Question

A train moving at a constant speed of 50.0 km/h moves east for 45.0 min, then in a direction 45.0° east of due north for 10.0 min, and then west for 55.0 min. What is the average velocity of the train during this run? (Let the +x-axis point towards the east and the +y-axis point towards the north.)l

Answer 1

Answer:

3.47km/h with a direction of 67.8 degrees north of west.

Explanation:

First we need to calculate the displacement on the X axis, so:

$d_{x1}=50.0km/h*45min(\frac{1h}{60min})=37.5km\\d_{x2}=50.0km/h*10min(\frac{1h}{60min})*cos(45^o)=5.90km\\d_{x1}=-50.0km/h*55.0min(\frac{1h}{60min})=45.8km\\D_x=37.5+5.90-45.8=-2.4km$

then on the Y axis:

$D_y=50.0km/h*10min(\frac{1h}{60min})*sin(45^o)=5.90km$

The magnitud of the displacement is given by:

$D=\sqrt{D_x^2+D_y^2} \\D=6.37km$

and the angle:

$\alpha =arctg(\frac{5.90}{2.41})=67.8^o$

that is 67.8 degrees north of west.

$v=\frac{D}{t}\\v=\frac{6.37km}{110min*\frac{1h}{60min}}\\v=3.47km/h$