Given:
The train first travels 450 km towards the east from Toronto, then 200 km towards the south, and finally 150 km [W 55 degrees S].
The time traveled is t = 12 hours.
To find The train's total displacement relative to Toronto.
The train's average velocity.
Explanation:
The vector diagram can be drawn as shown below
Let Toronto be at the origin.
Here, X-axis is in the East direction.
Y-Axis is in the North direction.
The displacement in the X-direction will be
[tex]\begin{gathered} d_x=450-150cos55^{\circ}\text{ } \\ =363.96km \end{gathered}[/tex]
The displacement in the Y-direction is
[tex]\begin{gathered} d_y=200+150\sin (55^{\circ})\text{ } \\ =\text{ 322.87 }km \end{gathered}[/tex]
The magnitude of resultant displacement will be
[tex]\begin{gathered} d=\sqrt[]{(d_x)^2+(d_y)^2} \\ =486.53km \end{gathered}[/tex]
The direction of displacement will be
[tex]\begin{gathered} \tan \theta=\frac{d_y}{d_x} \\ \theta=\tan ^{-1}(\frac{d_y}{d_x}) \\ =\tan ^{-1}(\frac{322.87}{363.96}) \\ =41.57\text{ degr}ees \end{gathered}[/tex]
(b) The train's average velocity will be
[tex]\begin{gathered} v=\frac{d}{t} \\ =\frac{486.53}{12} \\ =\frac{40.54km}{h} \end{gathered}[/tex]
Final Answer: (a) The magnitude of resultant displacement is 486.53 km.
The direction of resultant displacement is E 41.57 degrees S.
(b) The train's average velocity is 40.54 km/h.
