Respuesta :
Answer:
Explanation:
Given
Velocity of train [tex]\vec{v}=55 km/h\hat{i}[/tex] for [tex]33 min[/tex]
Position vector after [tex]33 min[/tex]
[tex]\vec{r_1}=55\times \frac{33}{60}\hat{i}[/tex]
After that it moves [tex]49^{\circ}[/tex] east of due north
Position vector after 43 min
[tex]\vec{r_{21}}=55\times \frac{43}{60}(\cos (49)\hat{j}+\sin (49)\hat{i})[/tex]
Position of Vector [tex]\vec{r_2}=55\times \frac{43}{60}(\cos (49)\hat{j}+\sin (49)\hat{i})+55\times \frac{33}{60}\hat{i}[/tex]
[tex]\vec{r_2}=(30.25+29.747)\hat{i}+(25.85)\hat{j}[/tex]
magnitude of final Position [tex]|r_2|=\sqrt{59.99^2+25.85^2}[/tex]
[tex]|r_2|=65.32 km[/tex]
average velocity [tex]v_{avg}=\frac{displacement}{time}[/tex]
[tex]v_{avg}=\frac{65.32}{28+43}=\frac{65.32}{71}[/tex]
[tex]v_{avg}=0.92 km/h[/tex]
Direction of average velocity will be similar to displacement
[tex]\tan \theta =\frac{25.85}{59.99}[/tex]
[tex]\theta =23.311^{\circ}[/tex]
