Question

2. A train is travelling at a speed
of 90 km h-!. Brakes are applied
so as to produce a uniform
acceleration of - 0.5 m s2. Find
how far the train will go before it
is brought to rest.​

Answer 1

Explanation:

v=u+at

0=25+(-0.5)t

-25=-0.5t

t=50 s

s=ut+½at²

s=25×50+1/2×-0.5×50×50

s=1250+(-625)

s=1250-625

s=625m

Answer 2

here,

Initial velocity(u)=90 km/h

converting u in m/s which is its SI unit

so,

u= (90×5/18)m/s = 25 m/s

final velocity(v) = 0 m/s { because brakes are applied and train is brought to rest}

acceleration(a)= -0.5 m/s^2

according to 3rd equation of motion,

2as= v^2-u^2

so using given values

2×(-0.5)×s = 0^2-25^2

-1s = -25×25{minus is cancelled by minus}

s=625 m = 0.625 km