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

Q78 / 100 · BITSAT 2024 (Memory Based Paper 2) · Physics

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:

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

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 $a=0$ $\Rightarrow 6 \mathrm{t}-12=0 \Rightarrow t=2 \mathrm{~s}$ At $\mathrm{t}=2 \mathrm{~s}, v=3(2)^2-12(2)+20=8 \mathrm{~m} / \mathrm{s}$

Hence, the correct answer is option: 'C'.

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

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