Question

③ solve for both ( x + y ) components of the displacement vector.
① ( d = 10m )
( \theta = 40^{circ} )
( dx = )
( dy = )
( cos 40=\frac{dlly}{y0} )
( cos 40\times10 = 7.66 )
( sin 40\times10 = )

Explanation:

Step1: Calculate the x - component

The formula for the x - component of a vector \(d\) with magnitude \(d\) and angle \(\theta\) is \(d_x = d\cos\theta\).
Given \(d = 10m\) and \(\theta=40^{\circ}\), then \(d_x=10\times\cos(40^{\circ})\).
Using a calculator, \(\cos(40^{\circ})\approx0.766\), so \(d_x = 10\times0.766 = 7.66m\).

Step2: Calculate the y - component

The formula for the y - component of a vector \(d\) with magnitude \(d\) and angle \(\theta\) is \(d_y = d\sin\theta\).
Given \(d = 10m\) and \(\theta = 40^{\circ}\), then \(d_y=10\times\sin(40^{\circ})\).
Using a calculator, \(\sin(40^{\circ})\approx0.643\), so \(d_y=10\times0.643 = 6.43m\).

Answer:

The x - component \(d_x = 7.66m\) and the y - component \(d_y = 6.43m\)