Help with Veocity vs. Time Graph Needed

[ryandamartini]

Homework Statement

The green line is the graph. I added in the redlines to form the shapes to find the total distance X traveled by the particle. I know I can figure out that the one side of the rectangle is 40m due to the fact that it is going 2m/s for 20 seconds at that part. The thing that is causing me so much trouble is not knowing what to do for the two triangles.

Homework Equations

The Attempt at a Solution

I tried finding the slope of the triangle and came out with 1.5/20... Not even close. ((y2-y1)/(x2-x1)). I am just really lost and need a nudge in the right direction to go. If I could get some pointers, I would be very happy :) The reason why I tried slope for the Aavg for the first 20 seconds was that I had read elsewhere online that that was the way to do it, but in my case, apparently not, hehe.

Thank You in advance for any help you might be able to offer, how big or small, I would be very grateful!
-Ryan

 

[MadPhysics08]

hey...

remember acceleration is equal to the changed of velocity with respect to time

 

[MadPhysics08]

so, your initial velocity could be at 2m/s and the final velocity is...
then... what do you think..
could be reading the graph is similar with reading the coordinates for (1,2)

 

[LowlyPion]

The green line is the graph. I added in the redlines to form the shapes to find the total distance X traveled by the particle. I know I can figure out that the one side of the rectangle is 40m due to the fact that it is going 2m/s for 20 seconds at that part. The thing that is causing me so much trouble is not knowing what to do for the two triangles.
...
I tried finding the slope of the triangle and came out with 1.5/20... Not even close. ((y2-y1)/(x2-x1)). I am just really lost and need a nudge in the right direction to go. If I could get some pointers, I would be very happy :) The reason why I tried slope for the Aavg for the first 20 seconds was that I had read elsewhere online that that was the way to do it, but in my case, apparently not, hehe.

Thank You in advance for any help you might be able to offer, how big or small, I would be very grateful!
-Ryan

First of all welcome to PF.

The first statement that I highlighted is inconsistent with the graph you've given. The "rectangle" that you are talking about is in velocity, time space and the area of the rectangle is distance. One side can be velocity and the other time are your choices. I presume that you mean that the area of it represents the distance of 40m. Except the rectangle you describe is not the one drawn. You only show a 1.5m/s x 20 s which would be only 30 m, not 40 m.

The second statement is apparently correct. The average Acceleration is Change in velocity over time. And that is the slope of that line. And it is (2 - .5)/ (20 - 0) sec = 1.5 / 20. Why do you think it is wrong?

 

[ryandamartini]

First of all welcome to PF.

The first statement that I highlighted is inconsistent with the graph you've given. The "rectangle" that you are talking about is in velocity, time space and the area of the rectangle is distance. One side can be velocity and the other time are your choices. I presume that you mean that the area of it represents the distance of 40m. Except the rectangle you describe is not the one drawn. You only show a 1.5m/s x 20 s which would be only 30 m, not 40 m.

The second statement is apparently correct. The average Acceleration is Change in velocity over time. And that is the slope of that line. And it is (2 - .5)/ (20 - 0) sec = 1.5 / 20. Why do you think it is wrong?

Ok I did get the avg acceleration for the first 20 seconds done by getting the slope which turned out to be right, evil online submission system >.>

As far as the rectangle.. uh how is it 1.5 * 20 seconds when it clearly levels off at 2(m/s) for the interval between 20 and 40 seconds, (40-20)=20seconds(2m/s)= 40 m. I drew the red line there for my purposes just to keep track of things.

RIght now I am working on the last triangle. The first triangle's X length I got is through the equation: X=Xo + volt + (1/2)at^2 where Xnot according to the book is at 0.5, the start of the shape and horizontal line of the triangle, Vnot is the initial velocity, the avg accel was 0.075m/s^2.

All together, X= 0.5+((0.5)(0))+((1/2)(0.075m/s)(20^2)) = 15.5m

Now I tried to find the X length of the far right triangle using the same equation.
initial velocity = 2 m/s and time at initial velocity =40 seconds, Xnot= 2, avg acceleration = -(1/5) m/s^2, time at end of triangle is 50s.

