What are the basics of physics regarding displacement, velocity, and time?

[roger]

Hi ,

I want to know the rigorous mathematical definition of displacement, which is applicable to physics ?

All we are taught is that its a vector quantity as opposed to a scalar such as distance.

And I have a velocity time graph, which again, I am not too sure on a few aspects of it.


For example, if a ball is dropped and bounces back up, and this repeats until the ball stops, each time the velocity decreases, the graph shows a straight sloped line going down to -3, then a staight line with same gradient from +2.

But what about the velocity in between 2 and -3 ?

and what meaning can be given to a vertical line on the velocity time graph ?

thanks

Roger

 

[HallsofIvy]

Hi ,

I want to know the rigorous mathematical definition of displacement, which is applicable to physics ?

All we are taught is that its a vector quantity as opposed to a scalar such as distance.

I don't believe for a moment that your text simply says that "displacement is a vector" without saying what vector! In order to have a "displacement" and object has to move from one point to another and the displacement vector is the vector from the beginning point to the end point.

And I have a velocity time graph, which again, I am not too sure on a few aspects of it.


For example, if a ball is dropped and bounces back up, and this repeats until the ball stops, each time the velocity decreases, the graph shows a straight sloped line going down to -3, then a staight line with same gradient from +2.

Is "-3" a time or velocity? "a straight sloped line going down to -3" is meaningless. Every line segment has two end points, every point on the graph has two coordinates. What are the time and velocity coordinates of the two endpoints of this line segment?

But what about the velocity in between 2 and -3 ?

Once again are those time or velocity values? What are the coordinates of the enpoints of the line segment you are referring to?

and what meaning can be given to a vertical line on the velocity time graph ?

No physical meaning! It might refer to an object "bouncing" as it hits the ground, changing velocity extremely fast. That can be "idealized", mathematically, as an instantaneous change in velocity but can't actually happen physically.

I suspect what you have is this: at some initial time, t= 0, you drop the ball. It's velocity at that time is 0. It accelerates with the acceleration due to 0: g= -9.81 m/s² and since acceleration is "change in velocity divided by change in time" that is the slope of the velocity versus time graph.

You don't say how long the ball drops or how far but presumably it drops long enough for the velocity to reach -3 m/s (and so the time dropped must have been (-3 m/s)/(-9.81 m/s²)= 0.3 seconds, approximately. Your graph should show a straight line, with slope -9.81, from (0,0) to to (0.3, -3) (That is, again, approximate. That line would have slope -10!).

Now, the ball bounces. Very, very quickly, in a time too brief to be seen on the graph, the ball's velocity changes. If this were a "perfectly elastic" collision, the ball's velocity would change from -3 to +3. This is not a perfectly elastic collision, you are told that the collision changes the ball's velocity from -3 to 2 (and the lost energy is absorbed in the ground). There is a "vertical" line from (0.3, -3) to (0.3, 2). If it makes you feel better, you can think of that line as not "vertical" but tilted very, very slightly- too slightly to be seen on the graph: say from (0.30581039755351681957186544342508, -3) to (0.30581039755351681957186544342509, +2) (that 0.30581039755351681957186544342508 is a more accurate value of (-3)/(-9.81)).

Since the acceleration is still that due to gravity, -9.81 m/s², that is still the slope of this new line segment. Remember that this is a graph of velocity agains time, not height. The ball's velocity changes suddenly to +2 but the ball itself is still on the ground at that instant.

thanks

Roger

You are welcome.

 

[Spastik_Relativity]

And I have a velocity time graph, which again, I am not too sure on a few aspects of it.
For example, if a ball is dropped and bounces back up, and this repeats until the ball stops, each time the velocity decreases, the graph shows a straight sloped line going down to -3, then a staight line with same gradient from +2.

But what about the velocity in between 2 and -3 ?

and what meaning can be given to a vertical line on the velocity time graph ?

I think I know what your talking about. Is it a graph that looks sort of like a zig-zag?

If so the velocity at the peak of the balls bounce changes direction therefore the sign infront of the velocity changes. There is no velocity (neglectible) between those two points. Similarly when the balls bounce of the ground teh direction of its velocity changes directions. The changes of direction are shown by the graph moving from positive to negative described by the straight lines. The vertical line means nothing. If one was to integrate it you would get nothing thus no displacement. Its just a change in direction.

 

[roger]

I don't believe for a moment that your text simply says that "displacement is a vector" without saying what vector! In order to have a "displacement" and object has to move from one point to another and the displacement vector is the vector from the beginning point to the end point.


Is "-3" a time or velocity? "a straight sloped line going down to -3" is meaningless. Every line segment has two end points, every point on the graph has two coordinates. What are the time and velocity coordinates of the two endpoints of this line segment?


Once again are those time or velocity values? What are the coordinates of the enpoints of the line segment you are referring to?


No physical meaning! It might refer to an object "bouncing" as it hits the ground, changing velocity extremely fast. That can be "idealized", mathematically, as an instantaneous change in velocity but can't actually happen physically.

I suspect what you have is this: at some initial time, t= 0, you drop the ball. It's velocity at that time is 0. It accelerates with the acceleration due to 0: g= -9.81 m/s² and since acceleration is "change in velocity divided by change in time" that is the slope of the velocity versus time graph.

You don't say how long the ball drops or how far but presumably it drops long enough for the velocity to reach -3 m/s (and so the time dropped must have been (-3 m/s)/(-9.81 m/s²)= 0.3 seconds, approximately. Your graph should show a straight line, with slope -9.81, from (0,0) to to (0.3, -3) (That is, again, approximate. That line would have slope -10!).

