# Calculating Particle's Position at t=13.7s

• mohemoto
In summary: That area is equal to the displacement of the particle at t=13.7 s.In summary, the particle starts at position x=6.00 m with initial velocity v=0 m/s. It then reaches a maximum velocity of 27.9 m/s after a total elapsed time of 20.5 s. To find its position at t=13.7 s, you can use the formula for displacement, which is the area under the velocity vs. time graph. By finding the velocity at t=13.7 s and using similar triangles, you can calculate the displacement of the particle at that time. Adding the initial position of x0=6.00 m will give you the final position.
mohemoto

## 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.

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.

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.

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.

Dear student,

Thank you for your question. To calculate the particle's position at t=13.7s, we can use the formula for position, s = at2/2 + v0t + s0, where a is the average acceleration, t is the time interval, v0 is the initial velocity, and s0 is the initial position.

First, we need to find the average velocity in the time interval of [0, 13.7]. To do this, we can use the formula for average velocity, v(avg) = delta d / t, where delta d is the change in distance and t is the time interval. In this case, delta d = s - s0, where s is the final position and s0 is the initial position. Therefore, v(avg) = (s - s0) / t.

Next, we can use the formula for average acceleration, a(avg) = delta v / t, where delta v is the change in velocity and t is the time interval. In this case, delta v = v - v0, where v is the final velocity and v0 is the initial velocity. Therefore, a(avg) = (v - v0) / t.

Now, we can substitute the values we know into the formulas. We know that s0 = 6.00m, v0 = 0m/s, t = 13.7s, and a(avg) = 1.36s. We also know that v = vmax = 27.9m/s. Therefore, we can calculate v(avg) = (v - v0) / t = (27.9 - 0) / 13.7 = 2.04m/s.

Finally, we can substitute all the values into the formula for position, s = at2/2 + v0t + s0. This gives us s = (1.36)(13.7)2/2 + (2.04)(13.7) + 6.00 = 159.9m. Therefore, the particle's position at t=13.7s is 159.9m.

I hope this helps to clarify the solution. Let me know if you have any further questions. Keep up the good work!

Best regards,

## 1. How do you calculate the position of a particle at a specific time?

To calculate the position of a particle at a specific time, you need to know the initial position, velocity, and acceleration of the particle. Then, you can use the equation x = x0 + v0t + 1/2at^2 to calculate the position.

## 2. How do you represent time in the equation for calculating particle position?

Time is represented by the variable t in the equation x = x0 + v0t + 1/2at^2. It is usually measured in seconds.

## 3. Can the position of a particle change over time?

Yes, the position of a particle can change over time. This is due to the particle's velocity and acceleration, which can cause it to move in a straight line or in a curved path.

## 4. What is the difference between position, velocity, and acceleration?

Position is the location of an object at a specific time. Velocity is the rate of change of position over time, and acceleration is the rate of change of velocity over time. In simpler terms, position tells us where an object is, velocity tells us how fast it's moving, and acceleration tells us how its speed is changing.

## 5. How can you use the equation for calculating particle position to predict its future position?

By plugging in different values for time (t), you can use the equation x = x0 + v0t + 1/2at^2 to calculate the position of a particle at different points in time. This can help you predict its future position if you know its initial position, velocity, and acceleration.

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