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M1, graphs, why does that not work out?

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1. I'm doing Q 4 in 2011 June M1 paper.

For part c they did:

(5 + V) / 2 * 20 = 400 - 310, which I understand totally fine, but why can't I seem to work out the area under the deceleration and use that instead of (400 - 310)

This is what I mean:

(5 + V) / 2 * 20 = area under graph when decelerating
(5 + V) / 2 * 20 = It's a trapezium, and the area works out to be exactly the same as the left hand side, which works out the be 0, which does not make sense to me. Can someone please help me understand if I want to actually have the area under the graph worked out in the equation instead of (400 - 310), how I can do it?

Question paper: http://qualifications.pearson.com/co...e_20110518.pdf
Mark Scheme: http://qualifications.pearson.com/co...s_20110817.pdf
2. (Original post by Reda2)
I'm doing Q 4 in 2011 June M1 paper.

For part c they did:

(5 + V) / 2 * 20 = 400 - 310, which I understand totally fine, but why can't I seem to work out the area under the deceleration and use that instead of (400 - 310)

This is what I mean:

(5 + V) / 2 * 20 = area under graph when decelerating
(5 + V) / 2 * 20 = It's a trapezium, and the area works out to be exactly the same as the left hand side, which works out the be 0, which does not make sense to me. Can someone please help me understand if I want to actually have the area under the graph worked out in the equation instead of (400 - 310), how I can do it?

Question paper: http://qualifications.pearson.com/co...e_20110518.pdf
Mark Scheme: http://qualifications.pearson.com/co...s_20110817.pdf
Huh? You're just saying that area under graph = area under graph...

You can't do it without using 400-310.
3. I just got it lol, I was just being weird. It's 400 = 310 + trapezium area.
4. Nice.

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