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JIPMER 2018
Exam
JIPMER
Year
2018
Subject
Physics
Chapter
Motion in One Dimension

JIPMER 2018 · Physics · Motion in One Dimension

Velocity is given by v=4 t(1-2 t) , then find time at which...

Velocity is given by v=4t(1−2t)v=4 t(1-2 t), then find time at which velocity is maximum

Options

  1. A0.25s0.25 s
  2. B1s1 s
  3. C0.45s0.45 s
  4. D4s4 s

Correct answer

A. 0.25s0.25 s

Step-by-step solution

To find value of time at which velocity is maximum, taking differentiation of v with respect to time ⇒dvdt=0\Rightarrow \quad \frac{\mathrm{dv}}{\mathrm{dt}}=0 Here, v=4t(l−2t)v=4 t(l-2 t) v=4t−8t2⇒ddt(4t−8t2)=0⇒4−16t=0⇒t=14 s=0.25 s\begin{array}{lc} & \mathrm{v}=4 \mathrm{t}-8 \mathrm{t}^2 \\ \Rightarrow & \frac{\mathrm{d}}{\mathrm{dt}}\left(4 \mathrm{t}-8 \mathrm{t}^2\right)=0 \\ \Rightarrow & 4-16 \mathrm{t}=0 \\ \Rightarrow & \mathrm{t}=\frac{1}{4} \mathrm{~s}=0.25 \mathrm{~s}\end{array} Again taking differentiation, ⇒d2vdt2=−16<0\Rightarrow \quad \frac{\mathrm{d}^2 v}{\mathrm{dt}^2}=-16 \lt 0 So, at t=0.25 s\mathrm{t}=0.25 \mathrm{~s} velocity is maximum.
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