Question

class 11
Q). A car travelling in a speed of 54km/h is brought to rest in 90second .find the distance travelled by the car before coming to rest?

Answer 1

Given :

Initial speed of car = 54km/h

Final speed of car = zero (rest)

Time interval = 90s

To Find :

Distance covered by car before it is brought to rest$.$

Conversion :

1 km/hr = 5/18 m/s

54 km/hr = 54 × 5/18 = 15 m/s

Concept :

First of all we need to find acceleration of car during the given interval of time after that we can calculate distance by using equation of kinematics.

Calculation :

◈ Acceleration of car :

➝ v = u + at

➝ 0 = 15 + 90a

➝ a = -15/90

➝ a = -1/6 m/s² [Retardation]

◈ Distance covered by car :

➝ v² - u² = 2as

➝ 0² - 15² = 2(-1/6)s

➝ -225 = -1/3 × s

➝ s = 225 × 3

➝ s = 675 m

Answer 2

u = 54 km/h = 15 m/s

t = 90 s

v = 0 m/s

a = v-u/t

» a = (0-15)/90

» a = -15/90

» a = -1/6

x = ut + 1/2at²

x = 15(90) - 1/2(1/6)(90²)

x = 1350 - 1/12(8100)

x = 1350 - 675

x = 675 m