Particle Motion; Graphical Method

[Alexanddros81]

Homework Statement

A particle moves in a straight line with the acceleration shown in the figure.
Knowing that it starts from the origin with $v_0 = -14ft/s$, plot the v-t and
x-t curves for 0 < t < 15 s and determine (a) the maximum value of the velocity
of the particle, (b) the maximum value of its position coordinate.

Homework Equations

The Attempt at a Solution

I have a question. To find the change in x from 2s to 5s do I split the time from 2s to 3s and from 3s to 5s in order to find the individual areas of the two triangles (see v-t graph) and add them up?

 

[tnich]

Yes, that would work. However, you are not calculating the areas under the various line segments correctly. When v is negative along a line segment, should the area under the curve be positive or negative? What is the area of a trapezoid?

 

[haruspex]

To find the change in x from 2s to 5s do I split the time from 2s to 3s and from 3s to 5s in order to find the individual areas of the two triangles (see v-t graph) and add them up?

You are not calculating any of the displacements correctly. You only seem to be taking the areas of the triangles and ignoring the rest of the area between the slope and the t axis.

 

[Alexanddros81]

Yes, that would work. However, you are not calculating the areas under the various line segments correctly. When v is negative along a line segment, should the area under the curve be positive or negative? What is the area of a trapezoid?

I have attached a corrected v-t diagram

when v is negative along a line segment, the area under rhe curve should be negative.
so in this case:

0 < t < 2s : $x_2 - x_0 = 1/2(2)(-6) = -6ft$
2s < t < 3s: $x_3 - x_2 = 1/2(1)(-8) = -4ft$
3s < t < 5s: $x_5 - x_3 = 1/2(2)(16) = 16ft$

By adding up the areas from 2s to 5s we get : 16ft + (-4ft) = 12ft

5s < t < 8s: $x_8 - x_5 = ((25+16)/2)(3) = 61.5ft$ (Area of trapezoid)

8s < t < 13s: $x_13 - x_8 = 1/2(5)(25) = 62.5ft$
15s < t < 13s: $x_15 - x_13 = 1/2(2)(-10) = -10ft$

By adding up the areas from 8s to 15s we get: 62.5ft + (-10ft) = 52.5ft

So to answer (a) the maximum value of the velocity of the particle is 25ft/s

Is the above correct?

 

[tnich]

I have attached a corrected v-t diagram
View attachment 223001

when v is negative along a line segment, the area under rhe curve should be negative.
so in this case:

0 < t < 2s : $x_2 - x_0 = 1/2(2)(-6) = -6ft$
2s < t < 3s: $x_3 - x_2 = 1/2(1)(-8) = -4ft$
3s < t < 5s: $x_5 - x_3 = 1/2(2)(16) = 16ft$

By adding up the areas from 2s to 5s we get : 16ft + (-4ft) = 12ft

5s < t < 8s: $x_8 - x_5 = ((25+16)/2)(3) = 61.5ft$ (Area of trapezoid)

8s < t < 13s: $x_13 - x_8 = 1/2(5)(25) = 62.5ft$
15s < t < 13s: $x_15 - x_13 = 1/2(2)(-10) = -10ft$

This looks mostly correct to this point. You are still calculating the trapezoid area incorrectly for 0 < t < 2s, but since the rest looks correct, I think you just forget to write down your new result for that part.

I don't follow your explanation below of why the maximum velocity is 25ft/s.

By adding up the areas from 8s to 15s we get: 62.5ft + (-10ft) = 52.5ft

So to answer (a) the maximum value of the velocity of the particle is 25ft/s.

Is the above correct?

It doesn't really require much explanation. Look at your velocity plot and find the maximum velocity.

 

[Alexanddros81]

No, I was calculating the area from $x_0$ to $x_2$ the wrong way.
So 0 < t < 2s : $x_2 - x_0 = (((-8) + (-14))/2) (2) = -22ft$

By looking at the plot the maximum velocity is 25ft/s.

 

[tnich]

No, I was calculating the area from $x_0$ to $x_2$ the wrong way.
So 0 < t < 2s : $x_2 - x_0 = (((-8) + (-14))/2) (2) = -22ft$

By looking at the plot the maximum velocity is 25ft/s.

Yes. Now how are you going to the find the maximum height?

 

[Alexanddros81]

Is my x-t graph correct? Also I haven't sketched from 0 to 2s cause I don't know how it should be.
Is the maximum value of its position coordinate 62.5ft?

 

[tnich]

View attachment 223043

Is my x-t graph correct? Also I haven't sketched from 0 to 2s cause I don't know how it should be.
Is the maximum value of its position coordinate 62.5ft?

When acceleration is constant, the position coordinate x looks like $x(t) = \frac 1 2 a t^2 + v_0 t + x_0$, so your plot should look like pieces of parabolas glued together (a different parabola for each constant acceleration). To find the maximum height, answer this question: what is the velocity when the height is at a maximum?

 

[haruspex]

View attachment 223043

Is my x-t graph correct? Also I haven't sketched from 0 to 2s cause I don't know how it should be.
Is the maximum value of its position coordinate 62.5ft?

Before trying to sketch the graph of position, calculate where it will be at the end of each period.
Since you know the velocity at start and end of each period, the easiest way is to use the average velocity for the period.
Post the numbers you get, preferably as a table showing the velocities too.

For maximum height, answer tnich's question.