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A jogger travels a route that has two parts. the first is a displacement upper a overscript right-arrow endscripts of 2.50 km due south, and the second involves a displacement upper b overscript right-arrow endscripts that points due east. the resultant displacement upper a overscript right-arrow endscripts + upper b overscript right-arrow endscripts has a magnitude of 4.00 km. (a) what is the magnitude of upper b overscript right-arrow endscripts, and (b) what is the direction of upper a overscript right-arrow endscripts + upper b overscript right-arrow endscripts as a positive angle relative to due south? suppose that upper a overscript right-arrow endscripts - upper b overscript right-arrow endscripts had a magnitude of 4.00 km. (c) what then would be the magnitude of upper b overscript right-arrow endscripts, and (d) what is the direction of upper a overscript right-arrow endscripts - upper b overscript right-arrow endscripts relative to due south?

Respuesta :

Define
[tex]\hat{i}[/tex] = unit vector due east.
[tex]\hat{j}[/tex] =  unit vector due north.

Therefore
[tex]\vec{A} = -2.5 \hat{j} \, km[/tex]
[tex]\vec{B} = B \hat{i} \, km[/tex]

Part (a)
[tex]\vec{A}+\vec{B} = B \hat{i} -2.5\hat{j}[/tex]
Because [tex]|vec{A}+\vec{B}| = 4 \, km[/tex], therefore
B² + (-2.5)² = 4²
B² = 9.75
B = 3.1225 km

Answer: [tex]|\vec{B}| = 3.1225 \, km[/tex]

Part (b)
As a positive angle (counterclockwise) measured relative to the southern direction, the direction of [tex]\vec{A}+\vec{B}[/tex] is
[tex]tan^{-1} \frac{3.1225}{2.5} =51.3^{o}[/tex]

Answer: 53°, measured counterclockwise from the south.

Part (c)
If [tex]|\vec{A}-\vec{B}| = 4\, km[/tex], then
(-B)²+(-2.5)² = 4²
B² = 16 - 6.25
B = 3.1225 km

Answer: [tex]|\vec{B}| = 3.1225 \, km[/tex]

Part (d)
The direction of [tex]\vec{A}-\vec{B}[/tex] relative to the southern direction is
tan⁻¹ (-3.1225/-2.5) = 51.3°, measured clockwise from south.

Answer: 51.3° measured clockwise from the south.

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