# How to determine a particle's position given the velocity-time graph and unknowns?

by mohemoto
Tags: determine, graph, particle, position, unknowns, velocitytime
 P: 8 1. The problem statement, all variables and given/known data 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 2. Relevant equations s = at2/2 + v0t + s0 a(avg)=delta v/t v(avg)=delta d/t 3. 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.
 HW Helper PF Gold P: 3,444 You don't show the graph that goes with this question. However, be aware that the displacement (not the position) of an object is the area under the v vs.t curve. So if you can figure out the area under the curve from t = 0 to t = 13.7 s, then that is the displacement in that time interval. To find the position, just add x0.
 P: 8 Oh I apologize; the graph is simply a straight diagonal line showing constant velocity, but the problem is, there are no numbers on the axes, so I need to find another way to find average velocity.
HW Helper
PF Gold
P: 3,444

## How to determine a particle's position given the velocity-time graph and unknowns?

You know that v=0 at t=0, that v=27.9 m/s at t=20.5 s and that the two points on the graph are connected with a straight line. This gives you a right triangle of base 20.5 s and height 27.9 m/s. Can you find the velocity at t=13.7 s? Hint: Think similar triangles. Once you have that velocity then you need to find the area of the smaller triangle that has base 13.7 s.

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