Question

a car travelling at a speed of 30 km per hour is brought to rest in distance of 8 metre by applying brakes if the same car is moving at a speed of 60 km per hour then it can be brought to rest with same breakes in

Answer 1

v² = u² + 2as

u1 = 30km/hr
= 30000/3600m/s
= 100/12 m/s

so
0 = (100/12)² - 2a(8)
=> a = (100/12)²/16


so
when u = 60km/hr

(200/12)² = 2(100/12)²/16×s
4/2×16 = s
32 = s

.
so it can be brought to the rest
after 32meter

Answer 2

FIRST CASE

u=30km/h
= 25/3m/s
v= 0m/s
s= 8m

3rd equation of motion

v²=u+ 2as
0 = (25/3)² + 2×a×8
0 = 625/9 + 16a
-625/9 × 1/16 = a
-625/144 m/s² = a

Now,


SECOND CASE

u= 60km/h
= 50/3 m/s
v=0m/s
a=-625/144m/s²
s=?

Third equation of motion

v²=u²+2as
0 = (50/3)² + 2×-625/144×s
0 = 2500/9 - 625/72s
625/72s = 2500/9
s = 2500/9 × 72/625
= 32


Therefore, car will stop after covering 32m.


- Abhinav Singh