Question #80589
If V sub x=9.80 units and v sub y= -6.40 units determine the magnitude and direction of vector V
Expert's answer
a) We can find the magnitude of the vector V ⃗ from the formula:
|V ⃗ |=√(V_x^2+V_y^2 )=√(〖9.80〗^2+(-6.40)^2 )=11.7 units.
b) We can find the direction of the vector V ⃗ from the formula:
tanθ=V_y/V_x ,
θ=tan^(-1)〖(V_y/V_x )=tan^(-1)〖((-6.40)/9.80)=〖-33〗^° 〗 〗.
However, we must add 〖360〗^° to obtain the correct answer:
θ=〖-33〗^°+〖360〗^°=〖327〗^°.
The angle between the vector V ⃗ and the positive x-axis is 〖327〗^° (counterclockwise).
|V ⃗ |=√(V_x^2+V_y^2 )=√(〖9.80〗^2+(-6.40)^2 )=11.7 units.
b) We can find the direction of the vector V ⃗ from the formula:
tanθ=V_y/V_x ,
θ=tan^(-1)〖(V_y/V_x )=tan^(-1)〖((-6.40)/9.80)=〖-33〗^° 〗 〗.
However, we must add 〖360〗^° to obtain the correct answer:
θ=〖-33〗^°+〖360〗^°=〖327〗^°.
The angle between the vector V ⃗ and the positive x-axis is 〖327〗^° (counterclockwise).
