A particle is moving in a straight line. The variation of position ' x ' as a function of time ' t ' is given as x=(t^3-6 t^2+20 t+15) m . The velocity of the body when its acceleration becomes zero is:

Correct answer: C. 8 m / s Displacement, x=t^3-6 t^2+20 t+15 Velocity, v= dx dt =3 t^2-12 t+20 Acceleration, a=(d v)/(d t)=6 t-12 When a=0 6 t-12=0 t=2 s At t=2 s, v=3(2)^2-12(2)+20=8 m / s

BITSAT 2024Paper 2

Exam: BITSAT
Year: 2024
Paper: Paper 2
Subject: Physics
Chapter: Motion in One Dimension

BITSAT 2024 · Paper 2 · Physics · Motion in One Dimension

A particle is moving in a straight line. The variation of position...

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:

Options

A $6 \mathrm{ m} / \mathrm{s}$
B $10 \mathrm{ m} / \mathrm{s}$
C $8 \mathrm{ m} / \mathrm{s}$
D $4 \mathrm{ m} / \mathrm{s}$

Correct answer

C. $8 \mathrm{ m} / \mathrm{s}$

Step-by-step solution

Displacement,

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

Velocity,

v=\frac{\mathrm{dx}}{\mathrm{dt}}=3 \mathrm{t}^2-12 \mathrm{t}+20\therefore

Acceleration,

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

When

a=0\Rightarrow 6 \mathrm{t}-12=0 \Rightarrow t=2 \mathrm{ s}

\mathrm{t}=2 \mathrm{ s}, v=3(2)^2-12(2)+20=8 \mathrm{ m} / \mathrm{s}

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