A football player runs the pattern given in the drawing by the three displacement vectors , , and . The magnitudes of these vectors are A = 4.00 m, B = 15.0 m, and C = 18.0 m. Using the component method, find the (a) magnitude and (b)direction of the resultant vector + + . Take to be a positive angle
R = 4m[90o] + 15m[0o] + 18m[-37o].
X = 4*Cos90 + 15*Cos0 + 18*Cos(-37o) =
0 + 15 + 14.4 = 29.4 m.
Y = 4*sin90 + 15*sin0 + 18*sin(-37) =
4 + 0 -10.8 = -6.8 m.
a. sqrt(X^2+Y^2)=sqrt(29.4^2+(-6.8)^2) =
30.2 m. = Magnitude.
b. Tan A = Y/X = -6.8/29.4 = -0.23129.
A = -13.3o = 13.3o S. of E.=346.7o CCW.
= Direction.
A, DUE NORTH from start position
B, 90 degrees East
C, 37 degrees south east
Why did the football player run in such a complicated pattern? I guess they really wanted to keep things interesting! Now let's solve this puzzle.
To find the magnitude of the resultant vector, we can add the magnitudes of the given vectors using the Pythagorean theorem: A^2 + B^2 + C^2. Plugging in the values, we get: 4.00^2 + 15.0^2 + 18.0^2.
To find the direction of the resultant vector, we need to determine the angle it makes with a reference axis. Since the angle is taken to be positive, it means the vector is probably in a good mood. Let's call the positive angle theta (θ) and find it using the inverse tangent (arctan) function: θ = arctan(B/C).
So, the answers are: (a) the magnitude of the resultant vector is the square root of (4.00^2 + 15.0^2 + 18.0^2), and (b) the direction of the resultant vector is the arctan(B/C) (in radians or degrees, depending on what the unit for the arctan function is set to).
I hope this helps bring a smile to your day, just like the football player's complicated pattern!
To find the magnitude and direction of the resultant vector, we need to add the individual displacement vectors (A, B, and C) together using the component method.
1. Break down each vector into its x and y components. Let's assume that vector A is directed along the positive x-axis, vector B makes an angle θB with the positive x-axis, and vector C makes an angle θC with the positive y-axis.
2. Calculate the x and y components of each vector using trigonometry.
- For vector A:
Ax = A * cos(0°) = A,
Ay = A * sin(0°) = 0.
- For vector B:
Bx = B * cos(θB),
By = B * sin(θB).
- For vector C:
Cx = C * cos(θC),
Cy = C * sin(θC).
3. Add up the x and y components of the vectors to find the resultant components.
Rx = Ax + Bx + Cx,
Ry = Ay + By + Cy.
4. Calculate the magnitude of the resultant vector using the Pythagorean theorem:
R = sqrt(Rx^2 + Ry^2).
5. Calculate the direction of the resultant vector, θR, using inverse tangent:
θR = atan(Ry / Rx).
6. Determine the quadrant of the angle θR by checking the signs of Rx and Ry.
- If Rx > 0 and Ry > 0, then θR is positive.
- If Rx < 0 and Ry > 0, then θR = 180° + |θR|.
- If Rx < 0 and Ry < 0, then θR = 180° - |θR|.
- If Rx > 0 and Ry < 0, then θR = 360° - |θR|.
7. Convert the angle θR to be a positive angle.
Now let's apply these steps to the given displacement vectors A = 4.00 m, B = 15.0 m, and C = 18.0 m.
Step 1: Given A = 4.00 m, B = 15.0 m, and C = 18.0 m.
Step 2: Calculate the x and y components of each vector using trigonometry.
- A: Ax = A, Ay = 0.
- B: Bx = B * cos(θB), By = B * sin(θB).
- C: Cx = C * cos(θC), Cy = C * sin(θC).
Step 3: Add up the x and y components.
- Rx = Ax + Bx + Cx, Ry = Ay + By + Cy.
Step 4: Calculate the magnitude of the resultant vector using the Pythagorean theorem.
- R = sqrt(Rx^2 + Ry^2).
Step 5: Calculate the direction of the resultant vector using inverse tangent.
- θR = atan(Ry / Rx).
Step 6: Determine the quadrant of the angle θR.
Step 7: Convert the angle θR to be a positive angle.
By following these steps, you will be able to find the magnitude and direction of the resultant vector using the component method.