Question

A car moving with a speed of 50 km/h , can be stopped by brakes after atleast 6m . if the same car is moving at speed of 100km/h , the minimum stopping distance is [AIEEE 2003] ​

Answer 1

Given, u=50km/hr=50×185=18250m/s

v=0m/s=6m

By third equation by motion,

v2=u2+2as

0=(18250×18250)+2(−a)(s)

-a=12192.90

a=(-16.07)

BY third equation of motion,

v2=u2+2as

⟹0=(100×185)2+2(−16.07)(s)

⟹0=771.60−32.14s

⟹s=24m

Answer 2

Given:

• U1= 50km/h
• V= 0km/h
• S1= 0.006km

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Now stopping distance:

$s= \dfrac{ {u}^{2} }{2a} \\ \implies \: 0.006 = \dfrac{ {50}^{2} }{2a} \\ \implies \: a = \dfrac{2500 \times 1000}{2 \times 6}$

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Given:

• u2= 100km/h
• s=?

$s = \dfrac{ {u}^{2} }{2a} \\ \implies \: s = \dfrac{100 \times 100}{2 \times \dfrac{2500 \times 1000}{2 \times 6} } \\ \implies \: s = 10000 \times 0.006\\ \implies \: s = 0.024km$

$\bold{\red{\large{s=0.024km}}}$