Let's try the problem two ways using different kinematic equations:

1)
x = x_o + v_ot + ¹/₂at²
x = 0 + 15*40 + ¹/₂(-0.5)*(40²) = 200 m

2)
v² = v_o² + 2a(x - x_o)
Set v_f = 0 ( the train comes to rest) and solve for x:
x = [v² - v_o² ]/2a + x_o = [0² - 15²]/2(-.5) + 0 = 225 m

Two different answers using two approaches; that can't be good.  Which is correct?  It turns out that the train actualy comes to rest after 30 seconds:
v = v_o + at
Set v_f = 0 and solve for t:
t = [v - v_o]/a = [0 - 15]/(-0.5) = 30 s.
Equation 1 assumes that the acceleration of the train was the same throughout the full 40 seconds, i.e., that the train stopped and then began to move backward, in this case 25 m in 10 seconds.

Was this a trick question?  No.  It serves to remind us not to use equations blindly.  Not all of the information in the problem is reflected in the equations.  Remember that one assumption we used when we derived the kinematic equations was that the acceleration is constant.  That is not the case in this problem.

Should we have known in advance which relationship would give us the correct answer?  No.  Indeed, if the numbers had been slightly different, then the first equation might have been the appropriate one, with the train not yet at rest at the end of 40 seconds.