If Vx= 8.00 units and Vy= -6.40 unis, determine (a)the magnitude and (b)direction of ->V(vector)
b) theta= ______ degrees below the positive x-axis.
I suppose you are the original poster, and not someone else. So I will give the answer on that basis. It would actually help if you use the same screen name all the time like most other students.
The magnitude of a vector is adding two orthogonal components (which are perpendicular to each other) to give the scalar value, much the same way as finding the hypotenuse of a right triangle. In fact, the formula is an application of Pythagoras theorem.
Magnitude
= √(vx²+vy²)
=√(8²+(-6.4)²)
=√(64+40.96)
=√(104.98)
=10.24 units
The angle can be visualized if you draw a graph and locate the point (8,-6.4) on the Cartesian plane. It will be below the x-axis, at 8 units to the right of the y-axis. The (negative) angle with the x-axis is the angle sought.
We look for tan(θ)=-6.4/8=0.8
We can find &theta using a calculator or tables to get -38.66°.
Hope this clarifies the answer.
Magnitude = √(vx²+vy²)
theta = tan-1(vy/vx)
You can post your answer for a check if you wish.
To determine the magnitude and direction of vector ->V, we can use the given components Vx and Vy.
(a) Magnitude of vector ->V:
The magnitude of a vector is given by the formula:
|V| = sqrt(Vx^2 + Vy^2)
In this case, Vx = 8.00 units and Vy = -6.40 units. Substituting these values into the formula, we have:
|V| = sqrt((8.00)^2 + (-6.40)^2)
= sqrt(64.00 + 40.96)
= sqrt(104.96)
≈ 10.24 units
Therefore, the magnitude of vector ->V is approximately 10.24 units.
(b) Direction of vector ->V:
The direction of a vector can be determined using trigonometry. We can use the components Vx and Vy to find the angle theta.
In this case, Vy is negative (-6.40 units), which means the vector points towards the negative y-axis. To find the angle theta below the positive x-axis, we need to find the reference angle, and then subtract it from 180 degrees.
The reference angle can be found using the formula:
tan(theta) = Vy/Vx
Substituting the given values, we have:
tan(theta) = (-6.40) / (8.00)
theta = arctan(-6.40/8.00)
Using a calculator, we find that:
theta ≈ -38.66 degrees
Since the angle is below the positive x-axis, we need to subtract the reference angle from 180 degrees:
b) theta ≈ 180 degrees - 38.66 degrees
≈ 141.34 degrees
Therefore, the direction of vector ->V is approximately 141.34 degrees below the positive x-axis.
mag= 10.24
theta= -38.66