X = 2 + ((2m/s)(40))+((0.5)(-0.2)(50^2))
//note this is the equation for constant acceleration.

so you end up getting -168, over and over again I keep getting large negative numbers.

Im just lost now, exact same procedure, and I am wrong every time. I've tried putting in 0 for Xo following an example in the book where Xo is the space between the horizontal line of the triangle and the X axis and keeps coming up the same. It seems the 40 and 50 seconds are just way too large with the steep slope to balance the equation

 

[LowlyPion]

Ok I did get the avg acceleration for the first 20 seconds done by getting the slope which turned out to be right, evil online submission system >.>

As far as the rectangle.. uh how is it 1.5 * 20 seconds when it clearly levels off at 2(m/s) for the interval between 20 and 40 seconds, (40-20)=20seconds(2m/s)= 40 m. I drew the red line there for my purposes just to keep track of things.

RIght now I am working on the last triangle. The first triangle's X length I got is through the equation: X=Xo + volt + (1/2)at^2 where Xnot according to the book is at 0.5, the start of the shape and horizontal line of the triangle, Vnot is the initial velocity, the avg accel was 0.075m/s^2.

All together, X= 0.5+((0.5)(0))+((1/2)(0.075m/s)(20^2)) = 15.5m

Now I tried to find the X length of the far right triangle using the same equation.
initial velocity = 2 m/s and time at initial velocity =40 seconds, Xnot= 2, avg acceleration = -(1/5) m/s^2, time at end of triangle is 50s.

X = 2 + ((2m/s)(40))+((0.5)(-0.2)(50^2))
//note this is the equation for constant acceleration.

so you end up getting -168, over and over again I keep getting large negative numbers.

Im just lost now, exact same procedure, and I am wrong every time. I've tried putting in 0 for Xo following an example in the book where Xo is the space between the horizontal line of the triangle and the X axis and keeps coming up the same. It seems the 40 and 50 seconds are just way too large with the steep slope to balance the equation

That looks a bit complicated. Why don't you go with your knowledge that the distance is the area under the V,t curve by time t?

Since the area of each block is .25 times 5 = 1.25 m isn't it simply a matter of adding up all the areas?

(8 + 24/2 + 32 +8) * 1.25 m = ?

 

[ryandamartini]

That looks a bit complicated. Why don't you go with your knowledge that the distance is the area under the V,t curve by time t?

Since the area of each block is .25 times 5 = 1.25 m isn't it simply a matter of adding up all the areas?

(8 + 24/2 + 32 +8) * 1.25 m = ?

Can you please e plain how you worked that out ?

Thank You,
Ryan

 

[LowlyPion]

Can you please e plain how you worked that out ?

Thank You,
Ryan

Isn't the area of each block = to distance?

What is the area of each block? That is your distance for each block.

How many blocks do you count.

area of a rectangle = Height x base
area of a triangle = 1/2 Height x base.

Add up all the areas.

It's that simple.

 

[ryandamartini]

Isn't the area of each block = to distance?

What is the area of each block? That is your distance for each block.

How many blocks do you count.

area of a rectangle = Height x base
area of a triangle = 1/2 Height x base.

Add up all the areas.

It's that simple.

Thank you for your insight. My physics professor is I assume trying to cause mental injury to his students then. Just one question. how is an area of a block 0.25*5 ? I had figured out the previous question while I was looking at the pic and you psoted before I could hehe.

Please forgive the lack of sleep and Thank You VERY much for this trick.. It will help immensely.

 

[LowlyPion]

Thank you for your insight. My physics professor is I assume trying to cause mental injury to his students then. Just one question. how is an area of a block 0.25*5 ? I had figured out the previous question while I was looking at the pic and you psoted before I could hehe.

Please forgive the lack of sleep and Thank You VERY much for this trick.. It will help immensely.

Each block is .25 m/s high on your graph.
Each block is 5 seconds wide.
Area = time x velocity = distance = .25 x 5 = 1.25 m

Good luck.

 

[ryandamartini]

Each block is .25 m/s high on your graph.
Each block is 5 seconds wide.
Area = time x velocity = distance = .25 x 5 = 1.25 m

Good luck.

Ah, Thank You so much for this. I was able to nail down all the rest of my problems like this.

PROPS!


-Ryan