Calculating Acceleration of a Braking Train

[Nanu Nana]

Homework Statement

A train that runs 140 km / h requires about 700m to stop. Calculate the average acceleration of the train while braking.

Homework Equations

v: x/t[/B]

The Attempt at a Solution

v= x/t
so t= x/v= 700m/ (38.89m/s)=18 s
Now to find acceleration a= v/t
a= 38,89m/s / 18s =2,160 m/s^2
They're asking the average so 2,160 m/s2 is our initial acceleration
therefore
0-2,160 m/s^2 / (2) = -1,08024
Is it the correct formula to use ? or is it wrong ?

 

[Chestermiller]

The equation you used for the velocity is correct only if the acceleration is zero (i.e., only if the velocity is constant).

 

[Nanu Nana]

Oh I see. I think I've found out which kinematic equation to use .

I tried this one
initial velocity = 140 km/h (38,89 m/s)
Final velocity = 0 m/s because its going to stop
(vx)^2=(u)^2 + 2x a x700 m
(38,89)^2=1400 x a
a= -1512,34/ 1400
a= -1,0802

 

[PeroK]

Oh I see. I think I've found out which kinematic equation to use .

I tried this one
initial velocity = 140 km/h (38,89 m/s)
Final velocity = 0 m/s because its going to stop
(vx)^2=(u)^2 + 2x a x700 m
(38,89)^2=1400 x a
a= -1512,34/ 1400
a= -1,0802

That's correct. Can you think of a way to confirm that answer is right?

 

[Nanu Nana]

My workbook says so