Question

A car is travelling with a speed of 30km/h . The driver applies the break and retards the car uniformly . The car is stopped in 5sec. find the retardation of the car and distance travelled before it is stopped after applying the break.​

Answer 1

Given :

• initial velocity of Car , u = 30 km/h

        u = 30 × 5/18 = 25/3  m/s

• final velocity of Car ,v = 0
• time taken by Car to stop , t = 5 s

To find :

• Retardation of Car
• Distance travelled by Car before coming to rest and after applying breaks.

Formulae to be used :

• First equation of motion

       v = u + a t

• second equation of motion

     s = u t + 1/2 a t ²

( where v is final velocity , u is initial velocity , a is acceleration , t is time taken, and s is distance travelled )

Solution :-

Let, acceleration of Car be 'a'

then,

Using first equation of motion

→  v = u + a t

→  0 = (25/3) + a (5)

→  -25/3 = 5 a

→  a = (-25/3) / 5

→ a = -5/3  m/s²

Therefore,

acceleration of Car is -5/3 m/s² and

▶Retardation of Car will be 5/3 m/s².

Now,

Let, distance travelled by Car before stopping and after applying breaks be 's'

Using second equation of motion

→ s = u t + 1/2 a t²

→ s = (25/3) (5) + 1/2 (-5/3) (5)²

→ s = (125/3) + (1/2) (-125/3)

→ s = 20.834 m

Therefore,

▶Distance travelled by Car is 20.834 m .