# Calculate the speed of a point based on a graph

• Davidllerenav
In summary, The average speed of the point during the movement time is 0.1m/s. The maximum speed of the point is between t=10s and t=16s, where the slope of the curve is steepest. The moment t0 in which the instantaneous speed is equal to the average speed averaged over the first t0 seconds can be calculated using the formula v=Δs/Δt, using s1=4m, s2=18m, t1=10s, and t2=16s.
Davidllerenav

## Homework Statement

A point moves rectilinearly in one direction. The figure shows the distance s traveled by the point as a function of time. Using the graph, find:
1. The average speed of the point during the movement time.
2. Maximum speed
3. The moment t0 in which the instantaneous speed is equal to the average speed averaged over the first t0 seconds.

v=Δs/Δt
a=(v-v0)/Δt

## The Attempt at a Solution

I tried the first part using the formula v=Δs/Δt, by looking at the picture I did it like this: v=(2.0m-0m)/20s-0s = 0.1m/s. Is it correct?
My problem is with the second and third part. How do I calculate the maximum velocity and how do I find the moment where the instantaneous velocity is equal to the average speed averaged over the first t0 seconds? Thanks.

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1]
I tried the first part using the formula v=Δs/Δt, by looking at the picture I did it like this: v=(2.0m-0m)/20s-0s = 0.1m/s. Is it correct?
Always do a sanity check.
How far did it travel in total? How long did it take?
Does 0.1m/s seem like it would get you that far in that time?

2] Well, velocity is literally s over t, and you want the biggest value. Where is the greatest delta s in the least delta t? Where on the graph does that occur?

Is there anything on the graph that will allow you to quantify (i.e. put numbers to) the deltas of s and t?

3] Have you studied slopes? What is the slope of the average velocity? Is there anywhere on the graph where the curve has the same slope as the average?

DaveC426913 said:
Well, velocity is literally s over t, and you want the biggest value. Where is the greatest delta s in the least delta t? Where on the graph does that occur?

Is there anything on the graph that will allow you to quantify (i.e. put numbers to) the deltas of s and t?
No, there is no more information than the one I've provided. I choose 2.0m and 20s because those are the total distance that the object traveled and the total time it spent, at least I think that the graph says.

Davidllerenav said:
No, there is no more information than the one I've provided. I choose 2.0m and 20s because those are the total distance that the object traveled and the total time it spent, at least I think that the graph says.
I was being too cryptic.
The axes will allow you to put numbers to the delta s and delta t.

Identify a segment of the curve where v is maximum. Find the x and y intercepts at each end of the segment.

DaveC426913 said:
1]

Always do a sanity check.
How far did it travel in total? How long did it take?
Does 0.1m/s seem like it would get you that far in that time?

Well, It traveled 2m in total, and it took 20 seconds, so if the speed is 0.1m/s and I multiplied it by the time 0.1m/s * 20 s = 2m, so yes it is correct.
DaveC426913 said:
2] Well, velocity is literally s over t, and you want the biggest value. Where is the greatest delta s in the least delta t? Where on the graph does that occur?
I think it would be from t=10s to t=16s, right? Because it goes from s=4 to s=18., right?
DaveC426913 said:
3] Have you studied slopes? What is the slope of the average velocity? Is there anywhere on the graph where the curve has the same slope as the average?
The slope would be its derivative?

Davidllerenav said:
Well, It traveled 2m in total, and it took 20 seconds, so if the speed is 0.1m/s and I multiplied it by the time 0.1m/s * 20 s = 2m, so yes it is correct.
Ad you didn't even need us.

Davidllerenav said:
I think it would be from t=10s to t=16s, right? Because it goes from s=4 to s=18., right?
Looks right. Mind those decimals though.

Davidllerenav said:
The slope would be its derivative?
Well, the slope can be directly calculated from s2-s1 over t2-t1.

That should help with both 2] and 3].

DaveC426913 said:
Looks right. Mind those decimals though.

That should help with both 2] and 3].
So, I would have use the same formula of ##v=\frac {\Delta s}{\Delta t}## but using s1 as 4 and s2 as 18, the same with ##t##, right? What decimals?
DaveC426913 said:
Well, the slope can be directly calculated from s2-s1 over t2-t1.

That should help with both 2] and 3].
But wouldn't it end up being the same as the speed I got on the first part? Since is the same formula.

Davidllerenav said:
So, I would have use the same formula of ##v=\frac {\Delta s}{\Delta t}## but using s1 as 4 and s2 as 18, the same with ##t##, right?
Yes.

Although...
Davidllerenav said:
What decimals?
You used the correct values in your first post - to get 0.1m/s, but you have since dropped the decimals in your s values.

Davidllerenav said:
But wouldn't it end up being the same as the speed I got on the first part? Since is the same formula.
Yes. That's what question 3 is asking for. At what point(s) on the slope is the instantaneous velocity equal to the average velocity?
You can calc the average velocity. Now find at least one point on the curve where the instantaneous v has the same slope. That's where you'll need to calc the derivative.

DaveC426913 said:
You used the correct values in your first post - to get 0.1m/s, but you have since dropped the decimals in your s values.
I see, so it would be ##s_1=4.0m##, ##s_2=18.0m## and ##t_1=10.0s##, ##t_2=16s##?
DaveC426913 said:
Yes. That's what question 3 is asking for. At what point(s) on the slope is the instantaneous velocity equal to the average velocity?
You can calc the average velocity. Now find at least one point on the curve where the instantaneous v has the same slope. That'swhere you'll need to calc the derivative.
This one I find it kind of difficult, I should take the derivative at the point that seams equat to de straight line that concets (0,0) to (20,2.0), right? If so, I think that there are two, t=10 and t=18, right?

