Finding velocity and position from an acceleration vs time graph

[Kot]

Homework Statement

The graph of acceleration versus time for an object in linear motion. Assuming the object starts from rest at the origin at t = 0, find the velocity and position of the object at times t = 1 through 10 seconds.

Homework Equations

$v=v_i + at$

The Attempt at a Solution

I searched online and found that the velocity is the area under the acceleration vs time graph, which is the integral of the acceleration vs time graph. Since the acceleration function wasn't given to me, I found the equation of the first part of the graph from (0 seconds to 2 seconds) by using basic algebra. I found that the line was $y=\dfrac{5}{2}x$. Would it be correct to take the integral of $y=\dfrac{5}{2}x$ since it would give me the area under the curve? I took the integral with the limit from 0 to 2 and got 5m/s. The distance traveled from 0 seconds to 2 seconds is 10m (I found this by using s=d/t). Is this the correct approach to solving this?

 

[Delphi51]

You ARE given the acceleration vs time graph!
You could use your approach of writing the equation for each segment of the given graph and integrate it. If you do this, make your initial equation a = (a function of time).

The usual approach is to find the area under the acceleration graph from time zero to time 1, then zero to time 2, etc. and mark these velocity points on another graph for v vs t. This is integration by hand and it may well be faster for this problem.

Once you have the v vs t graph, you know what to do to get x vs t.

 

[Kot]

I have all the velocity and position for the time intervals but now I am having trouble plotting it on a v vs t graph. From 0 to 2, the velocity is 5m/s, from 2 to 5 the velocity is 10m/s, from 5 to 6 the velocity is 4m/s, from 6 to 9 the velocity is 9m/s, and from 9 to 10 the velocity is 1.5m/s. Is the velocity graph supposed to have horizontal lines like the acceleration vs time graph or jagged edges connecting the different velocities?

 

[Delphi51]

From 0 to 2, the velocity is 5m/s

Oh, I think it is more complicated than that! If the velocity was constant for the first two seconds, its derivative (the acceleration) would be zero. But it isn't, the given graph shows the acceleration is nonzero and rapidly increasing. If you want help sorting it out, show some details of how you find the velocity at time 1 second.

 

[Kot]

I am not sure what you mean at 1 second (do you mean from 0s to 1s?). I found the velocity from 0s to 2s by taking the integral of $y=\dfrac{5}{2}x$ with 0 as the lower bound and 2 as the upper bound. I got 5 which is the area under the curve so I thought that the velocity was 5m/s.

 

[Delphi51]

I agree with 5 m/s at time 2 seconds. I think it might help avoid a misunderstanding if I suggest your y = 5/2x should actually be a = 5/2t, then after integrating v = 1.25t².

At time 1 second, you must find the area under the graph from 0 to 1 s. The brute force way is to note that it is a right triangle with base 1 s and height 2.5 m/s². A = 0.5*b*h = 0.5*1*2.5 = 1.25 m/s.
Or your way, v = 1.25t² = 1.25(1)² = 1.25 m/s.
This gives you two points on the velocity vs time graph. You could carry on this way but I think you will prefer to integrate again to get a function for x = (a function of time).

 

[Kot]

If I integrated the velocity function again I would get the displacement? I am still unsure how I would plot and connect the points on the velocity vs time graph.

 

[Delphi51]

Yes, you can just integrate again to get displacement (x). Remember the a = 5/2t, v = 1.25t² are only good for the first two seconds. The acceleration function changes completely at time 2 seconds.

 

[Kot]

Yes I am working on the rest at the moment. After I get all velocity and positions, how would I graph the data in a velocity vs time graph? I can plot the points but how will I connect them?

*edit I am working on the second part of the graph where acceleration is constant. I integrated the function a=5, and got v=5t. I plugged 3s into t and found that the velocity is 15m/s at 3 seconds. I know that the velocity is 5m/s at 2 seconds, where did the extra 10m/s come from? Did I miscalculate something?

 

[Delphi51]

Ah, you must use your knowledge of functions. The 0 to 2 s interval is x = 0.416*t³ so you sketch a curve that resembles a cubic function.

Lots of work, isn't it? For the second interval you will have to add a constant after integrating to make the displacement at time 2 the same as you got for time 2 from the x = 0.416*t³. I'm trying to work it by both methods and so far the answers aren't agreeing, so I've got a mistake!

 

[Kot]

Would that constant be 1? Since I evaluated the function at t=3, subtracting a 1 would give me t=2.

 

[Delphi51]

I didn't get that, but maybe its my mistake. What is your function for velocity in the 2 to 5 s interval?

