Question

0. When a car of mass 1200 kg is moving with a
velocity of 15 ms- on a rough horizontal road,
its engine is switched off. How far does the
car travel before it comes to rest if the
coefficient of kinetic friction between the road
and tyres of the car is 0.5? (g=10ms-2)​

Answer 1

Answer:

22.5 m

Explanation:

The stopping distance is given by

• S = u²/(2μg)

S = (15 m/s)²/(2 × 0.5 × 10 m/s²)

= (225 m²/s²) / (10 m/s²)

= 22.5 m

Answer 2

Answer:

s=22.5 m

Explanation:

As we know that friction force always oppose the motion and want to rest the object.

The friction force Fr= μ m g

There is only one force that is friction

Fr = m a

a= The retardation of car

m= mass

μ m g = m a

a = μ  g

Given that

μ = 0.5

So

a = 0.5 x 10 =5 m/s²

We know that the final speed of car will be zero.

v = 0

v² = u²  - 2 a s

Initial speed of car given u = 15 m/s

0² = 15²  - 2 x 5 x s

s=22.5 m

So after 22.5 m the car will stop.