Difference between revisions of "Skill Set 03.1"

Revision as of 10:14, 12 February 2009

Main Page > IB Physics Skill Sets > Skill Set 03.1

1. Find the components of the vector.

Since the angle given is the same as the trig angle, we can use 17^o in our equations.

32cos(17) = 30.60 m x
32sin(17) = 9.356 m y

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2. Find the components of the vector.

Converting the 23^o angle to a trig angle is done by subtracting it by 360, since the angle is just below the 0 or 360^o line. Using our new angle, 337^o, we can solve for the components.

15cos(337) = 13.81 m x
15sin(337) = -5.861 m y

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3. Convert 13.2 m/s x + 5.70 m/s y into an angle magnitude vector.

To get the angle:

tan^-1(5.7/13.2) = 23.4^o

To get the magnitude:

a² + b² = c²
5.7² + 13.2 ² = c²
c = 14.4 m

14.4 m @ 23.4^o

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4. Add and subtract the vector components.

All that's necessary to remember when doing these is to do the x and y components separately. For D+E:

16.2 + -13.7 = 2.5 m x
-3.5 + -4.2 = -7.7 m y

For E-D:

-13.7 - 16.2 = -29.9 m x
-4.2 - -3.5 = -.7 m y

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5. Question

Solution goes here

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