A train moving with constant acceleration?

U be the velocity at the beginning of 10 th second (i.e. End of 9th second) of its motion. Its velocity at the end of 10s will be U + a where a is its acceleration.

Distance moved in the 10 th second is 1*(U + a) / 2 {Average velocity * time and time = 1 s) (U + a) / 2 = 24 ft given Its velocity at the beginning of 12 th second (i.e. End of 11th second) is U + 2a Its velocity at the end of 12 s will be U + 3a where a is its acceleration. Distance moved in the 12 th second is1*{ U + 2a + U + 3a) * 1/2 (2U + 5a) / 2 = 18 U + a = 48 2U + 5a = 36 solving 68 ft/s and a = - 20ft/s.

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