-
Looking for HSC notes and resources? Check out our Notes & Resources page
velocity/time graphs - HELP!! (1 Viewer)
- Thread starter Finx
- Start date
boxhunter91
Member
- Joined
- Nov 16, 2007
- Messages
- 736
- Gender
- Male
- HSC
- 2009
You need to know what is being asked.
If your given a displacement/time graph ie (x vs t)
You can find when the particle is at rest when gradient function of the graph is dx/dt = 0
You should read solutions from success one.
See how they set it out. Will give you an understanding of what is being asked and intended for an appropriate answer.
Just have intense sessions of staring at them, get heaps of those questions and just do em and you'll improve, that's all anyone can say I reckon. Treat it like any other graph.
tonyharrison
Member
- Joined
- Mar 17, 2009
- Messages
- 366
- Gender
- Female
- HSC
- 2009
Displacement graphs:
0 = object is at the origin
>0 = object is at a distance from the origin in a positive direction
<0 = object is at a distance from the origin in a negative direction
The closer the line gets to the x-axis, the closer the object is to the origin
A question usually says, "an object is moving in a straight line along the axis". Imagine a 30cm ruler, the origin is at 15cm. If the curve is above the x-axis, the object is closer to 30cm. If the curve is below the x-axis, the object is closer to 0cm. (weird example i know, shhh)
Velocity graphs:
0 = object has 0 velocity (at rest)
>0 = object is moving in a positive direction
<0 = object is moving in a negative direction
NOTE: unlike a displacement graph, the x-axis DOES NOT equal the origin
If the curve is above the x-axis:
Positive gradient = speeding up in a positive direction
Negative gradient = slowing down in a negative direction
If the curve is below the x-axis:
Positive gradient = slowing down in a negative direction
Negative gradient = speeding up in a negative direction
NOTE: An object must stop before it changes direction (in a straight line), therefore velocity = 0
Displacement = area under the graph
Therefore, if object is at origin when t = 0, it returns to the origin when the area under the curve ABOVE the x-axis = area under the curve BELOW the x-axis
Acceleration Graphs:
0 = object is moving with a constant velocity (not speeding up or slowing down), OR at rest
Sorry, don't know much about these...especially how to draw them
If anyone can help, it would be much appreciated. Thanks!p>
0 = object is at the origin
>0 = object is at a distance from the origin in a positive direction
<0 = object is at a distance from the origin in a negative direction
The closer the line gets to the x-axis, the closer the object is to the origin
A question usually says, "an object is moving in a straight line along the axis". Imagine a 30cm ruler, the origin is at 15cm. If the curve is above the x-axis, the object is closer to 30cm. If the curve is below the x-axis, the object is closer to 0cm. (weird example i know, shhh)
Velocity graphs:
0 = object has 0 velocity (at rest)
>0 = object is moving in a positive direction
<0 = object is moving in a negative direction
NOTE: unlike a displacement graph, the x-axis DOES NOT equal the origin
If the curve is above the x-axis:
Positive gradient = speeding up in a positive direction
Negative gradient = slowing down in a negative direction
If the curve is below the x-axis:
Positive gradient = slowing down in a negative direction
Negative gradient = speeding up in a negative direction
NOTE: An object must stop before it changes direction (in a straight line), therefore velocity = 0
Displacement = area under the graph
Therefore, if object is at origin when t = 0, it returns to the origin when the area under the curve ABOVE the x-axis = area under the curve BELOW the x-axis
Acceleration Graphs:
0 = object is moving with a constant velocity (not speeding up or slowing down), OR at rest
Sorry, don't know much about these...especially how to draw them
If anyone can help, it would be much appreciated. Thanks!p>
ekoolish
Impossible?
Yeah i hate it when they ask you to graph the acceleration from the velocity or vice versa.
Displacement graphs...
(y = displacement, x = time)
Imagine the graph as a sorta, physical plane, the further away from the x-axis the line gets, the further away from its "start position" it gets. The closer it gets to the x-axis, the closer it gets BACK to where it started. (Ie the X-axis itself is the start point; essentially the x axis represents distance=0)
Velocity time graphs:
When iamgining our displacement graph you might have noticed that, if an object RUSHED away from where it started, it would need to move as far away from the X-axis, in a very short time. This.. would imply that it has covered alot of distance in a short time. What was the definition of velocity again? Distance/time with respect to direction.
