Respuesta :
Answer:
Magnitude of avg velocity, [tex]|v_{avg}| = 18.9 km/h[/tex]
[tex]\theta' = 56.85^{\circ}[/tex]
Given:
Constant speed of train, v = 79 km/h
Time taken in East direction, t = 27 min = [tex]\frac{27}{60} h[/tex]
Angle, [tex]\theta = 50^{\circ}[/tex]
Time taken in [tex]50^{\circ}[/tex]east of due North direction, t' = 29 min = [tex]\frac{29}{60} h[/tex]
Time taken in west direction, t'' = 37 min = [tex]\frac{27}{60} h[/tex]
Solution:
Now, the displacement, 's' in east direction is given by:
[tex]\vec{s} = vt = 79\times \frac{27}{60} = 35.5\hat{i} km[/tex]
Displacement in [tex]50^{\circ}[/tex] east of due North for 29.0 min is given by:
[tex]\vec{s'} = vt'sin50^{\circ}\hat{i} + vt'cos50^{\circ}\hat{j}[/tex]
[tex]\vec{s'} = 79(\frac{29}{60})sin50^{\circ}\hat{i} + 79(\frac{29}{60})cos50^{\circ}\hat{j}[/tex]
[tex]\vec{s'} = 29.25\hat{i} + 24.54\hat{j} km[/tex]
Now, displacement in the west direction for 37 min:
[tex]\vec{s''} = - vt''hat{i} = - 79\frac{37}{60} = - 48.72\hat{i} km[/tex]
Now, the overall displacement,
[tex]\vec{s_{net}} = \vec{s} + \vec{s'} + \vec{s''}[/tex]
[tex]\vec{s_{net}} = 35.5\hat{i} + 29.25\hat{i} + 24.54\hat{j} - 48.72\hat{i}[/tex]
[tex]\vec{s_{net}} = 16.03\hat{i} + 24.54\hat{j} km[/tex]
(a) Now, average velocity, [tex]v_{avg}[/tex] is given:
[tex]v_{avg} = \frac{total displacement, \vec{s_{net}}}{total time, t}[/tex]
[tex]v_{avg} = \frac{16.03\hat{i} + 24.54\hat{j}}{\frac{27 + 29 + 37}{60}}[/tex]
[tex]v_{avg} = 10.34\hat{i} + 15.83\hat{j}) km/h[/tex]
Magnitude of avg velocity is given by:
[tex]|v_{avg}| = \sqrt{(10.34)^{2} + (15.83)^{2}} = 18.9 km/h[/tex]
(b) angle can be calculated as:
[tex]tan\theta' = \frac{15.83}{10.34}[/tex]
[tex]\theta' = tan^{- 1}\frac{15.83}{10.34} = 56.85^{\circ}[/tex]
