Position at any one point on Velocity-Time Graph

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  • #1
A v-t graph depicts an object's movement along the positive x axis from 0-10 seconds. If the object at position x=xofc on the line at t= 6 seconds (depicted as point C on the graph), find its position at t= 10 seconds.

It appears the question is not referring to total displacement but just asking for where the object is exactly at 10 seconds. I assume this is straightforward and that calculating d=vt will get me the position at t=10 sec. In this case v = -15 so position is -150m.

I wonder if the question is more involved than I am assuming and that I need to do more. Any feedback would be great!


Answers and Replies

  • #2
d = vt only in the special case where v is constant. If your graph of v vs t is a straight, horizontal line, then you are in luck. If it has any non-zero slope, or if it is not straight, then this approach will not work.

Remember that the definition of v is dx/dt (if you have had calculus, the derivative of x vs t. If you have not had calculus, then the delta (change) in x over the delta (change) in time, for an infinitesimally small delta of time). That means that v is the slope of the x vs t graph (rise over the run). To reverse the process, you have to integrate v vs t to find x (or if you have not had calculus, find the area under the curve for the graph of v vs t).

I hope that made sense.
  • #3
Thanks. I have had calculus so I recall the antiderivative. So, x is the antiderivative of v. Meaning x(t) =∫ -15t dt. This is -15t^2/2 + C correct? I thought I had to know x at t=0 though to find C. If not then, x(10) is -15(100)/2 or -750m. ( I don't think this is right, though, my calc is rusty, but I think I'm close.

At any rate, this is algebra based physics, so while I would like to know if my calc is right (comments appreciated), I have to do the problem using algebra based physics.

I understand total displacement is the area under the graph, but I don't completely understand finding position at t=10 seconds. By position, do they mean total displacement after 10 seconds? In this case then I can calculate delta x=deltav/delta t for each time segment, add up the results and this will give me total displacement which adds up to +75m to the right. So, I would say that at t=10 seconds, the object is 75m to the right. (Total distance traveled is 105m, but I don't think that is what is being asked.)
Is this right? Was I just getting confused on the language used here by misinterpreting what is being meant by position? The graph is attached also now!

Thanks again.


  • #4
Thanks for posting the graph for the problem - this greatly affects the answer!

First: the equation for v as a function of t is NOT v = -15t. To use calc, you need to find an expression for v(t), then integrate. Between Point C and t = 10 sec, v(t) is a straight line, of the form y = mx + b. You would have to write this as v(t) = (slope) t + (y-intercept), and you can read the values for slope and intercept off the graph. After you find the correct expression, then you will use the definite integral from t = 6 to t = 10 sec., since this is the question that was asked (what is the total displacement between these two times?).

Second: Since you are in algebra/trig-based physics, go back to the definition of an integral as "the area under the curve", and use that to calculate the total displacement. (I think I recognize the style of graphics - if you are using Pearson's Knight book or MasteringPhysics, then I know they typically use displacement as the area under the graph of v vs t). Remember that areas above the x-axis are positive, and areas below the x-axis are negative. What is the total area under the curve (or the sum of the areas of the two triangles) between t = 6 seconds and t = 10 seconds? This gives you the displacement (or difference along the x-axis) from your starting position.

Please let me know if this helps.
  • #5
Thanks. It does help. I am very rusty in calculus obviously. But, I did figure it out. The expression as I see it is v(t)= -7.5t + 60 and the antiderivative is 60x-3.75x^2. Using the definite integral from t=6 to t =10 gives a result of zero. This makes perfect sense, of course, because just by looking at the graph it is clear that the sum of the two areas cancel each other out. This actually seemed to clear from the get-go, but the wording of the problem had me more confused than anything. I did not find it super clear that the question was asking for the displacement between 6 and 10 seconds. It is worded as such that I wasn't sure if it was the exact placement of the object at t=10, the total displacement or what that was being asked for. It seems you are interpreting the question as what is the total displacement between 6 and 10 seconds, in which case the question is quite straightforward. I don't know why physics questions have to include such ambiguities! If they want to know total displacement between 6 and 10 seconds, then just say so! My two cents, of course. I am finding these types of issues often as I work through physics and it just seems like such a game of words sometimes. Although, physics is clearly not my forte so it's probably my issue with not being familiar enough with how the physics world describes some things and not a problem with the writing, but well, I do think they could have just asked what the displacement is between 6 and 10 seconds. By asking to "find its (the object's) position at t=10 seconds" makes it a cloudy question in my opinion and does not make it super clear that total displacement between 6 & 10 seconds is what they are after.

THANKS for the help and apologize for my calculus idiocy. It's been a little while.
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  • #6
A v-t graph depicts an object's movement along the positive x axis from 0-10 seconds. If the object at position x=xofc on the line at t= 6 seconds (depicted as point C on the graph), find its position at t= 10 seconds
What does xofc mean?
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  • #7
sms - I am PROUD of the way you stuck with this problem and figured out not just the equations, but what they MEANT! Your understanding of physics is greater than what you are giving yourself credit for.

GOOD JOB!!!!!!

One word of caution - it looks like the question is asking for the actual location at t = 10 seconds. The problem gave you the actual location at t = 6.0 seconds, and you calculated the displacement between 6 sec and 10 sec to be 0. Tie it all in a final knot: what is the location at t = 10 seconds?

I hate to state the obvious, but the number one mistake I see students make in physics is forgetting to re-read the question to make sure they answer the question that was asked.

Good luck with your studies,
  • #8
Thanks, JF! I feel encouraged now. I will make you further proud in that I did reread the question (even before your suggestion to do so!) and did tie it all together. I answered the question as total displacement between 6-10 seconds is zero therefore position at t=0 is x subscript c (ie: the position as stated at t=6 sec). The question became more clear to me also after I went through it again. I guess it will just take a little time getting used to the physics texts and how things are worded but hopefully I will get better at deciphering problems with practice.