Question

A train comes to stop in 90 m at an acceleration of 3 m/s2 . Find the speed at which it was travelling

Answer 1

Answer:

Given that,

Acceleration a=−0.5m/s2

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=0.525

t=50 sec

Again, using equation of motion, 

S=ut+21at2

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=25×50−21×0.5×(50)2

s=625 m

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

Hence, A is correct.