Question

A train moving at 90 km/h takes 5 s to come to rest after the application of brakes. Assuming the retardation to be uniform, how far the train will move after applying the brakes?

Answer 1

Answer:

Correct option is

A

625 m

Given that,

Acceleration a=−0.5m/s

2

Speed v=90km/h=25m/s

Using equation of motion,

v=u+at

Where,

v = final velocity

u = initial velocity

a = acceleration

t = time

Put the value into the equation

Finally train will be rest so, final velocity,v=0

0=25−0.5t

25=0.5t

t=

0.5

25

t=50 sec

Again, using equation of motion,

S=ut+

2

1

at

2

Where, s = distance

v = final velocity

u = initial velocity

a = acceleration

t = time

Put the value into the equation

Where S is distance travelled before stop

s=25×50−

2

1

×0.5×(50)

2

s=625 m

So, the train will go before it is brought to rest is 625 m.

Hence, A is correct.

Explanation:

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Answer 2

Initial velocity ,u=90 km/h
Final velocity. ,v=0 km/h
Time =5second

First we change km/h to m/sec
Initial velocity,u=90 km/h
90.1000/3600 m/s
= 25 m/s
Final velocity,v =0 km/h
=0m/s
Time is in sec=5second
Acceleration =?
We know that
v=u+at
a=v-u/t
=0 m/s - 25 m/s / 5 sec
= -5 m/s2
Distance travelled by train=?
s=ut + 1/2at2
= 25 m/s.5s + 1/2.-5m/s2.5s.5s
= 187.5 metre