position vs time graph

[MCATMadness]

can you get any info about acceleration from a position vs time graph...


like for a straight line with constant positive slope (aka velocity)... can you say the a = 0 bc the velocity is constant? is that right?

can anyone elaborate?

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[TwoPaddles]

can you get any info about acceleration from a position vs time graph...


like for a straight line with constant positive slope (aka velocity)... can you say the a = 0 bc the velocity is constant? is that right?

can anyone elaborate?

This might help:
http://www.physicsclassroom.com/class/1dkin/u1l3a.cfm

 

[MCATMadness]

thnks! that website helped a lot!

one follow up question.... if you have a constant velocity (straight line slope) --> does that mean acceleration is 0 and constant? so if the slope/velocity is 3 m/s, then acceleration is 0 correct?

but if you have a velocity (say 40 m/s but traveling in a circle and constantly changing direction) --> then acceleration is not 0 and not constant....is this right?

any input? i get these ideas mixed up alot. thanks for the help!

 

[TwoPaddles]

thnks! that website helped a lot!

one follow up question.... if you have a constant velocity (straight line slope) --> does that mean acceleration is 0 and constant? so if the slope/velocity is 3 m/s, then acceleration is 0 correct?

but if you have a velocity (say 40 m/s but traveling in a circle and constantly changing direction) --> then acceleration is not 0 and not constant....is this right?

any input? i get these ideas mixed up alot. thanks for the help!

Well to have an acceleration you have to have changing direction. Be careful with the circle motion though. Because often times the circle questions involve a particle starting at one position and finishing where it started, meaning displacement = 0 --> which makes --> velocity = 0
Not sure about the first question because I always get them mixed up too.

 

[Pisiform]

thnks! that website helped a lot!

one follow up question.... if you have a constant velocity (straight line slope) --> does that mean acceleration is 0 and constant? so if the slope/velocity is 3 m/s, then acceleration is 0 correct?

any input? i get these ideas mixed up alot. thanks for the help!

Constant Velocity = Constant Zero Acceleration (line on x-axis of accelration time graph)

Slope Velocity (3m/s) = Constant Acceleration (but not zero) a horizontal line on the acceleration time graph anywhere but not on the x-axis

 

[MCATMadness]

im still kinda confused... i think its bc of my crappy wording on my previous question... let me try again

1. displacement vs time graph: you have a straight line increasing to the right --> slope is say, 3 m/s --> therefore velocity is a constant 3 m/s --> correct?

2. same graph: bc velocity is a constant 3 m/s --> acceleration must be 0--> correct?

acceleration = change in velocity/time
- if the velocity is a constant 3 m/s, this would yield a constant 0 acceleration along that line--> correct?
- when you say velocity is constant 3 m/s --> this means at every time point, the slope is the same number (i.e., 3 m/s) and therefore
a = 3-3/time = 0 m/s2

correct?


3. displacement vs. time graph: curved line going to the right --> this would be changing velocity --> and therefore a changing (not constant) acceleration --> correct?

thnx in advance for the help! aany additional clairifcation/help would be awesome

 

[Pisiform]

im still kinda confused... i think its bc of my crappy wording on my previous question... let me try again

1. displacement vs time graph: you have a straight line increasing to the right --> slope is say, 3 m/s --> therefore velocity is a constant 3 m/s --> correct?

YES ... you got it

2. same graph: bc velocity is a constant 3 m/s --> acceleration must be 0--> correct?

Yes .. you got it again

acceleration = change in velocity/time
- if the velocity is a constant 3 m/s, this would yield a constant 0 acceleration along that line--> correct?

Yes

- when you say velocity is constant 3 m/s --> this means at every time point, the slope is the same number (i.e., 3 m/s) and therefore
a = 3-3/time = 0 m/s2

correct?

YES


3. displacement vs. time graph: curved line going to the right --> this would be changing velocity --> and therefore a changing (not constant) acceleration --> correct?

If Disp/time is curved to the right then:

Velocity/time = straight line to the right (line from origin bent to right)
But acceleration would be constant (but not zero). It could be 1m/s2 at any time point or whatsoever. This is because the rate of change of velocity is same e.g.

At t = 1, v = 2 and t = 2, v = 4 -----t=4, v=8 ; t=5, v = 10

Slope = (4-2)/(2-1) = 2/1 = 2 ------ 10-8/5-4 = 2/1 = 2


thnx in advance for the help! aany additional clairifcation/help would be awesome

^^^ I hope you get it

 

[MCATMadness]

following up on your last point...

if the acceleration is non zero, constant number when you have a curved line on a displacement vs time graph --> then why cant we use linear motion equations on this part of the graph?

i thought that a curved line on a d vs. t graph represents a changing velocity --> therefore changing, non zero a which is not constant --> and this is why we cant use the LME to solve for varibles (bc a is not constant)

am i missing something again?

 

[Pisiform]

following up on your last point...

if the acceleration is non zero, constant number when you have a curved line on a displacement vs time graph --> then why cant we use linear motion equations on this part of the graph?

i thought that a curved line on a d vs. t graph represents a changing velocity --> therefore changing, non zero a which is not constant --> and this is why we cant use the LME to solve for varibles (bc a is not constant)

am i missing something again?

As you said, curved line on a d/t graph represent changing velocity BUT the velocity changing at the same rate. If the velocity is changing at the same rate, then its Gradient/slope would be constant throughout. And this gradient/slope represent acceleration

and I didn't get what u said about LME

 

[MCATMadness]

if you have a curved line on a d vs t graph, velocity is changing --> taking multiple tangent lines to the curve to yield slope (velocity) results in different velocities thru out the curved line

thus wouldnt acceleration be changing as well? if velocity is changing, acceleration must be changing, correct?

and if acceleration is changing --> we can not use the LME eqns (LME eqns are only used when acceleration is constant).

is this correct?

how could you have changing velocities (curved line on d vs t graph) but constant acceleration? you couldnt right?

 

[Pisiform]

if you have a curved line on a d vs t graph, velocity is changing --> taking multiple tangent lines to the curve to yield slope (velocity) results in different velocities thru out the curved line

NO it would not result in different velocity. Same rate means same velocity covered in same number of time. That's why it is known as Constant rate of change.

thus wouldnt acceleration be changing as well? if velocity is changing, acceleration must be changing, correct?

Acceleration would not change. Because the gradient of a straight line from a origin is always constant.

and if acceleration is changing --> we can not use the LME eqns (LME eqns are only used when acceleration is constant).

^ We can use LME because acceleration is constant (though not zero constant)

is this correct?

how could you have changing velocities (curved line on d vs t graph) but constant acceleration? you couldnt right?

You are a hard cookie hun !

I just FAILED

 

[MCATMadness]

can someone else weigh in?

if you have ek physics ....pages 8-9 explain that you can not use LME on the curved line of the d vs t graph bc acceleration is not constant... so ????

 

[Pisiform]

http://www.ngsir.netfirms.com/englishhtm/Kinematics.htm

This should help !