v after 3 ... 5 + (3*3) = 14
ave. v = (5 + 14) / 2 = 9.5
distance = 3 * 9.5 = ?
An object has an initial velocity of 5 m/s and a constant acceleration of 3 m/s. How far does the particle move during the time 0<=t<=3
A. 28.5 m
B. 27.0 m
C. 53.5 m
D. 33.0 m
E. 19.5 m
2 answers
or:
s(t) = (3/2)t^2 + 5t
s(0) = 0
s(3) = (3/2)(9) + 15 = 57/2
displacement = (57/2 - 0) or 28.5 m
s(t) = (3/2)t^2 + 5t
s(0) = 0
s(3) = (3/2)(9) + 15 = 57/2
displacement = (57/2 - 0) or 28.5 m