The Student Room Group

Displacement time graph

Screenshot_20211005-140910_Drive.jpg
For the fist 2s, the displacement time graph is not a stariaght line.
Since it's not a straight line, the object is accelerating and the velocity is not constant.
If we represent this in a velocity time graph for the first 2s, should we draw a straight line or a bit curved like the one in the displacement time graph?
20211005_142845.jpg
I'm a bit perplexed about this.
This is what I did.
because
Screenshot_20211005-142013_Google.jpg
Screenshot_20211005-143358_Drive.jpg
And in the ms it is a straight line.
(edited 2 years ago)
Reply 1
If the displacement curve is quadratic, the velocity is linear and the acceleration is constant. So the ms looks good.
The dispacement is ~quadratic on [0,2] amd [6,8], so the velocity is linear in each interval
The displacement is linear in [2,6], so the velocity is constant.
(edited 2 years ago)
Original post by mqb2766
If the displacement curve is quadratic, the velocity is linear and the acceleration is constant. So the ms looks good.
The dispacement is ~quadratic on [0,2] amd [6,8], so the velocity is linear in each interval
The displacement is linear in [2,6], so the velocity is constant.

thanks.
i didn't know about the quadratic part which is why i was bewildered.
Reply 3
Original post by Aleksander Krol
thanks.
i didn't know about the quadratic part which is why i was bewildered.

Its just suvat (constant acceleration) so
s = ut + 0.5at^2
displacement is a quadratic function of time.

What you drew was ~ok, but its simpler to assume suvat-type behaviour unless its obviously not. Here the only difference was it appears to be piecewise suvat on 0-2, 2-6, 6-8. Also, note the similar displacement behaviour on 0-2 and 6-8. The velocity should have some similarity as well on those intervals.
(edited 2 years ago)

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