The velocity graph of a particle moving along the x-axis is shown. The particle has zero velocity at t=0.00 s and reaches a maximum velocity, vmax, after a total elapsed time, ttotal. If the initial position of the particle is d0=8.29 m, the maximum velocity of the particle is vmax=47.9 m/s, and the total elapsed time is total=43.2 s, what is the particle's position at t=28.8 s? The graph that is provided with the question shows a constant slope on the
find the final distance when t=28.2s
The Attempt at a Solution
So since the slope of the vt graph is constant, I found v(avg) by dividing the final+initial by 2.
V(avg) = d2 - d1 / t
23.95m/s = d2 - 8.29m / 28.8s
d2 = 698m
I still got the answer wrong... I don't really care for the correct answer, I'm just worried that I'm really misunderstanding some basic concepts.