A particle starts from x_0 = 14m at t_0 = 0 and moves with the velocity graph shown
What is the object's position at t = 2 s?
Area under velocity time graph is the displacement of the particle.
The Attempt at a Solution
The part that confuses me is where the velocity goes negative. I think that I should assign negative value to the portion where v_x is negative.
So from t=1 to t=2: (1/2)(1*4) = 2m
from t=0 to t=1 it is negative: - (1/2)(1*4) = -2m
-2m+2m = 0m = no displacement?
Is this correct before I try using calculus to solve same problem? Thanks!
sorry I forgot the initial velocity = 14m/s.
So objects position after 2s is 0m + 14m = 14m
10 KB Views: 1,354