[Kinematics] Faster way?

Mar 2010
71
2

I know how to do it, but it's ridiculously long.
How do I do this the fast way, assuming there is one.
I found the area of all and removed the negative areas to get the distance.

\(\displaystyle \frac{\text{distance}}{\text{time}}\)

And correctly got A.
 

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MHF Hall of Honor
Mar 2010
2,340
821
Chicago
I know how to do it, but it's ridiculously long.
How do I do this the fast way, assuming there is one.
I found the area of all and removed the negative areas to get the distance.

\(\displaystyle \frac{\text{distance}}{\text{time}}\)

And correctly got A.
I don't think there's a faster way with the information you've been given. You just have to be fast with finding areas of triangles and rectangles (and trapezoid/trapezium) and seeing when certain things cancel each other out. :(

Now, if you'd been given the start position (call it a) and end position (call it b), then there would be an easier way. It would be (1/14)(b-a). (By the fundamental theorem of calculus.)

Edit: You could however eliminate B immediately, since there's no way a denominator of 14 could get simplified to 16. You could eliminate E by eyeballing it. But I don't see easy ways to eliminate C and D.
 
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Mar 2010
71
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I don't think there's a faster way with the information you've been given. You just have to be fast with finding areas of triangles and rectangles (and trapezoid/trapezium) and seeing when certain things cancel each other out. :(

Now, if you'd been given the start position (call it a) and end position (call it b), then there would be an easier way. It would be (1/14)(b-a). (By the fundamental theorem of calculus.)

Edit: You could however eliminate B immediately, since there's no way a denominator of 14 could get simplified to 16. You could eliminate E by eyeballing it. But I don't see easy ways to eliminate C and D.
Could you show me how you would do it?
 

undefined

MHF Hall of Honor
Mar 2010
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Could you show me how you would do it?
No problem.

So going from point A to B, the area is 0 because the positive and negative contributions are equal.

Then from B to C it's -18.

Then from C to the place where it crosses the x-axis, it's -6, making overall total of -24. Then from that point to D it's 24, making overall total 0.

Then from D to E it's 2 * (average of 8 and 12) = 20. So the answer is 20/14 = 10/7.
 
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Mar 2010
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Thanks, that's really helpful. I wasn't aware of the first part being that they would cancel out.