Question

A train is travelling at a speed of 25 m/s.retradation of 0.5m/Square, find how far
train will go before it is brought to rest
​

Answer 1

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.