ExamBro
ExamBro
JEE Mains · Physics · STD 11 - 2. 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 _______.

  1. A  \(4 \mathrm{~m} / \mathrm{s}\)
  2. B \(8 \mathrm{~m} / \mathrm{s}\)
  3. C \(10 \mathrm{~m} / \mathrm{s}\)
  4. D \(6 \mathrm{~m} / \mathrm{s}\)
Verified Solution

Answer & Solution

Correct Answer

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

Step-by-step Solution

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
Explore more questions on app
From JEE Mains
Explore more questions on app