Question

brakes are applied to a train travelling at 72km/hour passing over 200 m its velocity reduces to 36km/hour at the same time rate of retardation now how much further will it go before it is bought to rest

Answer 1

May be this answer you will feel too long but it's a good question
Answer:

67 m

Explanation:

To make the calculations esier, convert the speeds of the train from kilometers per hour to meters per second

72 kmhr⋅1000 m1km⋅1hr3600 s=20 m/s

SInce 36 km/hr=72 km/hr2, you get that

36 km/h=10 m/s

So, you know that it takes the train 200 m to reduce its speed from 20 to 10 m/s. This means that you can say

v2=v20−2⋅a⋅d , where

a - the decelleration of the train.

Rearrange to solve for a

a=v20−v22⋅d=(202−102)m2s22⋅200m=0.75 m/s2

Now, I assume that you meant to say how much further will the train go until it's brought to rest. Well, if the train decelerates at constant rate, and provided, of course, that the speed of the train is zero once it comes to a rest, you can say that

v2final=0=v2−2⋅a⋅dx

This means that it travels for an additional

dx=v22⋅g=102m2s22⋅0.75ms2=66.7 m

I'll leave the answer rounded to two sig figs, despite the fact that you only gave one sig fig for the 200-m distance

dx=67 m

You can test the result by using the total distance it travelled before being brought to rest and its initial speed, 20 m/s

02=202−2⋅0.75⋅(200+67)

400=2⋅0.75⋅267

400=400.5→ close enough, if you take into account rounding

For dx=66.7 m, you would get

400=400.05