Phys 1 with calc 1The rectangle shown in Figure P357 has sid

Phys 1 with calc

1.The rectangle shown in Figure P3.57 has sides parallel to the x and y axes. The position vectors of two corners are A = 8.00 m at 50.0° and B = 12.0 m at 29.0°.

(a) Find the perimeter of the rectangle.


(b) Find the magnitude and direction of the vector from the origin to the upper right corner of the rectangle.
________ at _______ ° counterclockwise from the +x axis.

Solution

First, we can only assume that the two \"position vectors\" give us opposite corners of the rectangle.

So... we know that one corner is 8.0 meters away from the origin at an angle of 50 degrees.

it\'s x and y coordinates would then be:
x = 8.0 cos 50 = 5.14 m
y = 8.0 sin 50 = 6.12 m
(5.14, 6.12)
now, the other corner is 12.0 meters away at an angle of 29 degrees, so:

x = 12.0 cos 29 = 10.49 m
y = 12.0 sin 29 = 5.81 m
(10.49, 5.81)
So, now we have the location of two of the corners, and we can figure out the other two as well
(5.14, 6.12) (?, ?)
(?,?) (10.49, 5.81)  

Now, since the top and bottom are parallel to the x- axis, we can get the y-values for the other two points:
(5.14, 6.12) (?, 6.12)
(?,5.81) (10.49, 5.81)

Likewise, the sides are parallel to the y-axis, so we can get the x-values:
(5.14, 6.12) (10.49, 6.12)
(5.14,5.81)  (10.49, 5.81)

So, now we can find the length of each side!
10.49 - 5.14 = 5.34 meters
6.12 - 5.81 = 0.31 meters

so, the parameter is 2lw,

perimeter = 2lw = 2*5.34*0.31 = 3.31m

part b

Now, for the b part, you already have the position of the upper right corner (10.49, 6.12). So, compute its distance from the origin:

a^2 + b^2 = c^2

10.49^2 + 6.12^2 = c^2
c = 12.14 m
Now the angle will need an inverse trig function, in this case, any will do. But the inverse tangent is easy to use. Tangent of an angle is the ratio of the opposite side (y value) to the adjacent side (x - value). So we know:

tan (angle) = y-value/x-value
tan (angle) = 6.12 / 10.49

So: angle = tan^-1 (6.12 / 10.49)

angle = 30.25