Well, we've satisfied the first condition; The STEEPER the line in the displacement graph the FASTER the object was moving, and thus the HIGHER the line in the velocity graph should be. So If we had an object, say a car that moved 5 meters away from its start point every 5 seconds, we know we have a constant velocity 5/5=1. Ie, on the velocity graph it'd be a straight line.
About direction; THis is rather simple once you've grasped the above concept, if the object MOVING AWAY FROM THE X AXIS, that means its velocity is in a POSITIVE direction; (in our car example, this means you draw a straight line at y=1). What about if it was returning TO the X axis? Well, that would basically mean its velocity is in the NEGATIVE direction, the direction OPPOSITE to the POSITIVE direction (which we defined as being away from the x axis). SO that if the car was travelling back to the X axis, we would draw a velocity line, that is at y=-1.
But in all these cases we've only delt with "straight" line velocities, ie; the velocity is never changing, its always a constant at y=1 or y=-1 in our above example. What if it was a diagonal line?
Acceleration
That's where acceleration comes in. Acceleration is the CHANGE in velocity. If the velocity was increasing at a constant rate, (ie if the velocity line looked like y=x, a diagonal) then that would mean that the acceleration must be a constant. This is best understood in the context of physics, but... Imagine i dropped a ball. As you can imagine the ball will fall down, its velocity would actually INCREASE, (it'd be getting faster and faster, that's why if i threw a ball of a skyscraper someone's neck would brake, rather than if i dropped it from a meter above them). IN this case, the velocity will be CONSTANTLY increasing. But by what rate? what is the RATE at which the velocity cahnges.
THAT, is what acceleration is. It is essentially the gradient of the velocity graph. On earth the acceleration due to gravity is 9.8m/s (or in that close vicinity). As you can see it is a constant. It is constantly 9.8. The acceleration graph of this ball dropped off a skyscraper would be a straight line at 9.8. Its velocity graph will have a gradient of 9.8, if i plotted its velocity against time, i'd notice that for every second in time, its velocity would increase by 9.8! It's displacement, is thus as you can imagine the distance it moves from my hand, always increasing.
Great!, now how do i get one graph out of another.
First notice the "hierachy" we have here. Displacement is the most basic with velocity following and then finally acceleration.
Know this; you cannot go from acceleration to velocity, or velocity to displacement, without having some "point" to reference the graph. This is for the same reason when you integrate you get a C. Unless you can find what the C is, you only know the SHAPE of a graph, not where it is.
Now, going from displacement to velocity to acceleration is quite easy. As we've stated above, velocity is the change in displacment. First consider where is my displacement graph STEEPEST. That' is the point where im moving FASTEST away from the start point, and thus corresponds to MAXIMUMS (or minimums) in my velocity graph.
Same principle applies to acceleration, find where the line is the steepest, and mark that as maximums/minimums.
IT is a MAXIMUM, if the "steep" line has a positive gradient (ie; it is getting AWAY from the start point), it is MINIMUM if the steep line has a negative gradient, ie its getting CLOSER to the starting point.
The rest is more common sense, and as an above poster put it, stare intensly at the graph until it coems to you. Or a more systematic approach is, IF the displacement graph doesn't seem to be moving, ie; time passes in which the graph doesn't CHANGE displacement, it sorta plaetu's out... THEN, you know that your velocity is 0. (pretty obvious eh, think of it ouside the context of math, if im not moving, obviously i have no speed).
SAme applies for acceleration, if my velocity doesn't change, obviously i havn't been accelerated. Think of a block sliding on ice. It will move at the same velocity until it is accelerated, or decelerated. (On ice because, friction would have a greater role to play anwhere else, and it isn't considered in Math.)
