Question

a car travelling with velocity 108km/h is brought to rest .And its acceleration is 3m/s² than calculate The distance covered by it​

Answer 1

Answer:

• Distance covered by Car is 150 metres.

Explanation:

Given,

• initial velocity of Car, u = 108 km/h = 108 × (5/18) = 30 m/s
• final velocity of Car, v = 0   [ since, it is brought to rest ]
• Acceleration of Car, a = -3 m/s²  [ negative acceleration because car is decelerating ]

To find,

• Distance covered by it before coming to rest while decelerating, s = ?

So,

Using Third equation of motion

→ 2 a s = v² - u²

→ 2 ( -3 ) s = ( 0 )² - ( 30 )²

→ - 6 s = - 900

→ s = 900 / 6

→ s = 150 m

therefore,

• Car covered a distance of 150 metres before coming to rest during decelerating.

Know Three equation of motion:

• First equation of motion

     v = u + a t

• Second equation of motion

     s = u t + 1/2 a t²

• Third equation of motion

     2 a s = v² - u²

[ where v is final velocity, u is initial velocity, t is time taken, a is acceleration, s is distance covered by a body ]