Answer:
Part a)
[tex]v = 7.57 km/h[/tex]
Part b)
[tex]\theta = 67.5 degree[/tex]North of East
Explanation:
Speed of train towards East = 60 km/h
displacement towards East is given as
[tex]d_1 = 40 km[/tex]
now it turns towards 50 degree East of North
so its distance is given as
[tex]d_2 = 20 km(sin50 \hat i + cos50\hat j)[/tex]
[tex]d_2 = 15.3 \hat i + 12.8 \hat j[/tex]
then finally it moves towards west for 50 min
[tex]d_3 = -50 \hat i[/tex]
Now the total displacement of the train is given as
[tex]d = d_1 + d_2 + d_3[/tex]
[tex]d = (40 + 15.3 - 50)\hat i + 12.8 \hat j[/tex]
[tex]d = 5.3\hat i + 12.8 \hat j[/tex]
now total time duration of the motion is given as
[tex]T = 40 min + 20 min + 50 min [/tex]
[tex]T = 1.83 h[/tex]
now average velocity is given as
[tex]v_{avg} = \frac{5.3\hat i + 12.8\hat j}{1.83}[/tex]
[tex]v_{avg} = 2.89\hat i + 6.99\hat j[/tex]
Part a)
magnitude of the average velocity is given as
[tex]v = \sqrt{v_x^2 + v_y^2}[/tex]
[tex]v = \sqrt{2.89^2 + 6.99^2}[/tex]
[tex]v = 7.57 km/h[/tex]
Part b)
Direction of the velocity is given as
[tex]tan\theta = \frac{v_y}{v_x}[/tex]
[tex]tan\theta = \frac{6.99}{2.89}[/tex]
[tex]\theta = 67.5 degree[/tex]North of East
