Question

A car moves at a speed of 60km/h, it is stopped by applying brakes which
produces a uniform acceleration of -0.6m/s 2 . How much distance will the
vehicle move before coming to stop?

Answer 1

Answer:

0  

hr

Km

​  

=40×1000/(60×60)  

s

m

​  

(=11.1111…).

a=  

“delta”t

“delta”v

​  

=  

t

(v  

f

​  

−v  

0

​  

)

​  

=  

t

(0−v  

0

​  

)

​  

,

t=  

a

−v  

0

​  

 

​  

=  

−0.6

−11.11

​  

=18.518…sec.,

d=v  

0

​  

×  

2

t

​  

=  

2

11.1111×18.518

​  

=102.9m

Explanation:

Answer 2

Distance covered by car before coming to stop = 231.58 m

Solution

Initial speed of car = 60 km/h

= 60 × (5/18)

= 16.67 m/s

Final speed of car = 0 m/s

Acceleration = -0.6 m/s²

We will use 3rd equation of motion :

◕ v² - u² = 2as

✒ 0² - (16.67)² = 2s × (-0.6)

✒ 0 - 277.89 = -1.2s

✒ -1.2s = -277.89

✒ s = -277.89/-1.2

✒ s = 231.58 m

∴ Distance covered by car before coming to stop = 231.58 m