A coyote chases a rabbit basic vector problem

[kirby27]

A coyote chasing a rabbit is moving 8.00 due east at one moment and 8.80 due south 3.80 later. Find the direction of the coyote’s average acceleration during that time (west of due south)?


i found Ax = -2.11 and Ay = -2.32. i used the theta = arctan(Ay/Ax) formula and got 47.7 but that isn't the right answer. am i using the wrong formula or am i in the wrong quadrant?

 

[PeterO]

A coyote chasing a rabbit is moving 8.00 due east at one moment and 8.80 due south 3.80 later. Find the direction of the coyote’s average acceleration during that time (west of due south)?i found Ax = -2.11 and Ay = -2.32. i used the theta = arctan(Ay/Ax) formula and got 47.7 but that isn't the right answer. am i using the wrong formula or am i in the wrong quadrant?

Actually you want arctan(Ax/Ay) since they asked for the angle West of South, not the direction South of West.

EDIT: A diagram would have helped.

 

[kirby27]

can anyone else confirm this is true?

 

[PeterO]

can anyone else confirm this is true?

I mean a diagram of the two vectors would have helped you see which angle you were after.

 

[rwisz]

It appears that you have correctly calculated the components of the coyote's acceleration, but the direction you have obtained is not the final answer. To find the direction of the coyote's average acceleration, we need to use the formula for the magnitude of a vector, which is given by the square root of the sum of the squares of its components. In this case, the magnitude of the acceleration vector is given by sqrt((-2.11)^2 + (-2.32)^2) = 3.12.

Next, we can use the formula for the direction of a vector, which is given by the inverse tangent of the y-component divided by the x-component. This will give us the angle in the coordinate plane, measured counterclockwise from the positive x-axis. In this case, the direction of the acceleration vector is given by arctan(-2.32/-2.11) = 48.7 degrees.

However, since the question asks for the direction of the acceleration vector west of due south, we need to subtract this angle from 180 degrees. This gives us a final direction of 180 - 48.7 = 131.3 degrees west of due south.

In summary, you have correctly calculated the components of the acceleration vector, but you need to use the correct formulas to find the magnitude and direction. Keep in mind that the direction of a vector is always measured counterclockwise from the positive x-axis. I hope this helps clarify the solution for you.