Rewrite the following vectors in terms of their magnitude and angle (counterclockwise from the +x direction).
1) A displacement vector with an x component of +9.5 m and a y component of -9.6 m.
magnitude:
2) angle:
3) A velocity vector with an x component of -96 m/s and a y component of +46 m/s.
magnitude:
4) angle:
5) A force vector with a magnitude of 58 lb that is in the third quadrant with an x component whose magnitude is 37 lb.
y component:
6) angle:
1. Disp. = X - yi -= 9.5 - 9.6i =
Sqrt(9.5^2+9.6^2) =
2. Tan A = (-9.6/9.5 = -1.01053.
A = -45.3o,Q4 = 45o S. of E. = 315o CCW.
3. V = -96 + 46i = Sqrt(96^2+46^2) =
4. Tan A = 46/(-96) = -0.47917.
A = -25.6o, Q2 = 25.6o N. of W. = 154.4o CCW.
1) To find the magnitude of the displacement vector, we can use the Pythagorean theorem:
magnitude = √(x^2 + y^2)
= √((9.5 m)^2 + (-9.6 m)^2)
= √(90.25 m^2 + 92.16 m^2)
= √(182.41 m^2)
≈ 13.51 m
2) To find the angle counterclockwise from the +x direction, we can use the inverse tangent function:
angle = tan^(-1)(y / x)
= tan^(-1)(-9.6 m / 9.5 m)
≈ -45.43 degrees
3) To find the magnitude of the velocity vector, we can use the Pythagorean theorem again:
magnitude = √(x^2 + y^2)
= √((-96 m/s)^2 + (46 m/s)^2)
= √(9216 m^2/s^2 + 2116 m^2/s^2)
= √(11332 m^2/s^2)
≈ 106.43 m/s
4) To find the angle counterclockwise from the +x direction, we use the inverse tangent function:
angle = tan^(-1)(y / x)
= tan^(-1)(46 m/s / (-96 m/s))
≈ -25.67 degrees
5) The y component can be found using the Pythagorean theorem and the angle:
y = magnitude * sin(angle)
= 58 lb * sin(180 - tan^(-1)(37 lb / 58 lb))
≈ -20.91 lb
6) The angle can be found using the inverse tangent function:
angle = 180 - tan^(-1)(y / x)
= 180 - tan^(-1)(-20.91 lb / 37 lb)
≈ 153.27 degrees
To rewrite the vectors in terms of their magnitude and angle, we can use the properties of right triangles and trigonometry. The magnitude of a vector can be found using the Pythagorean theorem, and the angle can be determined using inverse trigonometric functions.
Let's go through each question step by step.
1) A displacement vector with an x component of +9.5 m and a y component of -9.6 m.
Magnitude: We can find the magnitude using the Pythagorean theorem:
magnitude = sqrt(x^2 + y^2)
magnitude = sqrt((9.5 m)^2 + (-9.6 m)^2)
magnitude = sqrt(90.25 m^2 + 92.16 m^2)
magnitude = sqrt(182.41 m^2)
magnitude ≈ 13.51 m
2) Angle: To find the angle counterclockwise from the +x direction, we can use the inverse tangent function (arctan).
angle = arctan(y / x)
angle = arctan((-9.6 m) / (9.5 m))
angle ≈ -45.18 degrees
3) A velocity vector with an x component of -96 m/s and a y component of +46 m/s.
Magnitude: Using the Pythagorean theorem, we can calculate the magnitude:
magnitude = sqrt((-96 m/s)^2 + (46 m/s)^2)
magnitude = sqrt(9216 m^2/s^2 + 2116 m^2/s^2)
magnitude = sqrt(11332 m^2/s^2)
magnitude ≈ 106.44 m/s
4) Angle: We can find the angle using the inverse tangent function (arctan).
angle = arctan(y / x)
angle = arctan((46 m/s) / (-96 m/s))
angle ≈ -25.39 degrees
5) A force vector with a magnitude of 58 lb that is in the third quadrant with an x component whose magnitude is 37 lb.
Y component: Since the vector is in the third quadrant, the y component will be negative.
y component = - magnitude of the force = -58 lb
6) Angle: To find the angle counterclockwise from the +x direction, we can use the inverse cosine function (arccos).
angle = arccos(x / magnitude)
angle = arccos((37 lb) / (58 lb))
angle ≈ 50.09 degrees
5. X^2 + Y^2 = 58^2.
37^2 + Y^2 = 58^2. Y = ?.
6. Q3, -37 - Yi.
Calculate the angle.