# A train at a constant 52.0 km/h moves east for 21.0 min, then in a direction 50.0° east of due north for 24.0 min, and then west for 31.0 min. What are the (a) magnitude and (b) angle (relative to east) of its average velocity during this trip ?

Sep 22, 2015

Magnitude: $\text{12.0 km/hr}$
Angle: 28.4""^@

#### Explanation:

You know that average velocity is defined as the total displacement of an object divided by the total time needed for the object to complete its motion.

Displacement is independent of the path the object takes in its motion and is simply the distance from the starting point to the finish point.

Let's take the east direction to be the $x$-axis and the north direction to be the $y$-axis.

You know that your train starts its movement due east, i.e. to the right along the $x$-axis. The train then changes direction and moves ${50}^{\circ}$ east of due north.

This means that its new direction will be ${50}^{\circ}$ clockwise from the north direction, which is equivalent to

${90}^{\circ} - {50}^{\circ} = {40}^{\circ}$

counterclockwise from the east direction.

Fonally, it changes direction again to move due west, which is equivalent to moving to the left along the $x$-axis.

Start by calculating the displacement of the train along the $x$-axis. Convert the time of travel from minutes to hours first

21.0color(red)(cancel(color(black)("min"))) * "1 hr"/(60color(red)(cancel(color(black)("min")))) = "0.350 hr"

24.0color(red)(cancel(color(black)("min"))) * "1 hr"/(60color(red)(cancel(color(black)("min")))) = "0.400 hr"

31.0color(red)(cancel(color(black)("min"))) * "1 hr"/(60color(red)(cancel(color(black)("min")))) = "0.517 hr"

The displacement along the $x$-axis will be

$52.0 \text{km"/color(red)(cancel(color(black)("hr"))) * 0.350color(red)(cancel(color(black)("hr"))) + 52.0"km"/color(red)(cancel(color(black)("hr"))) * 0.400color(red)(cancel(color(black)("hr"))) * cos(40^@) + (- 52.0"km"/color(red)(cancel(color(black)("hr"))) * 0.517color(red)(cancel(color(black)("hr")))) = "7.25 km}$

The displacement along the $y$-axis will be given by

$52.0 \text{km"/color(red)(cancel(color(black)("hr"))) * 0.400color(red)(cancel(color(black)("hr"))) * sin(40^@) = "13.4 km}$

The total displacement will be

${d}_{\text{total}}^{2} = {d}_{x}^{2} + {d}_{y}^{2}$

${d}_{\text{total" = sqrt(d_x^2 + d_y^2) = sqrt(7.25^2 + 13.4^2) = "15.2 km}}$

The total time is equal to

${t}_{\text{total" = 0.350 + 0.400 + 0.517 = "1.267 hr}}$

The magnitude of the average velocity will thus be

$\overline{v} = {d}_{\text{total"/t_"total}}$

bar(v) = "15.2 km"/"1.267 hr" = color(green)("12.0 km/hr")

To get the angle of its average velocity relative to due east, use the tangent function

$\tan \left(\theta\right) = {d}_{y} / {d}_{x} \implies \theta = \arctan \left({d}_{y} / {d}_{x}\right)$

theta = arctan((7.25color(red)(cancel(color(black)("km"))))/(13.4color(red)(cancel(color(black)("km"))))) = color(green)(28.4""^@