Answer:
[tex]5.2\ \text{m/s}[/tex]
[tex]70^{\circ}[/tex] south of east
Explanation:
[tex]v_a[/tex] = 3 m/s
[tex]\theta_a[/tex] = [tex]20^{\circ}[/tex] north of east
[tex]v_b[/tex] = 6 m/s
[tex]\theta_b[/tex] = [tex]40^{\circ}[/tex] south of east = [tex]360-40=320^{\circ}[/tex] north of east
x and y component of [tex]v_a[/tex]
[tex]v_{ax}=v_a\cos \theta\\\Rightarrow v_{ax}=3\times \cos 20^{\circ}\\\Rightarrow v_{ax}=2.82\ \text{m/s}[/tex]
[tex]v_{ay}=v_a\sin\theta\\\Rightarrow v_{ay}=3\times \sin20^{\circ}\\\Rightarrow v_{ay}=1.03\ \text{m/s}[/tex]
x and y component of [tex]v_b[/tex]
[tex]v_{bx}=v_b\cos \theta\\\Rightarrow v_{bx}=6\times \cos 320^{\circ}\\\Rightarrow v_{bx}=4.6\ \text{m/s}[/tex]
[tex]v_{by}=v_b\sin\theta\\\Rightarrow v_{by}=6\times \sin320^{\circ}\\\Rightarrow v_{by}=-3.86\ \text{m/s}[/tex]
[tex]\Delta v=v_b-v_a\\\Rightarrow \Delta v=(4.6-2.82)\hat{i}+(-3.86-1.03)\hat{j}\\\Rightarrow \Delta v=1.78\hat[i}-4.89\hat{j}[/tex]
Magnitude
[tex]|\Delta v|=\sqrt{(-4.89)^2+1.78^2}\\\Rightarrow \Delta v=5.2\ \text{m/s}[/tex]
Direction
[tex]\theta=\tan{-1}|\dfrac{-4.89}{1.78}|\\\Rightarrow \theta=70^{\circ}[/tex]
The magnitude of the change in velocity vector is [tex]5.2\ \text{m/s}[/tex] and the direction is [tex]70^{\circ}[/tex] south of east.
The change in velocity will be [tex]\Delta V=5.2\ \frac{m}{s}[/tex] and the direction will be [tex]70^o[/tex] South to east.
Any quantity which is defined by its magnitude and direction both are called as the vector quantities.
Now the data given in the question will be given as:
[tex]V_a[/tex] = 3 m/s
[tex]\theta[/tex] = [tex]20^o[/tex] north of east
[tex]V_b[/tex] = 6 m/s
[tex]\theta[/tex] = [tex]40^o[/tex]south of east = 360-40=320 north of east
Now we will find the x and y component of [tex]V_a[/tex]
[tex]V_{ax}=V_acos\theta[/tex]
[tex]V_{ax}=3\times Cos20[/tex]
[tex]V_{ax}=2.82\ \frac{m}{s}[/tex]
[tex]V_{ay}=V_aSin\theta[/tex]
[tex]V_{ay}=3\times Sin20[/tex]
[tex]V_{ay}=1.03\ \frac{m}{s}[/tex]
Now we will find the x and y component of [tex]V_b[/tex]
[tex]V_{bx}=V_bcos\theta[/tex]
[tex]V_{bx}=6\times cos\320[/tex]
[tex]V_{bx}=4.6\ \frac{m}{s}[/tex]
[tex]V_{by}=V_bSin\theta[/tex]
[tex]V_{by}=6\times Sin320[/tex]
[tex]V_{by}=-3.86\ \frac{m}{s}[/tex]
Now change in velocity will be
[tex]\Delta V=V_b-V_a[/tex]
[tex]\Delta V=(4.6-2.82)i+(-3.86-1.03)j[/tex]
[tex]\Delta V=1.78i-4.89j[/tex]
The magnitude can be find out as follows:
[tex]\Delta V=\sqrt{(-4.89^2+(1.78^2)}[/tex]
[tex]\Delta V=5.2\ \frac{m}{s}[/tex]
The direction of the vector will be
[tex]\theta= tan^{-1}(\dfrac{-4.89}{1.78})[/tex]
[tex]\theta=70^o[/tex]
Thus the change in velocity will be [tex]\Delta V=5.2\ \frac{m}{s}[/tex] and the direction will be [tex]70^o[/tex] South to east.
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