Question

③ solve for both x + y components of the displacement vector.
@
( d = 10m )
( \theta = 40 ^ { circ } )
( d x = )
( d y = )

( d = 26 m )
( \theta = 70 ^ { circ } )
( d x = )
( d y = )

Explanation:

Step1: Recall the formula for vector components

The \(x -\)component of a vector \(d\) is \(d_x = d\cos\theta\) and the \(y -\)component is \(d_y = d\sin\theta\)

Step2: Calculate for part (a)

Given \(d = 10\space m\) and \(\theta=40^{\circ}\)

• \(d_x=10\cos(40^{\circ})\)

$$d_x = 10\times0.7660\approx7.66\space m$$

• \(d_y = 10\sin(40^{\circ})\)

$$d_y=10\times0.6428\approx6.43\space m$$

Step3: Calculate for part (b)

Given \(d = 26\space m\) and \(\theta = 70^{\circ}\)

• Since the vector is in the second - quadrant (assuming standard coordinate system where positive \(x\) is to the right and positive \(y\) is up), \(d_x=-d\cos\theta\)

$$d_x=- 26\cos(70^{\circ})=-26\times0.3420\approx - 8.89\space m$$

• \(d_y = d\sin\theta\)

$$d_y=26\sin(70^{\circ})=26\times0.9397\approx24.43\space m$$

Answer:

a. \(d_x\approx7.66\space m\), \(d_y\approx6.43\space m\)
b. \(d_x\approx - 8.89\space m\), \(d_y\approx24.43\space m\)