Question

A car travels with uniform velocity of 25 m s^-1 for 5 s. The brakes are then applied and the car is uniformly retarded and comes to rest in further 10 s. Find:

1. The distance which the car travels before the brakes are applied,
2. Retardation and
3. The distance travelled by car after applying the brakes.

Answer 1

(i) As we know, Distance = Speed × time

Initial velocity (u) = 25 m s^-1; Final velocity (v) = 0; time = 5 s

Substituting the values in the formula,

Distance = Speed × time

Distance = (25) × (5) m

∴ Distance = 125 m

(ii) Acceleration = `("Final velocity" - "Initial velocity")/"Time taken"`

∴ a = `("v" - "u")/"t"`

 = `(0 - 25)/10` ms^-2 

 = `(-5)/2` ms^-2

 = - 2.5 ms^-2 

∵ If v < u, then a is negative, and a is the retardation.

Therefore, retardation = 2.5 ms^-2

(iii) After applying brakes, the time taken to come to stop = 10 s

Let S' be the distance travelled after applying the brakes.

Initial velocity u = 25 m/s

Final velocity v = 0

Using the third equation of motion,

∴ v² - u² = 2as

We get,

⇒ (0) ² - (25)² = 2 (- 2.5) (S')

⇒ 625 = 5(S')

⇒ `625/5` = S'

⇒ S' = 125 m