Question

a train is moving in the speed of 120km per hour is brought to rest in 15 second by applying breaks find the magnitude of average retardation due to breaks and distance traveled by car after applying the break

Answer 1

Explanation:

Given,

u=108km/h=

60×60

108×1000

=30m/s

a=−1m/s

2

v=0m/s

From 3rd equation of motion,

2as=v

2

−u

2

−2×1×s=0

2

−30×30

s=

−2×1

−30×30

s=450m