A train that is 268.7 m long undergoes constant acceleration the moment the
last car (end of the train) is outside of the station, how far is the front of the
train from the station after 25.0 s if its initial speed before acceleration is
4.48 m/s and its final speed is 27.4 meters per second?
A 667 m
B 678 m
C 399 m
D 663 m

Initial velocity,  u = 4.9 m/sfinal velocity = v m/sinitial height,  h₁ =  245 m    final height, h₂ = 0a = -9.8  m/s²time taken = t secondss = ut + 0.5at²⇒  (h₂-h₁) = ut + 0.5at²⇒  0-245 = 4.9t + 0.5×(-9.8)×t²⇒ -245 = 4.9t - 4.9t²⇒ 4.9t² -4.9t -245 =0solving it, we get   t = 7.59sv  = u +  at = 4.9 -9.8×7.59 = 4.9 - 74.38 = -69.48 m/svelocity is  69.48 m/s downward