# 03.1

## Problem One

Find the components of this vector: A=_________x + _________y

To solve this problem you use: $A_x\ = (mag)cos\Theta$ $A_y\ = (mag)sin\Theta$

To solve: $A_x\ = (32)cos17$ $A_x\ = 30.60 m\hat{x}$ $A_y\ = (32)sin17$ $A_y\ = 9.35 m\hat{y}$ $A\ = 30.60 m\hat{x}\ + 9.356 m\hat{y}$

## Problem Two

Find the components of this vector: B=_________x + _________y

Again: $A_x\ = (mag)cos\Theta$ $A_y\ = (mag)sin\Theta$

So, we do the same thing as before. This time, don't forget that $\Theta\,$ must always be a trig angle! $\; 270^\circ\; + 23^\circ = 293^\circ$ $A_x\ = (15)cos293$ $A_x\ = 5.861 m\hat{x}$ $A_y\ = (15)sin293$ $A_y\ = -13.81 m\hat{y}$ $B\ = 5.861 m\hat{x}\ + -13.81 m\hat{y}$

## Problem Three

Convert this vector component vector to an angle magnitude vector. Draw it as an arrow, and label any angle it makes with its value, and find the magnitude.

C = 13.2 m/s x + 5.70 m/s y

We solve this problem like the first two, only in reverse. The first thing that you do is draw out the equation without lifting your pen. It should look like this: Using SOHCAHTOA, we can take the inverse tangent (the arctan) of these values to find the angle, and using the pythagorean theorem we can take the square root of x squared plus y squared to get the magnitude. $Mag = \sqrt{(x^2 + y^2)}$ $Mag = \sqrt{(5.70^2 + 13.2^2)}$

Mag = 14.4 m/s $Angle = \arctan\,{(\frac{O}{A})}$ $Angle = \arctan\,{(\frac{5.70}{13.2})}$
Angle = 23.4o

Angle Magnitude: 14.4 m/s @ 23.4o

## Problem Four

Given: $D\ = 16.2 m\hat{x}\ + -3.5 m\hat{y}$ $A\ = -13.7 m\hat{x}\ + -4.2 m\hat{y}$

Find:
D+E=_________x + _________y
E-D=_________x + _________y

To solve this problem, you add and subtract the vector components like basic arithmetic.

### Finding D+E

With this problem, you simply add the corresponding values. $+16.2 m\hat{x}\ + -3.5 m\hat{y}$ $-13.7 m\hat{x}\ + -4.2 m\hat{y}$
+___________________________ $+2.5 m\hat{x}\ + -7.7 m\hat{y}$ $D+E\ = 2.5 m\hat{x}\ + -7.7 m\hat{y}$

### Finding E-D

As with the previous problem, you subtract the D values from the E values to find the answer. $-13.7 m\hat{x}\ + -4.2 m\hat{y}$ $+16.2 m\hat{x}\ + -3.5 m\hat{y}$
-___________________________ $-29.9 m\hat{x}\ + -.7 m\hat{y}$ $E-D\ = -29.9 m\hat{x}\ + -.7 m\hat{y}$

## Problem Five

Calculate the sum of vectors A and B (From problem 1 and 2). Draw the resultant vector as an angle magnitude vector. (Label your angle and magnitude clearly – use an arrow) Do your work on the back of this sheet, and label and describe the three steps for doing so.

### Sum of vectors

Calculate the sum of the two vectors $30.6 m\hat{x}\ + 9.356 m\hat{y}$ $5.861 m\hat{x}\ + -13.81 m\hat{y}$
+___________________________ $36.461 m\hat{x}\ + -4.454 m\hat{y}$

### Draw the vector

Label the angle, sides, and vector. ### Calculate the Angle Magnitude

Do the inverse tangent and square root of the two distances squared. Write these values as angle magnitude. $Mag = \sqrt{(x^2 + y^2)}$ $Mag = \sqrt{(4.46^2 + 36.461^2)}$

Mag = 36.8 m/s $Angle = \arctan\,{(\frac{O}{A})}$ $Angle = \arctan\,{(\frac{4.46}{36.461})}$
Angle = 6.97o

Angle Magnitude: 36.8 m/s @ 6.97o