Calculate the speed of a point based on a graph

[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.
   

Homework Equations

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.

 

[DaveC426913]

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?

 

[Davidllerenav]

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.

 

[DaveC426913]

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.

 

[Davidllerenav]

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.

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?

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?

 

[DaveC426913]

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.

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.

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].

 

[Davidllerenav]

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?

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.

 

[DaveC426913]

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...

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.

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.

 

[Davidllerenav]

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 $t_1=10.0s$, $t_2=16s$?

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?

 

[DaveC426913]

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.

 

[Davidllerenav]

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?

 

[haruspex]

Sorry, what do you mean?

Look more carefully at the scale on the vertical axis. It does not say 10m and 20m.

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?

 

[Davidllerenav]

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 $s_2=2.0m$.

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.

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?

 

[haruspex]

Oh, I see. Then it would be $s_1=1.0m$ and $s_2=2.0m$.

Yes.

Maybe s=1.4? It is still an straight line.

Yes.

A tangent line?

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

 

[Davidllerenav]

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.

 

[haruspex]

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?

 

[Davidllerenav]

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?

 

[haruspex]

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?

 

[Davidllerenav]

A line is not a speed. What feature of the line would represent the average speed?

The slope.

 

[haruspex]

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?

 

[Davidllerenav]

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.

 

[haruspex]

That they have the same slope.

Yes, but more than that. They are ...?

 

[Davidllerenav]

Yes, but more than that. They are ...?

The same line. They are equal.

 

[haruspex]

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.)

 

[Davidllerenav]

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?

 

[haruspex]

Oh, sorry. It would be a tangent line, right?

Yes.
How can you use that to find such a time on the graph?

 

[Davidllerenav]

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.

 

[haruspex]

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.

 

[Davidllerenav]

No, you are looking for a time when the two lines are the same.

It would be time $t=10s$.

 

[haruspex]

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.

 

[Davidllerenav]

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.

 

[haruspex]

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.

 

[Davidllerenav]

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.

 

[haruspex]

It seems that t=16 would do.

Bingo.

 

[Davidllerenav]

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?