Skill Set 03.1

From TuHSPhysicsWiki
Revision as of 11:14, 12 February 2009 by PHerstedt (talk | contribs) (4. Question)
Jump to: navigation, search

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 17o in our equations.

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

Table of Contents


2. Find the components of the vector.

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

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

Table of Contents


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) = 23.4o

To get the magnitude:

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

14.4 m @ 23.4o

Table of Contents


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


Table of Contents


5. Question

Solution goes here

Table of Contents