Basic question re Motion (1 Viewer)

[akkatracker]

Something really basic here (I must be missing the obvious - first day back at school)

I've had to graph two graphs...
One velocity vs time and the other displacement vs time squared.
They are both straight line plots.
Assuming this should the gradients be the same (they were based off the same data) and shouldn't they both result in the acceleration?
The velocity vs time has a gradient of 1.8ms^2 and the displacement vs time has a gradient of 0.9ms^2
Another thing that's throwing me off is that if you rearrange the equation r = ut + 1/2 *at^2 you end up with a/2 = r/t^2
If this is the case the gradient of the displacement vs time graph wouldn't be the acceleration (however meters divided by time squared = ms^2)
Any help would be appreciated.

 

[InteGrand]

Something really basic here (I must be missing the obvious - first day back at school)

I've had to graph two graphs...
One velocity vs time and the other displacement vs time squared.
They are both straight line plots.
Assuming this should the gradients be the same (they were based off the same data) and shouldn't they both result in the acceleration?
The velocity vs time has a gradient of 1.8ms^2 and the displacement vs time has a gradient of 0.9ms^2
Another thing that's throwing me off is that if you rearrange the equation r = ut + 1/2 *at^2 you end up with a/2 = r/t^2
If this is the case the gradient of the displacement vs time graph wouldn't be the acceleration (however meters divided by time squared = ms^2)
Any help would be appreciated.

If (no idea why the HSC Physics syllabus uses r for displacement), then the gradient of the displacement vs. time-squared is (initial velocity must have been 0 for this graph to be a straight line, as the acceleration is constant (the acceleration is constant, since the v-t graph is a straight line)).

Velocity time is (assuming initial velocity 0), so the gradient is a. So the graphs have different gradients, differing by a factor of 2.

 

[akkatracker]

Thanks, that was what I suspected.

 

[akkatracker]

If (no idea why the HSC Physics syllabus uses r for displacement), then the gradient of the displacement vs. time-squared is (initial velocity must have been 0 for this graph to be a straight line, as the acceleration is constant (the acceleration is constant, since the v-t graph is a straight line)).

Velocity time is (assuming initial velocity 0), so the gradient is a. So the graphs have different gradients, differing by a factor of 2.

Actually one last thing:

You said the gradient of r vs time squared = 1/2 (a)

How come s/t^2 is equal to acceleration then?

 

[InteGrand]

How come s/t^2 is equal to acceleration then?

It's not. It's equal to . (So it still has units of .)