# 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 ?

##### 1 Answer

Magnitude:

Angle:

#### 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 *north direction* to be the

You know that your train starts its movement due east, i.e. **to the right** along the **east of due north**.

This means that its new direction will be *clockwise* **from the north direction**, which is equivalent to

#90^@ - 50^@ = 40^@#

*counterclockwise* from the **east direction**.

Fonally, it changes direction again to move *due west*, which is equivalent to moving **to the left** along the

Start by calculating the displacement of the train *along the*

#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

#52.0"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

#52.0"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_"total"^2 = d_x^2 + d_y^2#

#d_"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_"total" = 0.350 + 0.400 + 0.517 = "1.267 hr"#

The magnitude of the **average velocity** will thus be

#bar(v) = d_"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(theta) = d_y/d_x implies theta = arctan(d_y/d_x)#

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