Question

HELP WITH PHYSICS HOMEWORK PLEASE

Answer 1

how do i graph the formula Df=Di+Vit+1/2at^2

Answer 2

It's just a parabola.

Answer 3

but how do i graph it ? i don't understand how

Answer 4

I'm not sure what you mean by 'graphing' since your scales will be off. But, you should get something like this:

http://fooplot.com/plot/7usx07t9p8

Where your origin's y-coordinate is \(D_i\) and its x-coordinate is \(-\frac{V_i}{a}\).

Answer 5

This is for a lab . and for the question it says create a particle diagram that describes the motion of your truck. When i asked my teacher for help he just said to graph the equation that's why im confused.

Answer 6

Well, if you know the initial displacement, velocity, and acceleration, then you can graph it by using a calculator or by hand.

Answer 7

well , the lab was drawing back a toy truck 20 times. so for final velocity i have 20 diff answers same for accleration

Answer 8

Take an average or something of the like and then plot it using that.

Answer 9

so the average for velocity is 2.26 and acceleration is 1.28. i still dont understand how to graph it ):

Answer 10

You ever done a table plot and then graphing it on a graphing paper?

Answer 11

yea . What would be x & y?

Answer 12

Wait nvm i think i got it

Answer 13

First of all, what was your initial question to the teacher, to make him say for you to graph the equation?

second of all, What was the purpose of your lab? Was it to find the initial velocity and acceleration of the truck in respect to the distance of far you drew back the toy truck?

Answer 14

well the question says using the calculated values, generate a formula and create a particle diagram that describes the motion of the truck. He gave us the formula But, i didnt understand how to create a particle diagram. Then he said a particle diagram is just graphing the formula
and yes the point was to find the final velocity and acceleration of the truck when we drew it back

Answer 15

I'm guessing the toy truck was drew back by a cranking action like twisting a nob? Or it was sling shot action?

Answer 16

it's one of those toy trucks that you literally just draw it back and it clicks

Answer 17

You guys also measured how long it took to stop right?

Answer 18

yes

Answer 19

Alright. So it seems like you got the average time and the average distance the toy truck put out every time you pulled it back, from there you got your average velocity right?

Using those information we could find the initial acceleration of the toy truck from this equation

\[d = vt +\frac{ 1 }{ 2 }at^2\]

Then you would have the final velocity and acceleration of the toy truck.

For the particle diagram describing the motion of the truck, I'll have to think upon that a little more.

Answer 20

I have the final velocity and accleration . these are my known information Di=0m Df=2m Vi=0m/s T=measures (20 different times) A=(20 different acceleration) Vf=(20 diff V) there are 20 different velocity and acceleration because i drew the truck back 20 times. then to find the Vf i used an equation and to find acceleration i used an equation.

Answer 21

He did say that when i graph it it should be a parabola . But, i'm just confused on how to graph it

Answer 22

The only way it would be a parabola would be if you measuring the velocity with respect to time or acceleration with respect to time. Since the acceleration and velocity would increase after the cranking and you let the toy truck go, but after a certain amount of time it decelerates because of the friction of the ground and the fact that it ran out of energy to keep going further.

If we were to graph the toy truck's horizontal displacement versus time, it would be a straight line. Your earlier equation takes into account the horizontal displacement (Di) and the time (t). The initial velocity and acceleration were figured out from the data.

I don't see how it would be a parabola when we're measuring the horizontal displacement. That means it would go forward certain amount of distance, then stop and go back in reverse.

Answer 23

idk ... he said parabola.. oh forget it . thanks for you help

Answer 24

Meh. Re-ask the teacher for help.

Answer 25

Wait, what? If \(a>0\) it would be a parabola. Clearly, if we make \(V_i=0\), then, we're simply starting from the vertex \(t\ge 0\), which means that the car *doesn't* stop or go in reverse, at all.

Answer 26

You are clearly confusing your self. Earlier you stated that the y-coordinate values are the displacement with respect to the x-coordinates values in which you named to be "-v/a", then you stated a > 0 If you plugged in those values, we would get negative displacement. Even with what you just stated in your last response, by assuming we're starting in quadrant 1 of the graph.

Answer 27

No. I'm not confusing myself, I'm pretty sure as an EE and Physics double I know my general mechanics.

First off, the vertex of any parabola is:
\[
\frac{d}{dx}y=0=\frac{d}{dx}(ax^2+bx+c)=2ax+b
\]Which is:
\[
x=-\frac{b}{2a}
\]If we allow \(V_i\) to equal zero (*that is* \(b\) to equal zero), we have exactly what I have said before, that the vertex is at \(x=0\). Of course, in my case, I considered that changing variables was trivial and that it would be fairly obvious to readers that I had shown our independent variable to be \(t\) instead of \(x\).

There is no point anywhere where I state that there would be a negative displacement (I, nowhere, have stated that \(a>0\). necessarily, and, considering the experiment, it would make no difference what the value of \(a\) is IF we have \(V_i=0\), which is clearly the case).

Check your calculations before criticizing.

Answer 28

No. you are making a contradiction with your statements. You just tried to correct yourself.

CLEARLY, you stated your independent variable to be "-v/a" in your 2nd response to the opening question (scroll up) to find where the origin is. AND you did state "a > 0" in your 5th response.

And of course the origin starts at y = 0 and x = 0 where else would it start if it began at rest, considering there's no initial displacement.

You're putting too much emphasis on the location of the vertex of the parabola, when I was questioning upon the displacement over time with the graph. And yes it would be HALF of a parabola if we took only the t > 0, I agree with you on that part, but his teacher and yourself in your earlier statement plainly said "parabola". If my English is correct, I don't think that's one in the same type of image, would you?

Check what you're implying before making contradicting comments.