Question

You drive a beat-up pickup truck along a straight road for 8.4 km at 70 km/h, at which point the truck runs out of gasoline and stops. Over the next 30 min, you walk another 2.0 km farther along the road to a gasoline station. What is your average velocity from the beginning of your drive to your arrival at the gasoline station?

Answer 1

Answer:

v_avg = 37 km/h

Explanation:

To find the average velocity in the complete trajectory you use the following formula:

$v_{avg}=\frac{v_1+v_2}{2}$   ( 1 )

v1: velocity in the first part of the trajectory = 70 km/h

v2: velocity in the second part of the trajectory = ?

You calculate v2 by using the following equation for a motion with constant velocity:

$v_2=\frac{2.0km}{30min}*\frac{60min}{1h}=4\frac{km}{h}$

you replace the values of v1 and v2 in (1) and you obtain:

$v_{avg}=\frac{70km/h+4km/h}{2}=37\frac{km}{h}$

hence, the average velocity is 37 km/h

Answer 2

Answer:

v = 16.8 km/h

Explanation:

The average velocity can be calculated usign the following equation:

$v = \frac{\Delta x}{\Delta t}$

Where:

Δx: is the change in the displacement

Δt: is the change in the time

The total displacement is:

$x_{t} = 8.4 km + 2.0 km = 10.4 km$

The initial time is:

$t = \frac{x}{v} = \frac{8.4 km}{70 km/h} = 0.12 h$  

The total time is:

$t_{t} = 0.12 h + 30min*\frac{1 h}{60 min} = 0.62 h$  

Finally, by taking:

Δt = 0.62 h

Δx= 10.4 km

The average velocity is:

$v = \frac{10.4 km}{0.62 h} = 16.8 km/h$        

Therefore, the average velocity is 16.8 km/h.

I hope it helps you!