- #1

- 13

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I've tried finding the area under the lines but that doesn't seem to be it...

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- Thread starter slag1928
- Start date

- #1

- 13

- 0

I've tried finding the area under the lines but that doesn't seem to be it...

- #2

- 5

- 0

I've tried finding the area under the lines but that doesn't seem to be it...

Velocity=distance/ time. multiply by the time stamp at whatever point you are on on the graph.

- #3

- 50

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Using that, you should be able to figure out the position. (At least, that's how I do it.)

- #4

- 16

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What did you do when you tried finding the area under the graph?

- #5

- 13

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completing each time increment as a triangle and using 1/2b*h.

- #6

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completing each time increment as a triangle and using 1/2b*h.

Could you post a picture of the graph?

- #7

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maybe, lemme try

- #8

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well uh no actually. idk how D: could i get an e-mail? i think i could e-mail it

- #9

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- #10

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it's #7 i need help with!

- #11

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The blue graph right?

What ares did you use 1/2b * h on (what time)?

What ares did you use 1/2b * h on (what time)?

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- #12

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- #13

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What did you find for displacement?

- #14

- 13

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i got -54

- #15

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Okay, and for what parts did you use 1/2b*h?

- #16

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time interval 3-4 seconds have a constant velocity of -18 m/s so i added -18 to the first -18 getting -36. then i found the are from 4-5 by multiplying 1/2 the base of 1 second, by the height of -9, getting an additonal -4.5. Sum of all these? -54!

- #17

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i meant times the height! sorry i'm a confusing typer !

- #18

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please ignore the first time i said "this got -36" its a typo...

- #19

- 16

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time interval 3-4 seconds have a constant velocity of -18 m/s so i added -18 to the first -18 getting -36. then i found the are from 4-5 by multiplying 1/2 the base of 1 second, by the height of -9, getting an additonal -4.5. Sum of all these? -54!

You found the area for time 4-5s wrong.

You found the red part, but not the blue part. The blue part is a rectangle.

Also, I think for this problem you can assume displacement and distance are meant to be the same thing.

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