"You are given a vector in the xy plane that has a magnitude of 95.0
units and a y component of -55.0 units.
(a) What are the two possibilities for its x component?"
-55.0 units, eh? That means the y-component is pointed downward from
the horizontal or x-axis.
Two possible x-components? So the 95.0-unit vector is pointed downward
from the origin (0,0). That means the tip of the 95.0-unit vector is
either at the 3rd quadrant or at the fourth quadrant.
At the 3rd quadrant, the x-component is negative. At the 4th quadrant
it is positive.
Given:
Vector magnitude = 95.0 units
y-component = -55.0 units
So,
x-component = +,-sqrt[(95.0)^2 -(-55.0)^2] = +,-77.46
Therefore, the x-component is either +77.46 units or -77.46 units. ---answer.
-----------------
"(b) Assuming the x component is known to be positive, specify the
vector which, if you add it to the original one would give a resultant
vector that is 80.0 units long and points entirely in the -x
direction.
Magnitude=.....
Direction=.....° "
So x-component of the 95.0-unit vector is +77.46 units.
The resultant is
>>>80.0 units in magnitude
>>>"pointing entirely to the -x direction", meaning, the resultant is horizontal.
That means further then that resultant has an x-component of -80.0
units and zero y-component.
The unknown vector has
>>>u = x-component
>>>v = y-component
Resultant x-commponent = Sum of x-components of the 95.0-unit vector
and that of the unknown vector.
-80.0 = +77.46 +u
-80.0 -77.46 = u
u = -157.46 units ----***
Resultant y-commponent = Sum of y-components of the 95.0-unit vector
and that of the unknown vector.
0 = -55.0 +v
0 +55.0 = v
v = +55.0 units ----***
So,
Magnitude of unknown vector = sqrt[(-157.46)^2 +(55.0)^2]
= 166.8 units. --------answer.
Direction = arctan(55.0 / -157.46) = arctan(-0.3493) = -19.25 degrees
Or, since the "unknown" vector is in the 2nd quadrant,
direction = 180 -19.25 = 160.75 degrees ------------answer.
(Or, direction is N 70.75 deg W) |
