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# Mechanics 1 - Kinematic Graphs watch

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1. This image shows the question. I cannot think of any way to work out ii) without working out iii) first. Surely it must be possible though, otherwise it wouldn't be written in this order.

EDIT: I assume the time mentioned in ii) would be the total time of the graph?
2. So you know that the time for the first 3 sections of the graph is 60s, so you need to look at the last 3.
You know the flat 5th section is 4s and the area under the graph is the same as that calculated in part (i), and you can find the area under part 5.
3*4=12
Subtract the area under part 5 from the area under parts 4, 5 and 6 (from part (i) which I get as 30m?) to give the area under parts 4 and 6
30-12=18
Act like part 5 isn't between them and treat parts 4 and 6 like a triangle of area 18
Height = 3, 0.5*b*h=18
0.5*b*3=18
b=18/(0.5*3)
b=12s
Add that to the time from sections 1, 2 and 3 (60s) and the time from part 5 (4s)
12+60+4=76s
So it's 76 seconds
3. (Original post by wannabeme)
So you know that the time for the first 3 sections of the graph is 60s, so you need to look at the last 3.
You know the flat 5th section is 4s and the area under the graph is the same as that calculated in part (i), and you can find the area under part 5.
3*4=12
Subtract the area under part 5 from the area under parts 4, 5 and 6 (from part (i) which I get as 30m?) to give the area under parts 4 and 6
30-12=18
Act like part 5 isn't between them and treat parts 4 and 6 like a triangle of area 18
Height = 3, 0.5*b*h=18
0.5*b*3=18
b=18/(0.5*3)
b=12s
Add that to the time from sections 1, 2 and 3 (60s) and the time from part 5 (4s)
12+60+4=76s
So it's 76 seconds
Thank you so much! I got the area under 4,5 and 6 as 30 too, I then took 12 from it, but from here is when I started using bits from stage iii). I got 75.125 seconds overall, but this will be because I did it in the wrong order.

Thanks! (You can have a rep)

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Updated: September 25, 2011
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