An object begins moving at x= 0 from rest in the positive x direction with constant acceleration. It travels D meters in T seconds. Its final speed is V meters per second.

Question

What is the position x of the object at time t= 12 T? Express your answer in terms of D.

(The answer is D/4 I just don't know how to get there)

Answer 1

I am assuming that when you say t = 12 T you actually mean t = (1/2)T

Initial conditions: Final conditions

t0 = 0 t = T

v0 = 0 v = V

x0 = 0 x = D

a = constant

Since a is a constant,

a = Δv / Δt = V/T;

D = v(ave) x T where v(ave) = (V - 0)/2 = V/2 ==> D = (1/2) VT

General motion equation:

x = x0 + v0t + (1/2)a t^2 = (1/2) a t^2

substituting in the value a and t = (1/2) T get

x = (1/2) (V/T)[(1/2)T]^2 = (1/2) (V T)/4 = [(1/2)VT]/4

from above, D = (1/2)VT

so x = D/4