 

[Kot]

My function for velocity from 2 to 5 is v=5t. I integrated the function a=5.

 

[Delphi51]

Okay, so the whole thing is v = 5t + C
and v at time 2 is known to be 5
Find C.

 

[Kot]

I got -5 for C and the answer makes sense now. I am confused as to why you needed to do this, will I need to find C for the other intervals? Could you please explain where this displacement came from?

 

[Delphi51]

Right, and the v = 5t -5 will become x = 2.5t² -5t + C when you integrate again to get x. My mistake was forgetting that -5t!

And so on; yes another C at each integration in each interval.

 

[Kot]

I feel like I am beating a dead horse but I still do not understand why it is necessary to have the constant, what is causing the displacement?

 

[Delphi51]

Well, you always get a +C when you integrate. It is mathematically because f(t) + C has the same derivative as f(t). Physically, if you just used v = 5t then it wouldn't fit the problem: according to our formulas for the previous interval up to time 2, v = 5 at time 2. You need C = -5 to make v = 5t + C = 5 at time 2. Of the many functions whose derivative is a = 5, the v = 5t - 5 is the only one that fits the distance at the left boundary of the interval.

I have x = 3.33 at time 2, x = 40.83 at time 5 and x = 63 at time 6. These checked with the other method of counting squares under the graph to get the area.
I'm glad I don't have to finish the whole thing!

 

[Kot]

Thanks for the alternative explanation, I will try to do the rest of the graph now.

 

[Kot]

I finished calculating the velocities for t=1 to t=10. Everything seems to make sense until I reach the last interval t=9 to t=10. Here are my velocities at each t:
t=1s v=1.25m/s,
t=2s v=5m/s,
t=3s v=10m/s
t=4s v=15m/s
t=5s v=20m/s
t=6s v=24m/s
t=7s v=27m/s
t=8s v=30m/s
t=9s v=33m/s
t=10s v=34.5m/s

Looking at the graph, I see that the acceleration is rapidly slowing down but my calculations say that the velocities are still increasing. I integrated the function a=-3t+30 (the equation representing the line at that interval). After integrating the acceleration I got $v=-\dfrac{3}{2}t^2 +30t + C$ I then used what you taught me to find the C and got $v=-\dfrac{3}{2}t^2 +30t - \dfrac{231}{2}$. I then plugged in 9s and got 33m/s which checked with my previous interval but 10s gives me 34.5m/s. Is this correct or did I go wrong somewhere in my calculation?

 

[Delphi51]

Good morning! I agree with all of those velocities!

Looking at the graph, I see that the acceleration is rapidly slowing down but my calculations say that the velocities are still increasing.

This makes perfect sense. As long as the acceleration is positive, the speed is increasing. When a =0, the speed doesn't change. Using the "velocity is area under acceleration" approach, the area is positive (increasing velocity) as long as the area is above the a=0 axis.

Yes, 34.5 makes sense. The area under the acceleration graph from time 9 to 10 is a triangle with area A =.5*b*h = .5*1*3 = 1.5, so the velocity increases by 1.5 from 33 to 34.5 in that interval.
Your function calculations seem to be very reliable; I would have mistakes if I wasn't doing the problem both ways!

 

[Kot]

I finished calculating the position during each time interval. Is this distance (m) the distance traveled in that time interval or is it the distance traveled from t=0s to the time interval? For example at 5s I got 40.83m and at 6s I got 14.83m. It doesn't make sense if I just plot these numbers into a position-time graph. I assume that the distance I calculated is specific to that interval (I have to add all the distances from each interval to get the total distance.) Is this correct?

 

[Delphi51]

If you are finding the area under the graph for the interval 5 to 6 s (either by integrating from 5 to 6 or by using a triangle area calc), then you must add that onto the distance at 5 m to get the total distance at 6 m. If you have been integrating and carefully setting the value of the C constant each time as we did with velocity, then your function should give the total distance. The C effectively makes the function cumulative.

I have x = 40.83 m at time 5, and 63 m at time 6 using x = -t³ + 15t² - 30t + 45.

 

[Kot]

For the interval t from 5 to 6 I got $x=-\dfrac{t^3}{3}+\dfrac{15t^2}{2}-30t+40.83$, I double checked my work and couldn't find any mistakes. Where did the denominator for your first two terms go and how did you get 45 as a constant?

*edit I was able to finish the problem and got an answer that made sense. Although the calculations were a bit tedious... Thank you for the help :)

 

[Delphi51]

I bungled it! No doubt you are correct.