A car, moving with a speed of 50 ~km / hr , can be stopped by brakes after at least 6 ~m . If the same car is moving at a speed of 100 ~km / hr , the minimum stopping distance is

Correct answer: C. 24 ~m Energy = Work done by force (F) ⇒ (1)/(2) ~m .( 50)^2=( F)( 6) F= 2500 ~m 2 × 6 For v=100 ~km / hr (1)/(2) · m(100)^2=(F)(S) aligned & (1)/(2) ~m(100)^2=( 2500 ~m 2 × 6 ) S & S=(100 × 100 × 6 × 2)/(2500 × 2)=24 ~m aligned

JEE Main 2003

Exam: JEE Main
Year: 2003
Subject: Physics
Chapter: Motion in One Dimension

JEE Main 2003 · Physics · Motion in One Dimension

A car, moving with a speed of 50 ~km / hr , can be stopped by...

A car, moving with a speed of

50 \mathrm{~km} / \mathrm{hr}

, can be stopped by brakes after at least

6 \mathrm{~m}

. If the same car is moving at a speed of

100 \mathrm{~km} / \mathrm{hr}

, the minimum stopping distance is

Options

A $12 \mathrm{~m}$
B $18 \mathrm{~m}$
C $24 \mathrm{~m}$
D $6 \mathrm{~m}$

Correct answer

C. $24 \mathrm{~m}$

Step-by-step solution

Energy = Work done by force

(\mathrm{F}) ⇒ \frac{1}{2} \mathrm{~m} .(

50)^2=(

\mathrm{F})(

6) \quad \Rightarrow

\mathrm{F}=\frac{2500 \mathrm{~m}}{2 \times 6}

For

\mathrm{v}=100 \mathrm{~km} / \mathrm{hr} \frac{1}{2} \cdot \mathrm{m}(100)^2=(\mathrm{F})(\mathrm{S})

\begin{aligned}& \Rightarrow \frac{1}{2} \mathrm{~m}(100)^2=\left(\frac{2500 \mathrm{~m}}{2 \times 6}\right) \mathrm{S}\\ & \Rightarrow \mathrm{S}=\frac{100 \times 100 \times 6 \times 2}{2500 \times 2}=24 \mathrm{~m} \end{aligned}

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