Question

A train is traveling at a speed of 90 km/h Brakes are appelied so as to produce a uniform acceleration of -0.5 m/s. find how far the train will go before it is brought to rest?​

Answer 1

Answer:

final velocity --0 m/s

initial velocity - 90 km/ h -- 25 m/s

Accelaration -- -0. 5 m/s

Distance -s

v square = u square + 2as

0 square = 25 whole square+ 2 ✖-0. 5 whole square ✖s

0=625-15

0=625-s

s=625m

Answer 2

Answer:

u(initial velocity)=90km/hr or90/60 ×1000/60 m/s

v(final velocity)=0 m/s

acc = -0.5m/s

Using v=u+at(first equation of motion )

0=90000/3600 + -.5t

or t =90000/(3600×5)

or t =5 second

Now,

using s=ut+1/2at² (third equation of motion)

s= 90000/3600×5+1/2(-.5)×5²

or s= 118.75 m

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