Question

A car moving with speed of 50km pr hr
Can be stopped by brakes after at least
6 min. If the same car is moving at a speed of 100 km pr hr, the minimum
Stopping distance is

A) 6 min
B) 2 min
C) 18min
D) 24min

Answer 1

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$\underline{\huge{\textbf{Solution}}}$:


Given,

Initial velocity of the Car $u = 50\: kmph$

Distance travelled before coming to rest = $6\: km$

If the same car is moving at a speed of 100 km pr hr, the minimum stopping distance is ?

$\underline{\large{\textbf{Step-1}}}$:
Find Initial velocity and final velocity of the car in terms of mps

We have,
$1: kmph = \frac{5}{18}\: mps$

lnitial velocity of the car = $50 \times \frac{5}{18}\: mps= \frac{125}{9}$

Final velocity = $0\: mps$
(Final Velocity will be zero when the breaks are applied, so it is 'Retardation)

$\underline{\large{\textbf{Step-2}}}$:
Find the retardation acting on the car

Using the relation,
$v^{2}-u^{2} = 2as$

when v = 0,
$a = - \frac{u^{2}}{2s}$

Here,
-ve sign just indicates the fact that velocity is decreasing.

$a = \frac{u^{2}}{2s}$ -------------[1]

Substituting Values

$\implies a = \dfrac{(\frac{125}{9})^{2}}{2 \times 6}$

$\implies a = \dfrac{(\frac{125}{9})^{2}}{12}$

$\underline{\large{\textbf{Step-3}}}$:
Find the stopping distance when the speed of the car is 100 kmph

Firstly, Express 100 kmph to mps

$Speed of the car = 100 \times \frac{5}{18}\: mps = \frac{250}{9}\: mps$

Now,
From [1],

$Stopping\: Distance (S) = \frac{u^{2}}{2a}$

$\implies S = \dfrac{(\frac{250}{9})^{2}}{2 \times (\frac{(\frac{125}{9})^{2}}{12})}$

$\implies S = 6 \times 4 = 24\: m$

Therefore,
If the car is moving with 100 kmph , then Minimum stopping distance is 24 min.

This Answer is in Option - D

So Option - D is correct

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