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4 votes
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?

User Havij
by
4.1k points

2 Answers

4 votes

Answer:

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

User David Moores
by
3.6k points
3 votes

Answer:

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!

User Ophelia
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3.7k points