- #1

mohemoto

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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, total. If the initial position of the particle is x0=6.00 m, the maximum velocity of the particle is vmax=27.9 m/s, and the total elapsed time is total=20.5 s, what is the particle's position at t=13.7 s?

Given:

t1=0s

t2=20.5s

d1=6.00m

v1=0m/s

v2=27.9m/s

## Homework Equations

s = at2/2 + v0t + s0

a(avg)=delta v/t

v(avg)=delta d/t

## The Attempt at a Solution

I was able to calculate the average acceleration to be 1.36s (a=27.9/20.5), and I know what the formula for average velocity is. However, what I do not know is how to calculate the average velocity in the time interval of [0, 13.7]. I take it that I cannot just divide the maximum velocity by 2, because that is over the entire time interval and it would not make sense. Could anybody offer any help? And just to be sure, once I find average velocity, it's just a matter of substituting everything into the formula for position, correct?

Thank you.