Davidllerenav said:
I see, so it would be ##s_1=4.0m##, ##s_2=18.0m## and ##t_1=10.0s##, ##t_2=16s##?
Okay...
You're put some decimals in. Good start.
Now let's work on getting them in the right place.

DaveC426913 said:
Okay...
You're put some decimals in. Good start.
Now let's work on getting them in the right place.
Sorry, what do you mean?

Davidllerenav said:
Sorry, what do you mean?
Look more carefully at the scale on the vertical axis. It does not say 10m and 20m.
Davidllerenav said:
it goes from s=4 to s=18
Looks to me that the speed has already dropped a little before s=1.8. You could use s=1.6, but the line does not conveniently pass through a grid intersection there. Where would be better?

For the third part, pick some arbitrary time t and mark on the curve where it is at that time. What line could you draw that shows the average speed up to then?

haruspex said:
Look more carefully at the scale on the vertical axis. It does not say 10m and 20m.
Oh, I see. Then it would be $s_1=1.0m$ and $s_2=2.0m$.
haruspex said:
Looks to me that the speed has already dropped a little before s=1.8. You could use s=1.6, but the line does not conveniently pass through a grid intersection there. Where would be better?
Maybe s=1.4? It is still an straight line.
haruspex said:
For the third part, pick some arbitrary time t and mark on the curve where it is at that time. What line could you draw that shows the average speed up to then?
A tangent line?

Davidllerenav said:
Oh, I see. Then it would be ##\$s_1=1.0m## and ##s_2=2.0m##.
Yes.
Davidllerenav said:
Maybe s=1.4? It is still an straight line.
Yes.
Davidllerenav said:
A tangent line?
No.
In terms of the point on the curve, how much time has elapsed and how much distance has been covered?

haruspex said:
No.
In terms of the point on the curve, how much time has elapsed and how much distance has been covered?

If I pick, for example, ##t=10s##, the time elapsed would be 10 seconds and the distance would be 4.0 m.

Last edited:
Davidllerenav said:
If I pick, for example, ##t=10s##, the time elapsed would be 10 seconds and the distance would be 4.0 m.
Right, so what line can you draw with that slope?

haruspex said:
Right, so what line can you draw with that slope?
A line that goes from the origin to the point (10, 4.0), It would be the average speed on the interval, right?

Davidllerenav said:
A line that goes from the origin to the point (10, 4.0), It would be the average speed on the interval, right?
A line is not a speed. What feature of the line would represent the average speed?

haruspex said:
A line is not a speed. What feature of the line would represent the average speed?
The slope.

Davidllerenav said:
The slope.
Right.
And what line through the point has a slope representing the instantaneous speed?
If the average and instantaneous speeds are the same, what can you say about these two lines?

haruspex said:
Right.
And what line through the point has a slope representing the instantaneous speed?
If the average and instantaneous speeds are the same, what can you say about these two lines?
That they have the same slope.

Davidllerenav said:
That they have the same slope.
Yes, but more than that. They are ...?

haruspex said:
Yes, but more than that. They are ...?
The same line. They are equal.

Davidllerenav said:
The same line.
Yes.
How can you use that to find such a time on the graph?
(But I notice you did not explicitly answer this question: what line through the point has a slope representing the instantaneous speed? You only described its slope.)

haruspex said:
Yes.
How can you use that to find such a time on the graph?
(But I notice you did not explicitly answer this question: what line through the point has a slope representing the instantaneous speed? You only described its slope.)
Oh, sorry. It would be a tangent line, right?

Davidllerenav said:
Oh, sorry. It would be a tangent line, right?
Yes.
How can you use that to find such a time on the graph?

haruspex said:
Yes.
How can you use that to find such a time on the graph?
I can just select any time but using the same slope, right? Something like ##s=mx+b## and b would be 0.

Davidllerenav said:
I can just select any time but using the same slope, right? Something like ##s=mx+b## and b would be 0.
No, you are looking for a time when the two lines are the same.

haruspex said:
No, you are looking for a time when the two lines are the same.
It would be time ##t=10s##.

Davidllerenav said:
It would be time ##t=10s##.
The line from the origin to the curve at t=10s is the same as the tangent at t=10s? I don't think so.

haruspex said:
The line from the origin to the curve at t=10s is the same as the tangent at t=10s? I don't think so.
But I think there isn't any line from the origin that would be equal to the tangent at t=10s. At leaks it looks like that.

Davidllerenav said:
But I think there isn't any line from the origin that would be equal to the tangent at t=10s. At leaks it looks like that.
Right, so t=10s is not the point we're after - that was just an example.
Find a point where the two lines to that point, as we have discussed, would be the same.

haruspex said:
Right, so t=10s is not the point we're after - that was just an example.
Find a point where the two lines to that point, as we have discussed, would be the same.
It seems that t=16 would do.

Davidllerenav said:
It seems that t=16 would do.
Bingo.

haruspex said:
Bingo.
Ok! So for the second question, I'll have to calculate the maximum speed (slope) at the interval from t=10 sto t=14s, where the distance goes from s=1.0m to s-1.4m.
For the third, I must say that the instantaneous speed at t=16 is equal to the average speed from t=0s to t=16s.
That's all, right?

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