A basketball player runs down the court, following the path indicated by the vectors A, B, and C in the figure. The magnitudes of these three vectors are: A = 12.0 m, B = 17.0 m, and C = 7.0 m. Let the +x-axis point to the right and the +y-axis point to the far side of the court.
Find the magnitude and direction of the net displacement of the player using the graphical method of vector addition.
Repeat the previous step, but using the component method. Compare your results with those obtained in part (a).
magnitude m
direction ° (counterclockwise from the +x axis)
i have been working for 6 hours on these things and cant figure out how to do them
without the figure, I can help. a) add the vectors graphically head to tail.
b) break up each vector into x and y components...then add them combining x's and y's.
The magnitude is 18.6067
To find the net displacement of the basketball player using the graphical method of vector addition, we need to draw a vector diagram.
1. Take a piece of graph paper and draw the x and y axes. Label the positive direction of the x-axis as right and the positive direction of the y-axis as the far side of the court.
2. Start by drawing the vector A, which has a magnitude of 12.0 m. Place the tail of the vector at the origin (0,0) and draw an arrow pointing in the direction indicated. Label the arrow with the magnitude 12.0 m.
3. Next, draw the vector B, which has a magnitude of 17.0 m. Place the tail of this vector at the endpoint of vector A and draw an arrow pointing in the direction indicated. Label the arrow with the magnitude 17.0 m.
4. Finally, draw the vector C, which has a magnitude of 7.0 m. Place the tail of this vector at the endpoint of vector B and draw an arrow pointing in the direction indicated. Label the arrow with the magnitude 7.0 m.
5. Measure the length of the arrow from the origin to the endpoint of the last vector. This length represents the magnitude of the net displacement of the player. Record this value.
6. Use a protractor to measure the angle that the net displacement vector makes with the positive x-axis. This angle represents the direction of the net displacement in a counterclockwise direction from the positive x-axis. Record this angle.
To find the net displacement of the player using the component method, you will need to break down the vectors A, B, and C into their x and y components and then add the corresponding components together.
1. To find the x and y components of vector A, we can use trigonometry. The angle between vector A and the positive x-axis can be found using the inverse tangent function, tan^(-1)(y-component/x-component). Assuming that vector A is directed at an angle of θA from the positive x-axis, the x-component of vector A would be A * cos(θA) and the y-component would be A * sin(θA).
2. Repeat the same process for vectors B and C.
3. Add up the x-components and y-components separately to find the net x and y components.
4. Use the Pythagorean theorem to find the magnitude of the net displacement vector: sqrt(x-component^2 + y-component^2).
5. Use the inverse tangent function to find the angle that the net displacement vector makes with the positive x-axis: tan^(-1)(y-component/x-component).
Compare the magnitude and direction obtained using the graphical method with those obtained using the component method to verify your results.