A particle is moving in a straight line. The variation of position ' x ' as a function of time ' t ' is given as x= (t3-6 t2+20 t+15) m. The velocity of the body when its acceleration becomes zero is _______.

(C) 10 \mathrm{~m} / \mathrm{s}

JEE Mains · Physics · STD 11 - Motion in straight line

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 _______.

Pick an option to check your answer.

Verified Solution

Correct Answer

(B) \(8 \mathrm{~m} / \mathrm{s}\)

Detailed explanation

\( x=t^3-6 t^2+20 t+15 \)

\( \frac{d x}{d t}=v=3 t^2-12 t+20\)

\( \frac{d v}{d t}=a=6 t-12\)

When \(\mathrm{a}=0\)

\(6 \mathrm{t}-12=0 ; \mathrm{t}=2 \mathrm{sec}\)

At \(\mathrm{t}=2 \mathrm{sec}\)

\(\mathrm{v}=3(2)^2-12(2)+20\)

\(\mathrm{v}=8 \mathrm{~m} / \mathrm{s}\)

Same subject

List\(-I\) (Layer of atmosphere)	List\(-II\) (Approximate height over earth's surface)
\((A)\) \(F_1\) - Layer	\((I)\) \(10\,km\)
\((B)\) \(D-\) Layer	\((II)\) \(170-190\,km\)
\((C)\) Troposphere	\((III)\) \(100\,km\)
\((D)\) \(E-\)layer	\((IV)\) \(65-75\,km\)

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