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

v_avg = 37 km/h

Step-by-step explanation:

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

`v_(avg)=(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=(2.0km)/(30min)*(60min)/(1h)=4(km)/(h)`

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

`v_(avg)=(70km/h+4km/h)/(2)=37(km)/(h)`

hence, the average velocity is 37 km/h

Answer 2

v = 16.8 km/h

Step-by-step explanation:

The average velocity can be calculated usign the following equation:

`v = (\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 = (x)/(v) = (8.4 km)/(70 km/h) = 0.12 h`

The total time is:

`t_(t) = 0.12 h + 30min*(1 h)/(60 min) = 0.62 h`

Finally, by taking:

Δt = 0.62 h

Δx= 10.4 km

The average velocity is:

`v = (10.4 km)/(0.62 h) = 16.8 km/h`

Therefore, the average velocity is 16.8 km/h.

I hope it helps you!