a locomotive of mass 200000kg is travelling at a speed of 54km/hr on a
straight and levelled track,if a braking force of 250N/tonns is
applied,how far will the locomotive travel before being brought to
rest

Hello and thank you for your question.

I get the same answer as the commenter touf-ga.  
Here are the steps that I followed in deriving my answer

mass = 200,000 kg = 200 tonnes
initial velocity = 54,000 / 3600 = 15 m/sec
force 250 N/tonne x 200 tonnes = 50,000 N

Because F = ma,   a = F/m  =  50,000/200,000 = .25 m/sec*sec negative
acceleration to bring it to a stop.

v^2 = 2*a*x  so x = v^2 / 2a     x = (15*15) / .5    x = 450 meters
   
[touf-ga is also correct that strictly speaking the final velocity v =
0, it's the initial velocity is 15, and a = -.25 so strictly speaking
v^2 = 2ax + [v initial]^2
0 = 2 * (-.25)x + (15 * 15)
x = (15*15) / .5
which will give you the same result]

Search terms used:
I knew the Newton formulas so no research was needed.

Thanks again for bringing us your question.

Sincerely,
Google Answers Researcher
Richard-ga

i got the answer on my own but i just wanted to be sure,because @my
college we were arguing on who's ans. was cool.so i'm rating you guys
high.
Thanks

I'm not sure what this N/tonns unit is - never have heard of it, but I
will show you the method of solving, and you can put in whatever
number you want for the braking force.

If it's what I am assuming, which is 250 N of braking force per ton of
train, then we would get a braking force of 250 N/ton * 200 tons =
50,000 N

Before we go further, I am going to convert the speed into
easier-to-handle units of meters per second.

54 km/hr * 1000m/km * 1 hr/3600sec = 15 m/sec.

The principal equations come from Newton's laws.

First, F=m*a, or FORCE = MASS * ACCELERATION

In this case, our force is the braking force.
mass is the mass of the train
acceleration is what we want to figure out.

50,000 N = 200,000 kg * acceleration

acceleration = 0.25 m/sec^2

NOTE that since we are braking, our "acceleration" is actually a
DEceleration.  So, we are going to put a negative sign in front of it
to show that the train is decelerating.

Therefore, our acceleration is -0.25 m/sec^2.

Next, we use one of the equations of motion, which goes as follows:

Vf = Vo + at 

where Vf = final velocity.  In this case, our final velocity is when
the train is at rest, or 0.

Vo = initial velocity.  In this case, 15 m/sec

a = acceleration, which we already determined to be -0.25 m/sec^2

t = time of acceleration...that's what we'll solve for.

(Vf - Vo) / a = t
(0 m/s - 15m/s) / -0.25 m/sec^2 = t
t = 60 sec.

Next, we use another equation which goes as follows:

delX = Vo*t + 0.5*a*t^2

where delX is the distance traveled through the acceleration (what you
want to know)
Vo = initial velocity ( = 15 m/s)
t = time (= 60 sec)
a = acceleration (= -0.25m.sec^2)

delX = (15 m/sec)*(60 sec) + (1/2)*(-0.25)*(60^2)
delX = 900m + (-450m)
delX = 450 m

Distance traveled is 450 meters or 0.45 kilometers.

If I was mistaken on my interpretation of your braking force, then you
can fill in the numbers yourself using the same method I used.

Cheers!

-touf-ga