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Graphs and gradients

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Hello IBS, i was wondering about two things:

What does the slope of a distance-time squared graph represent?

Is this the actual acceleration or do i have to multiply it?

And is it bad that my intercept with the y-axis is 8? :S

Using:

S = (at2)/2

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d/t2=v/t because v=d/t

So yes, your slope is your acceleration.

Y intercept just means that when you started timing, or whoever starting timing they were already at a displacement of 8m or whatever unit.

That point would be be 0,8

The slope of this graph (doing a derivative, assuming t2 is your x value so I'm just going to change it to x...)

f(x)=ax/2

[(2a)-ax(0)]/22

2a/4

0.5a=slope.

I'm hoping I didn't fail this part.

I don't know :(

I've got slope equaling a and 0.5a! This isn't working this late at night, I'm sorry if this further confuses you :P

I'm going to say yes simply because a slope is y/x which is d/t2 which is v/t which is a...

Edited by Drake Glau

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d/t2=v/t because v=d/t

So yes, your slope is your acceleration.

Y intercept just means that when you started timing, or whoever starting timing they were already at a displacement of 8m or whatever unit.

That point would be be 0,8

The slope of this graph (doing a derivative, assuming t2 is your x value so I'm just going to change it to x...)

f(x)=ax/2

[(2a)-ax(0)]/22

2a/4

0.5a=slope.

I'm hoping I didn't fail this part.

I don't know :(

I've got slope equaling a and 0.5a! This isn't working this late at night, I'm sorry if this further confuses you :P

I'm going to say yes simply because a slope is y/x which is d/t2 which is v/t which is a...

Gotcha, but when we get the slope, do we have to multiply/divide by 2?

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