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JEE Main 20127 MayOnline
Exam
JEE Main
Year
2012
Date
7 May
Mode
Online
Subject
Physics
Chapter
Laws of Motion

JEE Main 2012 · 7 May · Online · Physics · Laws of Motion

A car of mass 1000 ~kg is moving at a speed of 30 m / s . Brakes...

A car of mass 1000 kg1000 \mathrm{~kg} is moving at a speed of 30 m/s\mathrm{m} / \mathrm{s}. Brakes are applied to bring the car to rest. If the net retarding force is 5000 N5000 \mathrm{~N}, the car comes to stop after travelling d md \mathrm{~m} in t st \mathrm{~s}. Then

Options

  1. Ad=150,t=5d=150, t=5
  2. Bd=120,t=8d=120, t=8
  3. Cd=180,t=6d=180, t=6
  4. Dd=90,t=6d=90, t=6

Correct answer

D. d=90,t=6d=90, t=6

Step-by-step solution

Given: mass of car m=1000 kgm=1000 \mathrm{~kg}u=30 m/sv=0 m/s retarding force f=5000 N∴ retardation, −a=50001000=5 m/s2\begin{aligned}& u=30 \mathrm{~m}/ \mathrm{s}\\ & v=0 \mathrm{~m}/ \mathrm{s}\\ & \text{ retarding force }f=5000 \mathrm{~N}\\ & \therefore \text{ retardation, }-a=\frac{5000}{1000}=5 \mathrm{~m}/ \mathrm{s}^2 \end{aligned}Byequation,By equation,v^2-u^2=2asas0−(30)2=−2×5\timesd∴d=90010=90 m\begin{aligned}& 0-(30)^2=-2 \times5 \timesd \\ & \therefore d=\frac{900}{10}=90 \mathrm{~m} \end{aligned}andanda=\frac{v-u}{t} \therefore t=\frac{v-u}{a}=\frac{0-30}{-5}=6 \mathrm{~s}
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