Question

a delivery truck drives for 2 hrs at 50 miles/phr North and then 3 hours at 30 miles/phr North what was the magnitude and direction of the average velocity of the delivery truck?

Answer 1

We are asked to determine the average velocity of the given truck. To do that we need to divide the total distance over the total time.

Since the distance is the product of the velocity by the time we have that the total distance is:

`d=v_0t_0+v_ft_f`

Where:

`\begin{gathered} v_0,v_f=\text{ initial and final velocity} \\ t_0,t_f=\text{ initial and final time} \end{gathered}`

The total time is the sum of the initial and final time:

`t=t_0+t_f`

Now, the average velocity is the quotient between the total distance and the total time:

`v=(d)/(t)=(v_0t_0+v_ft_f)/(t_0+t_f)`

Now, we plug in the values:

`v=((50mph)(2h)+(30mph)(3h))/(2h+3h)`

Solving the operations we get:

`v=38mph`

Therefore, the magnitude of the average velocity is 38 mph.

The direction of the average velocity is the same as the direction of the moving object therefore, the direction must be North.