Motion in a Straight Line - Result Question 25
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26. A car moving with a speed of $40 km / h$ can be stopped by applying brakes at least after $2 m$. If the same car is moving with a speed of $80 km / h$, what is the minimum stopping distance?[1998]
======= ####26. A car moving with a speed of $40 km / h$ can be stopped by applying brakes at least after $2 m$. If the same car is moving with a speed of $80 km / h$, what is the minimum stopping distance?[1998]
3e0f7ab6f6a50373c3f2dbda6ca2533482a77bed:content/english/neet-pyq-chapterwise/physics/motion-in-a-straight-line/motion-in-a-straight-line—result-question-25.md (a) $8 m$
(b) $6 m$
(c) $4 m$
(d) $2 m$
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Answer:
Correct Answer: 26. (a)
Solution:
- (a) $v^{2}-u^{2}=2 a s$
$ \begin{aligned} \Rightarrow a & =\frac{v^{2}-u^{2}}{2 s} \\ & =-\frac{u_1^{2}}{2 s} \end{aligned} $
For same retarding force $s \propto u^{2}$
$\because \frac{s_2}{s_1}=\frac{u_2^{2}}{u_1^{2}} \Rightarrow \frac{s_2}{s_1}=(\frac{80}{40})^{2}=4$
$\therefore s_2=4 s_1=8 m$
If $F$ is retarding force and $s$ the stopping distance, then $\frac{1}{2} m v^{2}=F S$
For same retarding force, $s \alpha v^{2}$
$ \therefore \quad \frac{s_2}{s_1}=(\frac{v_2}{v_1})^{2}=(\frac{80 km / h}{40 km / h})^{2}=4 $
$\therefore \quad s_2=4 s_1=4 \times 2=8 m$
