# [Kinematics] Faster way?

• Jun 1st 2010, 10:36 PM
Cthul
[Kinematics] Faster way?
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.
• Jun 1st 2010, 10:46 PM
undefined
Quote:

Originally Posted by Cthul
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.
• Jun 1st 2010, 11:03 PM
Cthul
Quote:

Originally Posted by undefined
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?
• Jun 1st 2010, 11:14 PM
undefined
Quote:

Originally Posted by Cthul
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.
• Jun 1st 2010, 11:44 PM
Cthul
Thanks, that's really helpful. I wasn't aware of the first part being that they would cancel out.