(y = displacement, x = time)
Imagine the graph as a sorta, physical plane, the further away from the x-axis the line gets, the further away from its "start position" it gets. The closer it gets to the x-axis, the closer it gets BACK to where it started. (Ie the X-axis itself is the start point; essentially the x axis represents distance=0)
Velocity time graphs:
When iamgining our displacement graph you might have noticed that, if an object RUSHED away from where it started, it would need to move as far away from the X-axis, in a very short time. This.. would imply that it has covered alot of distance in a short time. What was the definition of velocity again? Distance/time with respect to direction.
Well, we've satisfied the first condition; The STEEPER the line in the displacement graph the FASTER the object was moving, and thus the HIGHER the line in the velocity graph should be. So If we had an object, say a car that moved 5 meters away from its start point every 5 seconds, we know we have a constant velocity 5/5=1. Ie, on the velocity graph it'd be a straight line.
About direction; THis is rather simple once you've grasped the above concept, if the object MOVING AWAY FROM THE X AXIS, that means its velocity is in a POSITIVE direction; (in our car example, this means you draw a straight line at y=1). What about if it was returning TO the X axis? Well, that would basically mean its velocity is in the NEGATIVE direction, the direction OPPOSITE to the POSITIVE direction (which we defined as being away from the x axis). SO that if the car was travelling back to the X axis, we would draw a velocity line, that is at y=-1.
But in all these cases we've only delt with "straight" line velocities, ie; the velocity is never changing, its always a constant at y=1 or y=-1 in our above example. What if it was a diagonal line?
Acceleration
That's where acceleration comes in. Acceleration is the CHANGE in velocity. If the velocity was increasing at a constant rate, (ie if the velocity line looked like y=x, a diagonal) then that would mean that the acceleration must be a constant. This is best understood in the context of physics, but... Imagine i dropped a ball. As you can imagine the ball will fall down, its velocity would actually INCREASE, (it'd be getting faster and faster, that's why if i threw a ball of a skyscraper someone's neck would brake, rather than if i dropped it from a meter above them). IN this case, the velocity will be CONSTANTLY increasing. But by what rate? what is the RATE at which the velocity cahnges.
THAT, is what acceleration is. It is essentially the gradient of the velocity graph. On earth the acceleration due to gravity is 9.8m/s (or in that close vicinity). As you can see it is a constant. It is constantly 9.8. The acceleration graph of this ball dropped off a skyscraper would be a straight line at 9.8. Its velocity graph will have a gradient of 9.8, if i plotted its velocity against time, i'd notice that for every second in time, its velocity would increase by 9.8! It's displacement, is thus as you can imagine the distance it moves from my hand, always increasing.
Great!, now how do i get one graph out of another.
First notice the "hierachy" we have here. Displacement is the most basic with velocity following and then finally acceleration.
Know this; you cannot go from acceleration to velocity, or velocity to displacement, without having some "point" to reference the graph. This is for the same reason when you integrate you get a C. Unless you can find what the C is, you only know the SHAPE of a graph, not where it is.
Now, going from displacement to velocity to acceleration is quite easy. As we've stated above, velocity is the change in displacment. First consider where is my displacement graph STEEPEST. That' is the point where im moving FASTEST away from the start point, and thus corresponds to MAXIMUMS (or minimums) in my velocity graph.
Same principle applies to acceleration, find where the line is the steepest, and mark that as maximums/minimums.
IT is a MAXIMUM, if the "steep" line has a positive gradient (ie; it is getting AWAY from the start point), it is MINIMUM if the steep line has a negative gradient, ie its getting CLOSER to the starting point.
The rest is more common sense, and as an above poster put it, stare intensly at the graph until it coems to you. Or a more systematic approach is, IF the displacement graph doesn't seem to be moving, ie; time passes in which the graph doesn't CHANGE displacement, it sorta plaetu's out... THEN, you know that your velocity is 0. (pretty obvious eh, think of it ouside the context of math, if im not moving, obviously i have no speed).
SAme applies for acceleration, if my velocity doesn't change, obviously i havn't been accelerated. Think of a block sliding on ice. It will move at the same velocity until it is accelerated, or decelerated. (On ice because, friction would have a greater role to play anwhere else, and it isn't considered in Math.)