Now, the ball bounces. Very, very quickly, in a time too brief to be seen on the graph, the ball's velocity changes. If this were a "perfectly elastic" collision, the ball's velocity would change from -3 to +3. This is not a perfectly elastic collision, you are told that the collision changes the ball's velocity from -3 to 2 (and the lost energy is absorbed in the ground). There is a "vertical" line from (0.3, -3) to (0.3, 2). If it makes you feel better, you can think of that line as not "vertical" but tilted very, very slightly- too slightly to be seen on the graph: say from (0.30581039755351681957186544342508, -3) to (0.30581039755351681957186544342509, +2) (that 0.30581039755351681957186544342508 is a more accurate value of (-3)/(-9.81)).

Since the acceleration is still that due to gravity, -9.81 m/s², that is still the slope of this new line segment. Remember that this is a graph of velocity agains time, not height. The ball's velocity changes suddenly to +2 but the ball itself is still on the ground at that instant.



You are welcome.

Thanks for the help, I understand the tilted line is very slight, but I wanted to know if we were to plot the period between (0.3, -3), and (0.3, 2) why would it be slanted and not vertical ?

if it was slanted would it mean the ball deccelerates to zero and then accelerates to 2 m/s ? If so, how does the ball do this ?

Also, shouldn't there be some discontinuities on the graph, i.e. when the ball is in contact with the ground and the top of its trajectory, for both of which there is no motion ?( for a split second)

thankyou
Roger

 

[HallsofIvy]

quote= Roger]Thanks for the help, I understand the tilted line is very slight, but I wanted to know if we were to plot the period between (0.3, -3), and (0.3, 2) why would it be slanted and not vertical ?[/quote]

The point I made before is that there is NO "period" between (0.3,-3) and (0.3,2). A graph of a line containing those two points would be a vertical line but that is not physically possible. Strictly speaking "period" refers to time- you are really asking about the "period" between 0.3 and 0.3! Since you say you understand that the line is very slightly tilted, you should understand that (0.3, -3) and (0.3, 2) are not on that line. It is possible that one is- the line might contain (0.3, -3) and (0.300000001, 2) or it might contain (0.2999999991,-3) and (0.3, 2)- but it can't contain both- that would have to be vertical line, reflecting an instantaneous acceleration which just isn't physically possible.

if it was slanted would it mean the ball deccelerates to zero and then accelerates to 2 m/s ? If so, how does the ball do this ?

Now, a "mathematical model" doesn't have to be 100% accurate to a physical situation. Physically, if we have a moment when the ball's speed is changing (imagine a "slow motion" film of the ball- watching it squash down as its speed changes from -3 to 0, then opening back up as its speed changes from 0 to 2)- physically, that takes some time (as I said before, perhaps from 0.3 to 0.300000001 or from 0.29999991 to 0.3)- if we wanted to be very accurate, we would have to try to draw a line showing a very slight slant- not vertical. Of course, probably we can't draw that accurately anyway! So, we either show a vertical line from (0.3, -3) to (0.3, 2) or, as you suggest, and a mathematician might prefer, making the velocity graph discontinuous there- not drawing the vertical line at all but leaving a gap between (0.3, -3) and (0.3, 2).

 

[roger]

I understand , but there are a few things which are I am still unclear about.

irrespective of which of the two points is on the tilted line, that line would represent a decceleration to zero and then an acceleration to 2. But why should it accelerate ?

And if the ball is changing direction , surely there is a point ''in between'' when the ball isn't moving at all ?

And should the discontinuity be a gap or a a small line segment at y=0 ?

Thanks

Roger

 

[HallsofIvy]

I understand , but there are a few things which are I am still unclear about.

irrespective of which of the two points is on the tilted line, that line would represent a decceleration to zero and then an acceleration to 2. But why should it accelerate ?

Do you understand what "acceleration" means- any time there is a change of velocity, there is an acceleration. In ordinary speech we think of "acceleration" as meaning an increase in speed, "deceleration" as a decrease. In physics, we use acceleration for both- increasing speed is positive acceleration, decreasing speed is negative acceleration. In this case the ball accelerates because it can't keep going in the same direction at the same speed- there's a floor in the way! Here, it really is an "acceleration"- speeding up- because the speed is initially -3 and it increases to +2. Physically, what happens is, as I said before, the ball hits the floor and squeezes up- that impact "squeezes" the ball slightly- it's velocity goes from -3 to 0 but its Kinetic Energy goes to 0 (remember that Kinetic Energy depends upon the square of the speed, it's not a vector quantity so direction doesn't matter). Some of that energy goes into "potential energy" in the squeeze. Once the velocity of the ball goes to 0, the ball starts to rebound- the potential energy stored in the ball changes into kinetic energy so the speed goes back up- but the velocity is the opposite way. Actually here, since the speed only goes up to 2, not 3, this is not a "perfectly elastic collision": some of that energy is irreversibly lost to heat.

And if the ball is changing direction , surely there is a point ''in between'' when the ball isn't moving at all ?

YES, YES, YES! Since the velocity goes form -3 to +2 there is a moment when the velocity is 0!

And should the discontinuity be a gap or a a small line segment at y=0 ?

A Mathematician would much prefer that there be a gap in the velocity graph from (0.3,-3) to (0.3,2). Some people (and graphing calculators!) just can't avoid drawing a vertical line. Either way represents a discontinuity in the time- velocity graph. Which, as I said earlier, is non-physical. There has to be some very short time interval between when the ball first hits the floor and when it leaves the floor- to be completely physical you should show a slight "tilt"- say from (0.3,-3) to (0.30000001, 2)- but you might have to make your graph really, really, large to show that!

Thanks

Roger

Once again, you are welcome.