Finding distance from velocity-time graph with piecewise linear motion

[liamporter1702]

Hi guys, having some trouble with this question as I cannot see where I am going wrong. Very simple question but I must be misunderstanding something.

Homework Statement

The following table shows how the velocity of a car varied with time:

Velocity (km/h) 0 19 35 48 48 27 0
Time (s) 0 3 6 9 12 15 18

Assuming these speed values to be joined by straight lines, plot a graph of speed against time and determine the total distance travelled.

Homework Equations

s = dxt
Area under graph = distance travelled

The Attempt at a Solution

I assumed this question would be as simple as finding the area under the graph to get the total distance travelled. So I went about drawing my graph along with converting all the values for speeds into (m/s). I have attached a picture of what my graph looks like in case anyone would like to see it (its a bit messy). I'm sure there is a very simple explanation to where I am going wrong but I'm just not seeing it myself.

Thanks in advanced guys!

 

[BvU]

So what did you get and why do you think it is wrong ?

 

[liamporter1702]

Sorry forgot to put that info in my original post. I keep getting 139.97m by using the area under the graph rule. I do (13.33x9)/2 + (13.33x3) + (13.33x6)/2 = 139.97. Yet my questions sheet gives me an answer of 147m.

 

[BvU]

Problem I have with your plot is not that it's messy (it really isn't). I have the same numbers but my plot looks quite diferent!

Problem I have is that you use different step sizes on the time axis! the 3 sec from 9-12 looks a lot longer than from 12-15.

This also suggests a linear constant acceleration from 0-9, which is not correct (otherwise the speeds would look x, 2x, 3x for t = 0,6,9 and they don't).

What do you do to determine the area under the graph ? 9 x 13.3 /2 + 3 x 13.3 + 6 x 13.3 /2 ?

[edit] well this guess of mine crossed your addendum.

Draw it with equidistant time intervals and you'll be OK.

 

[liamporter1702]

My plot was just a quick sketch as I was unsure how the graph would look, would you recommend drawing it out properly on graph paper?

I used that to get that area under graph based on splitting the area into two triangles and a rectangle. I then used 1/2 x base x height to get the area of the two triangles and added these onto the area of the rectangle.

 

[BvU]

Either that, or you go to $s=s_0 + v_0 \ t + {1\over 2} a \ t^2$, where $a_{\ t-3\ \rightarrow\ t} = {v(t) - v(t-3) \over t - (t-3)}$ in steps of 3 sec

But: did you pick up that 0-9 sec cannot be lumped together because the accelerations in these three sections are not the same ?
Idem 12-18 sec.

 

[liamporter1702]

I think that is where I have gone wrong yes. I have lumped together 0-9 seconds, thank you for pointing that out! I am currently working through it again but this time I have drawn the graph out properly and I am splitting the area beneath the graph into triangles, trapeziums and a rectangle, then adding all these areas together, rather than lumping parts together. Fingers crossed it works this time!

 

[BvU]

Let us know if you don't get 147.5 m !

By the way, do you realize you are doing exactly the same thing as this s = s0 + v0 t + 1/2 a t^2 , when you do it by adding triangles and rectangles ?

 

[liamporter1702]

I did get it this time! But if that equation is a faster way to get the answer then I'll try that also. What is that equation? Is it a SUVAT equation because I don't think I've seen that one. what does s0 and v0 represent?

 

[BvU]

s0 and v0 are distance and speed at start.
So first interval: 0 and 0. acceleration a is (5.28 - 0 )/3 = 1.76 m/s². s = 0 + 0 x 3 + 1/2 x 1.76 x 3² = 7.92 m .
Then second interval: 7.92 m and 5.28 m/s , a = (9.72-5.28)/3 = 1.48 m/s².
s = 7.92 m + 5.28 * 3 + 1/2 * 1.48 * 3² = 30.42 m
the 5.28 * 3 is a rectangle area and the 1/2 * 1.48 * 3² is a triangle area.
etc etc