Use the data to construct a graph etc.

[Alexanddros81]

Homework Statement

Serway Physics Section 2.4 Acceleration
(a) Use the data in Problem 5 to construct a smooth graph of position
versus time. (b) By constructing tangents to the x(t) curve, find the instantaneous
velocity of the car at several instants. (c) Plot the instantaneous velocity versus time
and, from this information, determine the average acceleration of the car.
(d) What was the initial velocity of the car?

Homework Equations

The Attempt at a Solution

[/B]
Is part (a) correct?
Any hints for part (b) ? For example if I take a tangent to x(t) at time 1.0s what is the instantaneous velocity at t=1.0 s?

 

[verty]

Try to find a formula for the graph. Use the turning point form: $y = a(x - b)^2 + c$

PS. All the help I can give.

 

[NateTheGreatt77]

Think about what the tangent gives you. What is the rise/run of the tangent line telling you about the position of the car vs time?

 

[Alexanddros81]

Try to find a formula for the graph. Use the turning point form: $y = a(x - b)^2 + c$

PS. All the help I can give.

Hi. Can you give me an example or point to an example?

 

[Alexanddros81]

I have found this web-page that tells you how to find the equation by knowing the x and y coordinates.

http://www.mathsmutt.co.uk/files/qe2.htm

If the equation is of the form $y=a(x-b)^2+c$ then:

for point $x=2.0 , y=9.2$ and $b=0$ (since it is the axis of symmetry) we can solve the quadratic:

$9.2=a(2.0)^2+c => 9.2=4a+c$ (1)

for point $x=3.0, y=20.7$ and $b=0$ => $20.7=a(3.0)^2+c => 20.7=9a+c$ (2)

By subtracting (1) from (2) we find that a=2.3 and also that c=0

So the quadratic equation is: $y=2.3x^2$

How do I proceed next?

 

[Alexanddros81]

Ok. So now I find the derivative of $y=2.3x^2$

$\frac {dy} {dx} = \frac {d} {dx} 2.3x^2 = 4.6x$

So at the point where x = 1 , y'(1) = 4.6

So at x= 1 the instantaneous slope of the curve will be 4.6

So at the point where x = 2 , y'(2) = 9.2

So at x= 2 the instantaneous slope of the curve will be 9.2

So at the point where x = 3 , y'(3) = 13.8

So at x= 3 the instantaneous slope of the curve will be 13.8

So at the point where x = 4 , y'(4) = 18.4

So at x= 4 the instantaneous slope of the curve will be 18.4

The instantaneous slopes found for each x is the instantaneous velocity for that x.

Is the above correct?

 

[ZapperZ]

@Alexanddros81 : You need to start over, and follow what @NateTheGreatt77 is suggesting.

1. What is the meaning of "tangent lines"?

2. What is the meaning of "instantaneous velocity" with respect to a x-t graph?

3. If I pick a point on that line, can you find the slope of the curve at that point? This is where you need to know what #1 means. You do NOT have to know the equation of the line that you have drawn, especially when it wasn't given to you.

4. Once you know the slope of the curve at that point, how does this relate to the "instantaneous velocity" at that point?

From the way this has been described, you are expected to do this "by hand", i.e. manually.

This exercise is a test on whether you have understood what a x-t curve represents, and what it can tell you.

Zz.

 

[Alexanddros81]

@Alexanddros81 : You need to start over, and follow what @NateTheGreatt77 is suggesting.

1. What is the meaning of "tangent lines"?

2. What is the meaning of "instantaneous velocity" with respect to a x-t graph?

3. If I pick a point on that line, can you find the slope of the curve at that point? This is where you need to know what #1 means. You do NOT have to know the equation of the line that you have drawn, especially when it wasn't given to you.

4. Once you know the slope of the curve at that point, how does this relate to the "instantaneous velocity" at that point?

From the way this has been described, you are expected to do this "by hand", i.e. manually.

This exercise is a test on whether you have understood what a x-t curve represents, and what it can tell you.

Zz.

From what I can tell from the graph by taking the tangent of the black line at point x (or t) = 2s for example
I can say that :

$v_x=\frac {18-0} {3-1} = +9m/s$

Is this correct?

