SATHEE: Motion in a Straight Line - Result Question 25

Motion in a Straight Line - Result Question 25

26. A car moving with a speed of $40 k m / h$ can be stopped by applying brakes at least after $2 m$ . If the same car is moving with a speed of $80 k m / h$ , what is the minimum stopping distance?[1998]

(a) $8 m$

(b) $6 m$

(c) $4 m$

(d) $2 m$

Show Answer

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 $s \propto u^{2}$

$∵ \frac{s_{2}}{s_{1}} = \frac{u_{2}^{2}}{u_{1}^{2}} \Rightarrow \frac{s_{2}}{s_{1}} = (\frac{80}{40})^{2} = 4$

$∴ s_{2} = 4 s_{1} = 8 m$

If $F$ is retarding force and $s$ the stopping distance, then $\frac{1}{2} m v^{2} = F S$

For same retarding force, $s α v^{2}$

$∴ \frac{s_{2}}{s_{1}} = (\frac{v_{2}}{v_{1}})^{2} = (\frac{80 k m / h}{40 k m / h})^{2} = 4$

$∴ s_{2} = 4 s_{1} = 4 \times 2 = 8 m$

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