Velocity-Time to Position-Time

[STEMucator]

Homework Statement

Using the velocity time graph, create a position time graph and record the total distance traveled.

V-T Graph : http://gyazo.com/672bd621cbd9f7ab9dcba1250268ee7e

Homework Equations

Since we have constant velocities for certain time intervals we need to use $Δd = v Δt$

The Attempt at a Solution

So I made a table of intervals and displacements and got these values ( Note that i converted the minutes to hours already ) :

t = 0h to t = 0.17h implies ##Δd = 12 km $t = 0.17h to t = 0.67h implies$Δd = 40 km $t = 0.67 to t = 0.75 implies$Δd = 6 km $t = 0.75 to t = 0.8 implies$Δd = 3 km $t = 0.8 to t = 1.17 implies$Δd = 20.4 km ##

Putting these together I got this graph using Wolframs' Listplot[] function : http://gyazo.com/6563dc287b98356467a1669ec4023a81

Have I done this correctly? Is that actually what the position time graph will look like?

 

[TSny]

Your position graph is not correct. I think it will help if you list the total distance traveled (from t = 0) at the end of each time interval.

 

[HallsofIvy]

I can see the first attachment (the graph in the problem) but get an error message when I click on the second (the graph of your solution), so I can't tell if your graph is correct. But what you write is certainly NOT correct. For example, you state, without showing how you got it, that in the first 17 minutes the train will have gone 12km.

According the the "v versus t" graph, for the first 10 min, the train ran at 70 km/h. 10 min is 10/60= 1/6 hr so at 70 km/h it will have gone 70/6= 11 and 2/3 km. For the next 30 (10 to 40) minutes the train ran at 80 km/hr so for 7 minutes, from 10 minutes to 17 minutes, the train went (7/60)(80)= 56/6= 9 and 1/3 km for at total of11 2/3+ 9 1/3= 21 km.

 

[STEMucator]

I can see the first attachment (the graph in the problem) but get an error message when I click on the second (the graph of your solution), so I can't tell if your graph is correct. But what you write is certainly NOT correct. For example, you state, without showing how you got it, that in the first 17 minutes the train will have gone 12km.

According the the "v versus t" graph, for the first 10 min, the train ran at 70 km/h. 10 min is 10/60= 1/6 hr so at 70 km/h it will have gone 70/6= 11 and 2/3 km. For the next 30 (10 to 40) minutes the train ran at 80 km/hr so for 7 minutes, from 10 minutes to 17 minutes, the train went (7/60)(80)= 56/6= 9 and 1/3 km for at total of11 2/3+ 9 1/3= 21 km.

This confused me.

I got the exact same answer as your first calculation. For the first 10 mins ( Or roughly 0.17 hours ) at 70 km/h the displacement would be ( roughly ) 12 km ( rounding up from 11.7 km ).

For the next 30 mins ( Or roughly 0.5h ) the train ran at 80 km/h. Then you said this which confused me :

so for 7 minutes, from 10 minutes to 17 minutes, the train went (7/60)(80)= 56/6= 9 and 1/3 km for at total of11 2/3+ 9 1/3= 21 km.

Why 7 minutes? Why from 10 to 17 and not 10 to 40? What does that have to do with the displacement from 10 mins to 40 mins? Without very much thought it's apparent that the displacement during the time interval from 10 mins to 40 mins at 80 km/h is 40 km.

Also Tsny said this :

Your position graph is not correct. I think it will help if you list the total distance traveled (from t = 0) at the end of each time interval.

So I should ADD the distances together to find my positioning. So let's say i did the first calculation and I got 12 km at t = 10, then I did the second one and got 40 km at t = 40. Would that mean at t = 40 the truck is at position 52 km?

EDIT : I followed that idea and I produced this P-T graph which makes a lot more sense than it did before :

http://gyazo.com/1634e96f5de561188639409d7bcf6931

 

[TSny]

So I should ADD the distances together to find my positioning. So let's say i did the first calculation and I got 12 km at t = 10, then I did the second one and got 40 km at t = 40. Would that mean at t = 40 the truck is at position 52 km?

EDIT : I followed that idea and I produced this P-T graph which makes a lot more sense than it did before :

http://gyazo.com/1634e96f5de561188639409d7bcf6931

That looks much better. I think you would want to include a point on the graph representing the initial position at t = 0.

 

[STEMucator]

That looks much better. I think you would want to include a point on the graph representing the initial position at t = 0.

Yes, good idea.

Thank you very much for your help.