to FinD :-

• The position when velocity is zero.

$\red{\frak{Given}}\begin{cases}\sf x = 2t^3 -3t^2-36t + 100 \\\sf All \ quantities \ are \ in \ SI \ unit .\end{cases}$

We need to find the position when velocity is 0 . We will differnciate both sides with respect to t , First order of differenciation will give Velocity . Subsequently second order of differenciation will give acceleration.

• Differenciating both sides wrt t :-

$\sf\dashrightarrow x = 2t^3 -3t^2-36t + 100 \\\\\sf\dashrightarrow \dfrac{dx}{dt}= \dfrac{ d( 2t^3 -3t^2-36t+100}{dx} \\\\\sf\dashrightarrow \dfrac{dx}{dt}= (2)(3)t^{3-1} - (3)(2)t^{2-1} -36t^{1-1} + 0 \\\\\sf\dashrightarrow \dfrac{dx}{dt}= 6t^2 - 6t -36\\\\\ \qquad\qquad\tiny{\red{\sf When \ velocity \ is \ 0}} \\\\\sf\dashrightarrow 0 = 6( t^2 -t - 6 ) \\\\\sf\dashrightarrow t^2 - t - 6 = 0 \\\\\sf\dashrightarrow t^2 -3t + 2t -6 = 0 \\\\\sf\dashrightarrow t(t-3)+2(t-3) = 0 \\\\\sf\dashrightarrow (t-3)(t+2) = 0 \\\\\sf\dashrightarrow \underset{\blue{\sf Values \ of \ t }}{\underbrace{\boxed{\pink{\frak{ t = 3 , (-2) }}}}}$

Here t = -2 will be neglected , since time can't be negative . On putting t = 3 , we have ,

$\sf\dashrightarrow x = 2t^3 -3t^2-36t + 100 \\\\\sf\dashrightarrow x = 2(3)^3 - 3(3)^2 -36(3) + 100 \\\\\sf\dashrightarrow x = (2)(27)-(3)(9)-108+100 \\\\\sf\dashrightarrow x = 54 - 27 - 8 \\\\\sf\dashrightarrow x = 54 - 35 \\\\\sf\dashrightarrow \boxed{\pink{\frak{ x_{(At \ t \ = \ 3)} =19 m }}}$

Hence the position when its Velocity is 0 is 19 m .