Last edited:
tonyharrison
Member
- Joined
- Mar 17, 2009
- Messages
- 366
- Gender
- Female
- HSC
- 2009
Haha you so obviously do physics. Me too =)
The only problem is, like you addressed, in physics, all the examples talk of CONSTANT velocity or CONSTANT acceleration, but in the maths examples, the graphs are usually spastic curvy lines.
Like there's a question i was trying to do, where it had the graph of the velocity of an object as a curvy line, then asking you to graph it as acceleration/time.
Do you know how to do them? The only thing i manage to understand is when the accleration is 0..but apart from that...
HSC 2006: Question 7 (b) (iii)
http://www.boardofstudies.nsw.edu.au/hsc_exams/hsc2005exams/pdf_doc/maths_05.pdf
The only problem is, like you addressed, in physics, all the examples talk of CONSTANT velocity or CONSTANT acceleration, but in the maths examples, the graphs are usually spastic curvy lines.
Like there's a question i was trying to do, where it had the graph of the velocity of an object as a curvy line, then asking you to graph it as acceleration/time.
Do you know how to do them? The only thing i manage to understand is when the accleration is 0..but apart from that...
HSC 2006: Question 7 (b) (iii)
http://www.boardofstudies.nsw.edu.au/hsc_exams/hsc2005exams/pdf_doc/maths_05.pdf
It's a little complex, but the principles don't change.
At the begining of the graph, from points 0 to 1, we notice that the velocity graph is stagnant. There has been no acceleration done upon this object.
From 1 to 2 we notice a steep fall in the velocity, caused by some negative acceleration.
At point 2. there is an inflexion, What has happened is the ACCELERATION has changed direction. It has encountered a minimum value, and from this point on from 2 to 3; the velocity is changing direction (it is still in heavily under negative velocity, just the level of negativness is decreasing. (think if the ball i dropped off a skyscrapper were to fall onto a trampoline, at first, the ball would still be moving downwards, but the springy trampoline is SLOWING it down).
At point 3, the "trampoline" has completely offset the force of gravity on the ball. Its velocity is constant now. and therefore its acceleration is 0. Until the point 5. At which point.... i dunno this very "slow to react trampoline"... decides to restore its original form, casuing the velocity to rapidly increase from its previous value of -2, back to 0. (this would mean positive acceleration)
Last edited:
tonyharrison
Member
- Joined
- Mar 17, 2009
- Messages
- 366
- Gender
- Female
- HSC
- 2009
Yeah ok that's kinda what i drew. EXCEPT wouldn't at 2, the acceleration be 0 because the velocity is 0 and therefore the object must've stopped so it's not acceleration?
So if a curve has a positive gradient BELOW the x-axis, what does that mean?
Thanks so much btw!
So if a curve has a positive gradient BELOW the x-axis, what does that mean?
Thanks so much btw!
Last edited:
It is true the velocity is 0, But that doesn't mean the acceleration is 0 also. Acceleration is the change in velocity, therefore; velocity is 0 only when there is no change in acceleration.
If the object is at rest, it does not mean that that there is no acceleration acting upon it.
Consider if i throw a ball up, the ball's velocity moves in the upward direction, then stops at the maximum point, and then falls back down. Through this whole process the acceleration is 9.8. Even though at the maximum point, the velocity IS 0.
If a curve has a positive gradient below the X-axis it could be describing a few things.
If its a displacement graph, it means that the object is returning to the starting position.
If its a velocity graph, that means the object is travelling AWAY from the starting position, but the speed at which its doing this is decreasing
If its an acceleration graph, it means that the acceleration is in the negative direction, as in the object is slowing down, but, the rate at which its slowing down is get lower. (ie, the derivative of the acceleration is positive)
tonyharrison
Member
- Joined
- Mar 17, 2009
- Messages
- 366
- Gender
- Female
- HSC
- 2009
Ooook yeah i think it's making sense now.
Just to confirm: acceleration graphs DO NOT show direction?
Thanks so much!
Just to confirm: acceleration graphs DO NOT show direction?
Thanks so much!