You are given a vector in the xy plane that has a magnitude of 90.0 units and, a y component of -41.0 units. Assuming the x component is known to be positive, specify the vector, V, if you add it to the original one, would give a resultant vector that is 89.0 units long and points entirely in the -x direction.

In your question where as the given vector in the xy plane that has a magnitude of 90 units and a,y component of -41 units. So the vector V, base on my calculation and understanding in the problem, the value if it is (8.88, 41)

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aristocles aristocles · Answer 2 · 2019-09-09T20:11:44-04:00

Answer:

\(\vec A = 80.11 \hat i - 41\hat j\)

\(\vec V = -169.1\hat i + 41\hat j\)

Explanation:

Magnitude of the vector is 90 units

Y component of the vector is -41 units

Now we know that

\(\vec A = x\hat i + y\hat j\)

now the magnitude of vector A is given as

\(A = \sqrt{x^2 + y^2}\)

\(90 = \sqrt{x^2 + 41^2}\)

\(90^2 = x^2 + 41^2\)

\(x = 80.11 units\)

so the vector is given as

\(\vec A = 80.11 \hat i - 41\hat j\)

now Another vector V is added in it such that the resultant is 89 units and along - X direction

so we have

\(\vec R = \vec A + \vec V\)

\(-89\hat i = (80.11\hat i - 41\hat j) + \vec V\)

\(-89\hat i - 80.11 \hat i + 41 \hat j = \vec V\)

\(\vec V = -169.1\hat i + 41\hat j\)