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## Homework Statement

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

Given:

t1=0s

t2=43.2s

v1=0m/s

v2=47.9m/s

d1=8.29m

find the final distance when t=28.2s

## Homework Equations

a(avg)=delta v/t

v(avg)=delta d/t

## 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.