 

[ZapperZ]

From what I can tell from the graph by taking the tangent of the black line at point x (or t) = 2s for example
I can say that :

$v_x=\frac {18-0} {3-1} = +9m/s$

View attachment 233582

Is this correct?

Yes, that is correct. You are on the right track.

Zz.

 

[Alexanddros81]

@ZapperZ :

Hi. I have some questions...
The tangent line is drawn to the left of point t=2s up to where the y-axis (or x) is equal to zero (and t=1s)
So do I draw to the right of point t=2s up to where the t=3s? Should it be symmetry?

 

[ZapperZ]

@ZapperZ :

Hi. I have some questions...
The tangent line is drawn to the left of point t=2s up to where the y-axis (or x) is equal to zero (and t=1s)
So do I draw to the right of point t=2s up to where the t=3s? Should it be symmetry?

You can draw the line arbitrarily long or short. It doesn't matter. All you care about is that it is a tangent line to that point. You can use ANY two points on the tangent line to calculate its slope. It will give you the same answer to matter which two points you pick.

Zz.

 

[Alexanddros81]

@ZapperZ
Thanks.
That was the info i didn't know.
Any web-page or paper describing this?

Alex

 

[ZapperZ]

@ZapperZ
Thanks.
That was the info i didn't know.
Any web-page or paper describing this?

Alex

Describing what?

Zz.

 

[Alexanddros81]

I meant the info:

You can draw the line arbitrarily long or short. It doesn't matter. All you care about is that it is a tangent line to that point. You can use ANY two points on the tangent line to calculate its slope. It will give you the same answer to matter which two points you pick.

 

[ZapperZ]

I meant the info:

This is math. It comes from your math class where you study straight line graphs: y = mx + b.

But also think about it. It is a straight line, meaning it has a constant slope all throughout the line. So why should a slope you found at one point be different than the slope at another point? If they differ, this is not a straight line.

So if you want to find the slope of a straight line, ANY two points on that line will give you the slope.

Zz.

 

[Alexanddros81]

@ZapperZ

Thanks.
I am coming back to math and physics after some time of absence from them.

 

[jedishrfu]

Something that may help you in future studies is the Geogebra app:

https://en.wikipedia.org/wiki/GeoGebra

 

[verty]

Sorry @Alexanddros81, I forgot about question b. I was thinking it was just a and then c.

 

[Alexanddros81]

For a more precise measurment of the instantaneous velocity for part (b)
I used Geogebra app.

I know that the function is given by $y = 2.3x^2$
I find the derivative to be $y'(x) = 4.6x$

Now I find the $y'(2) = 9.2$ . This is the slope at point t=2

Also I know that at t=2.0, x=9.2 or at x=2.0, y=9.2

So now I can calcuate the equation for the tangent line:

$y-y_1 = m(x-x_1)$ (1) where m is the slope and $x_1=2.0$ and $y_1=9.2$

So (1) gives y=9.2x-9.2

By adding the tangent line to the graph in Geogebra we get figure:

Is there a function in geogebra that when we input the point (2,9.2) on the graph to draw the tangent line automatically?

 

[ZapperZ]

I don't understand what you are trying to do here. If you managed to fit the data to an analytical equation, then the value of the slope at any point on the line is easily obtained. So why go through all that tedious exercise in finding the tangent line?

Please remember that the initial idea of using tangent lines here is to do this MANUALLY, i.e. without any kind of fitting program. If you were given this question in an exam, in class, and you are prohibited to use any electronic device other than a calculator, can you do this problem? My reading of the way the question is presented is that this is what they want you to do.

Zz.

 

[Alexanddros81]

So here is part (b):

Also tangent line at point x(or t) at 5s : $v_x = \frac {80-34} {6-4} = v_x = 23 m/s^2$

 

[ZapperZ]

OK, so now do (c) and (d).

Zz.

 

[Alexanddros81]

Ok so now for part (c) I plot the instantaneous velocities vs time.

And joining the points with a line I get velovity vs time graph.
If we take the slope of the red line we find the average acceleration.

So $a_{avg} = \frac {23-0} {5-01} = 4.69 m/s^2$

And for (d) at t=0 the velovity is (according to the graph): -0.4m/s
or
we could say is roughly zero?

Am I correct?

 

[ZapperZ]

Yes, that should be correct.

Zz.