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How to draw a position-time graph from a velocity-time graph?

by -Dragoon-
Tags: draw, graph, positiontime, velocitytime
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-Dragoon-
#1
Dec28-10, 03:13 AM
P: 292
And vice-versa? I'm having quite a hard time at doing this. Any tips? My book isn't exactly "layman" friendly. Thanks so much for all the help, I appreciate it.
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soothsayer
#2
Dec28-10, 04:15 AM
P: 398
The trick is in derivatives!

velocity is the derivative of position, dx/dt=v
So all you need to do is draw the derivative graph of position-time and you will have the velocity-time graph!

Here's a site with pretty trivial examples that might help, it gets trickier as the position-time graph gets more complicated, you have to be able to look at the functions for the graphs, take the derivative and then graph, or just estimate it.
http://www.physics247.com/physics-ho...ion-graphs.php

If you're going backward from velocity-time to position-time, you have to take the anti-derivative of the graphs, it's bascially just working backward; finding the position graph whose derivative would make the velocity-time graph you are looking at. It's a little more difficult to do that way.

Mathematically (if you're given the functions), an example would be
x(t) = 3t+5
v(t) = dx(t)/dt = 3
the value for the velocity function is simply the slope of the position function, 3, in this case, it's a linear function. For non-linear functions, it changes
x(t) = 2t2
v(t) = dx(t)/dt = 4t
Now the derivative is a function.

Hopefully that's all the info you need and more. Sorry if the derivative explanation was unneeded, didn't want to make any assumptions about your math level.
-Dragoon-
#3
Dec28-10, 01:29 PM
P: 292
Thanks for the reply, soothsayer!

The problem is, I am only doing grade 11 physics at the moment and haven't even touched calculus (which is next semester) so I have no idea what derivatives are.

But, my book basically tells me the first thing I need to do is find the displacement for every part of the graph.
Example: say I have a velocity-time graph, and the velocity is 50km/h [forward] in 0.10 h. And now, I would need to find the displacement for every 0.01h so I could start constructing the position time-graph. How would I be able to do that?

Thanks in advance.

soothsayer
#4
Dec28-10, 04:55 PM
P: 398
How to draw a position-time graph from a velocity-time graph?

Ok, as long as the graphs are simple, you can do them without calculus.

So we have a velocity-time graph where velocity is 50km/hr as you say, this velocity is sustained for 0.1 hr. So if the object in question travels at 50km/hr for 0.1 hr, the total displacement = 50 x 0.1 = 5 km, now, this only tells us the endpoints of the position-time graph, x=0 at t=0 and x=5km at t=0.1 hr, to know the rest, we have to look at the velocity-time graph bit by bit. So, pick arbitrary time values for t (a good one would be, as you mentioned, every 0.01 hrs) and figure out the displacement up to that point by multiplying those time values by the velocity.

t = 0 hr, x = 50 x 0 = 0 km
t = 0.01 hr, x = 50 x 0.01 = 0.5 km
t = 0.02 hr, x = 50 x 0.02 = 1 km
....
t = 0.09 hr, x = 50 x 0.09 = 4.5 km
t = 0.1 hr, x = 50 x 0.1 = 5 km

You'll see as you start to plot these points for position and time on a graph, that it creates a very simple linear function, so you can simply draw a straight line from the starting point (0,0) to the end point (0.1, 5) and that will be a perfect representation of the position-time function! This is the case every time velocity is constant, such as in this case, it never changes from 50km/hr

So the method for constructing these position-time graphs from velocity is not hard, simply multiply different time values by the velocity to come up with a displacement. The displacement would be the value of the y coordinate and the time value you have chosen is the corresponding x coordinate value.
-Dragoon-
#5
Dec28-10, 07:46 PM
P: 292
That clarified many things, thanks soothsayer.

Okay, now I have two graphs I'm going to have convert to position-time graphs, just want to see if I did it right:
1. The first graph: http://img189.imageshack.us/i/kinema...citytimeg.jpg/
The first line is 70km/h, but the time-graph is in minutes so I'll have to divide that by 60: 70/60 = 1.16 km/m. The change in time is 5 minutes for constant velocity, so the truck travels about 5.9 km south in 5 minutes at a constant velocity before it accelerates. Then the truck begins to travel at 90km/h for 30 minutes. Dividing 90km by 60 to get 1.5 km/m, and multiplying it by 30 minutes to find that the truck traveled a distance of 45km south. Total displacement so far is 50.9km south. Now the truck begins to slow down and travels at a constant speed of 80km/h which needs to be converted to km/m, which becomes 1.333 km/m. Change in time is 30 minutes again, so multiplying that by 1.333 gives us a distance of 40km traveled by the truck. The total displacement for this graph is 90.9km.

So I made the position-time graph on paint after deducing the above:
http://img214.imageshack.us/i/kinematics555graph.jpg/
I made the slope of the first line (which is 70km/h) the least steep, while the second line which is 90km/h the most steep, and finally the third less steep than the second but more steep than the first. Does this look right?

2. Now for the second graph: http://img9.imageshack.us/i/kinemati...citytimeg.jpg/
The truck travels at 70km/h for 10 minutes, since the time graph is in minutes I need to convert 70km/h into km/m and once again have a value of 1.16km/m. Therefore, the truck will travel 11.6 km in 10 minutes. Then the truck begins to pick up speed and travels at 80km/h for 30 minutes. The truck will travel a distance of 40km during this time. Total displacement so far 51.6 km. The truck then loses speed and travels at 75km/h for 5 minuets, which converts to 1.25km/m multiplied by 5 minutes, which shows that the truck traveled 6.25 km over these 5 minutes. The truck then loses more speed and travels at 60km/h, which is 1km/m in 2.5 minutes. The truck travels 2.5 km during this time. Total displacement so far is 60.35 km. The truck then deaccelerates to 50km/h for the rest of the distance which is 22.5 minutes. 0.83km/m(22.5 minutes) yields a distance of 18.75 km. The total displacement is 79.1 km.

So, like I did with the first graph, I made a second one based on the deductions above on paint: http://img31.imageshack.us/i/kinematics666.jpg/
I made the slopes more or less steep depending on the speed the truck traveled during those times.

If I did anything wrong, what can I fix? Thanks for all the help thus far.
soothsayer
#6
Dec28-10, 08:30 PM
P: 398
Those both look perfect! Yeah, that's exactly the way to do it.
-Dragoon-
#7
Dec28-10, 10:23 PM
P: 292
Quote Quote by soothsayer View Post
Those both look perfect! Yeah, that's exactly the way to do it.
Wow, thanks again soothsayer for all the help you've been.

Just one more question, if I have another question that is bugging me can I post it in this thread or do I start a new one?
soothsayer
#8
Dec28-10, 10:46 PM
P: 398
I would start a new thread, unless the second question is related to this one, then you can post it here.


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