Question

A train is travelling at a speed of 90km h-1 breakes are applied so as to produce a uniform acceleration of -0.5ms-2.find how far the. Train will go before is brought to rest

Answer 1

Explanation:

v=90kmph

=90×5/18=25m/s

v^2-u^2=2as

625-0=2×0.5×s

s=625m

Answer 2

Answer:

initial velocity=90km/h=90000metres/ 3600seconds

=25m/s

acceleration=-0.5m/s²

final velocity=0m/s

2as=v²-u²

2x-0.5m/s²xs=(0m/s)²-(25m/s)²

-1 m/s²xs=0m/s²-625m/s²

-1 m/sxs=-625m/s²

s=(-625m/s²)/-1 m/s²

s=625 metres