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28. A basketball player runs down the court, following the path indicated by the vectors A, B, and C in Figure 3-43. The magnitudes of these three vectors are A=10.0 m, B=20.0 m, and C=7.0 m. Find the magnitude and direction of the net displacement of the player using the graphical method and the component method of vector addition. Compare your results.

28. A basketball player runs down the court, following the path indicated by the vectors-example-1
User Solomon Rutzky
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1 Answer

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Given:

Magnitude of vector A = 10.0 m

Magnitude of vector B = 20.0 m

Magnitude of vector C = 7.0 m

Let's find the magnitude and direction of the net displacement using the component method of vector addition.

The vector of magnitude r which makes an angle θ with the positive x-axis in vector form is:


R=r(\cos \theta i+\sin \theta j)

Vector A:

From the figure, vector A makes an angle of -90° with the x-axis, thus, we have:


\begin{gathered} A=10.0(\cos (-90)i+\sin (-90)j) \\ \\ A=10.0(0i+(-1)j) \\ \\ A=-10.0j \end{gathered}

Vector B:

Vector B makes an angle of 45° with a magnitude of 20.0 m. Thus, we have:


\begin{gathered} B=20.0(\cos 45i+\sin 45i) \\ \\ B=20.0(0.5\sqrt[]{2}i+0.5\sqrt[]{2}j) \\ \\ B=10.0\sqrt[]{2i}+10\sqrt[]{2}j \end{gathered}

Vector C:

Vector C makes an angle of -30° with a magnitude of 7.0 m. Thus, we have:


\begin{gathered} C=7.0(\cos (-30)i+\sin (-30)j) \\ \\ C=3.5\sqrt[]{3}i+3.5j \end{gathered}

To find the net displacement, apply the formula:

Net displacement = A + B + C

Hence, we have the net diplacement below:


\begin{gathered} A+B+C \\ \\ (-10.0j)+(10.0\sqrt[]{2}i+10.0\sqrt[]{2}j)+(3.5\sqrt[]{3}i+3.5j) \\ \\ =10.0\sqrt[]{2}i+3.5\sqrt[]{3}i-10.0j+10.0\sqrt[]{2}j+3.5j \\ \\ =20.202i+0.642j \end{gathered}

To find the magnitude of the net displacement, we have:


\begin{gathered} \text{ magnitude=}\sqrt[]{20.202^2+0.642^2} \\ \\ \text{ magnitude=}\sqrt[]{408.12+0.412} \\ \\ \text{magnitude = }\sqrt[]{408.53} \\ \\ \text{magnitude = }20.212 \end{gathered}

To find the direction, we have:


\begin{gathered} \tan ^(-1)((y)/(x)) \\ \\ =\tan ^1((0.642)/(20.202)) \\ \\ =1.8\degree \end{gathered}

Therefore, the magnitude of the net displacement is 20.21 m while the direction is at 1.8 degrees.

ANSWER:

Magnitude = 20.21 m

Direction = 1.8°

User Harini
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