Question

u is coefficient of friction between tyres and road the minimum stopping distance for a car of mass M moving with velocity v

Answer 1

Answer:

We know that Friction is μN.

We can write N as mg or weight of an object as the motion is parallel to the ground.

So, now the frictional force can be calculated as:

Frictional Force = μmg

We know that Force = ma

⇒ ma = μmg

'm' gets cancelled on both sides and we get,

⇒ a = μg

But it is retardation case. So we get,

⇒ a = -μg

Now we know that according to Kinematics,

⇒ v² - u² = 2as

Since the object is going to stop at the end, the final velocity 'v' can be written as 0.

⇒ 0² - u² = 2 ( -μg ) s

⇒ -u² = -2μg s

⇒ u² = 2μg s

⇒ s = u² / 2μg

Hence minimum distance for stopping is u² / 2μg.

Hope it helped :)

Answer 2

$\huge\bf\green {Hey \: there}$


▶ Question - What is the minimum stopping distance for a car of mass 'm' moving with velocity 'v' along a level road if the coefficient of static friction between the tyres and the road is 'μ' ?


▶Explanation :-


Initial velocity (u) = v


The car will be stopped by friction


Final velocity (v) = 0


Acceleration = - μg


Using the third equation of motion, we get :-


⇒v^2 - u^2 = 2as


⇒ 0 - (v)^2 = 2as


⇒- v^2 = - 2μgs


⇒ v^2 = 2μgs


⇒ s = v^2 / 2μg


∴ Minimum stopping distance = v^2 / 2μg


$\mathbb {\huge {\fcolorbox{yellow}{green}{Hope It Helps}}}$