SATHEE: Laws Of Motion Question 99

Laws Of Motion Question 99

Question: A car is moving along a straight horizontal road with a speed $v_{0}$ . If the coefficient of friction between the tyres and the road is $μ$ , the shortest distance in which the car can be stopped is[MP PET 1985; BHU 2002]

Options:

A) $\frac{v_{0}^{2}}{2 μ g}$

B) $\frac{v_{0}}{μ g}$

C) ${(\frac{v_{0}}{μ g})}^{2}$

D) $\frac{v_{0}}{μ}$

Show Answer

Answer:

Correct Answer: A

Solution:

Retarding force $F = m a = μ R = μ m g$

$∴$ $a = μ g$ Now from equation of motion $v^{2} = u^{2} - 2 a s$

$\Rightarrow 0 = u^{2} - 2 a s$

therefore $s = \frac{u^{2}}{2 a} = \frac{u^{2}}{2 μ g}$

$∴$ $= \frac{v_{0}^{2}}{2 μ g}$

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Laws Of Motion Question 99

Question: A car is moving along a straight horizontal road with a speed v0 . If the coefficient of friction between the tyres and the road is μ , the shortest distance in which the car can be stopped is[MP PET 1985; BHU 2002]

Options:

Answer:

Solution:

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Question: A car is moving along a straight horizontal road with a speed $v_{0}$ . If the coefficient of friction between the tyres and the road is $μ$ , the shortest distance in which the car can be stopped is[MP PET 1985; BHU 2002]