Find the magnitude and direction of the net displacement of the player

[matt@USA]

Homework Statement

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 10, 20 and 7 m . Assume the positive x-axis is directed to the right.

There is a 45degree angle outside of where A and B meet, and there is a 30degree angle outside of where B and C meet.

I have found what Dx and Dy are. Dx and Dy are 20m, and 0m. In order to find the degree of the equation you would have to take the tan^-1(Dy/Dx). My question is, how do you solve this when Dy = 0? The answer is not 0 degrees, I know this for a fact. Thanks for all the help!

 

[Rayquesto]

if dy is zero, then the resultant angle should also be zero.

You must have added the components incorrectly. It's hard for me to imagine your details. Basically, I think youre trying to say that the angle vector A makes with the horizontal is 45 degrees, the angle vector B makes with the horizontal is 30 degrees, but what about the angle that vector c makes with the horizontal? what's that?

 

[matt@USA]

No, I know my answers are correct because I had to solve for the absolute value of vectors A, B, and C, and they were correct. Imagine vector A going south. Then vector B heads off to the North East. Then vector C heads to the south east.

 

[Rayquesto]

so dy= -10 + .707(20) + -.5(7)=.64
and dx=.707(20) + (sqrt3)/2 times 7=20.2

resultant theta=tan-1(.64/20.2)= 1.81degrees.

the only thing I'm scepticle about is the who idea of the 30 degrees south of east you gave. It could actually mean 60 degrees above horizontal, but I can't see the problem in your book or the source you got it from.

 

[rwisz]

I would approach this problem by first understanding the concept of net displacement. Net displacement is the overall change in position or location of an object, taking into account both magnitude and direction. In this case, the basketball player is moving in a specific path indicated by the vectors A, B, and C.

To find the net displacement, we can use vector addition to add the individual displacements of A, B, and C. This can be visualized as drawing a vector from the starting point to the end point of each vector and then connecting all three vectors together. The resulting vector is the net displacement.

The magnitude of the net displacement can be found using the Pythagorean theorem, which states that the square of the hypotenuse of a right triangle is equal to the sum of the squares of the other two sides. In this case, the magnitude of the net displacement would be the square root of (Dx^2 + Dy^2).

To find the direction of the net displacement, we can use trigonometric functions such as tangent (tan). However, as mentioned in the problem, when Dy = 0, we cannot use the formula tan^-1(Dy/Dx) as it would result in an undefined value. In this case, we can use the formula tan^-1(Dx/0) which would give us a direction of 90 degrees or due east.

In summary, to find the magnitude and direction of the net displacement of the basketball player, we would use vector addition to add the individual displacements of A, B, and C. The magnitude can be found using the Pythagorean theorem, and the direction can be found using trigonometric functions. In the special case where Dy = 0, we can use the formula tan^-1(Dx/0) to find the direction.