Question

A vector has the components Ax=29m and Ay=14m. (A) What is the magnitude of this vector? (B) What angle does this vector make with the positive x axis?

Answer 1

The magnitude of the vector is approximately 32.24m and the angle it makes with the positive x-axis is approximately 25.00°.

Step-by-step explanation:

To find the magnitude of the vector, we can use the Pythagorean theorem, which states that the magnitude of a vector is equal to the square root of the sum of the squares of its components. In this case, the magnitude A is given by:

A = sqrt(Ax² + Ay²)

Substituting the given values, we have:

A = sqrt((29m)² + (14m)²)

A = sqrt(841m² + 196m²)

A = sqrt(1037m²)

A ≈ 32.24m

To find the angle that the vector makes with the positive x-axis, we can use the inverse tangent function. In this case, the angle θ is given by:

θ = atan(Ay / Ax)

Substituting the given values, we have:

θ = atan(14m / 29m)

θ ≈ 25.00°

Answer 2

`|A|=32.2048438m`

`\alpha =25.76932762$^(\circ)$`

Step-by-step explanation:

In order to find the magnitude of the vector we need to use pythagorean theorem. As you can see in the picture I attached you, the components Ax and Ay represents the legs and the the hypotenuse is the magnitude of the vector A. Hence:

`|A|=\sqrt{(Ax^(2))+(Ay^(2))  } =\sqrt{(29)^(2)+(14)^(2)  }=32.20248438m`

Now in order to find the angle
`\alpha` we only need to use the trigonometry identity of tangent:

`tan(\alpha )=(Ay)/(Ax)`

Isolating
`\alpha`

`\alpha =arctan((Ay)/(Ax) )=arctan((14)/(29) )=25.76932762$^(\circ)$`