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JEE Main 2005
Exam
JEE Main
Year
2005
Subject
Physics
Chapter
Motion in One Dimension

JEE Main 2005 · Physics · Motion in One Dimension

A car starting from rest accelerates at the rate f through a...

A car starting from rest accelerates at the rate ff through a distance SS, then continues at constant speed for time tt and then decelerates at the rate f/2f / 2 to come to rest. If the total distance traversed is 15 S15 \mathrm{~S}, thenA car starting from rest accelerates at the rate f through a distance S , then continues at constant speed for time tA car starting from rest accelerates at the rate f through a distance S , then continues at constant speed for time t

Options

  1. AS=ftS=f t
  2. BS=1/6ft2\mathrm{S}=1 / 6 \mathrm{ft}^2
  3. CS=1/2ft2\mathrm{S}=1 / 2 \mathrm{ft}^2
  4. DNone of these

Correct answer

D. None of these

Step-by-step solution

A car starting from rest accelerates at the rate f through a distance S , then continues at constant speed for time t
S=ft122v0=2Sf\begin{aligned}& \mathrm{S}=\frac{\mathrm{ft}_1^2}{2}\\ & \mathrm{v}_{0}=\sqrt{2 \mathrm{Sf}} \end{aligned}During retardation S_2=2 S During constant velocity 15 S−3 S=12 S=v0t⇒S=ft272\begin{aligned} & 15 \mathrm{~S}-3 \mathrm{~S}=12 \mathrm{~S}=\mathrm{v}_{0} \mathrm{t}\\ & \Rightarrow \mathrm{S}=\frac{\mathrm{ft}^2}{72} \end{aligned}$
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