Physics Question - Average Velocity?

Question

A bus traveled in a straight line (east) for 3.2hours, with an average velocity of 88km/h. The bus stopped for 20 minutes, and then went south for 2.8hours at an average velocity of 75km/h.

What is the average velocity for the ENTIRE trip (including the time spent while stopped)?

I wouldn't normally ask but when you have a substitute that knows nothing about what you're learning it's just hard to understand it!!

Thanks in advance....

Answer 1

The person above gets the right idea but what he did was calculating the speed, not velocity.

velocity has direction. Speed does not.

average velocity = total displacement / total time

displacement = velocity * time
x = vt

find the distance the bus covers for 3.2hr at 88km/h
x = 88 * 3.2
x = 281.6km East

the bus stopped for 20 minutes or 1/3 hr. Then the bus goes South for 2.8hrs at 75km/hr
x = 75*2.8
x = 210km South

To find the total displacement of the bus, use pythagarian theorem
a^2 + b^2 = c^2
210^2 + 281.6^2 = c^2
c =~ 351.28 km

the total time is 3.2hr + 1/3hr + 2.8hr = 19/3 hr

aveg. velocity = 351.28km / (19/3)hr
aveg. velocity =~ 55.46 km/r

to find the angle, use trig
theta = tan^-1 (210 / 281.6)
theta = 36.71 degrees

ans: 55.45 km/hr at 36.71 degrees South of East

Answer 2

You need to separate this into TOTAL TIME and TOTAL DISTANCE.

Lets find out the total time:
3.2 hours + 20 minutes + 2.8 hours

20 minutes = 0.333333 hours

so
3.2 + 0.33333 + 2.8 = 6.33333 hours TOTAL

For TOTAL DISTANCE we need to incorporate a formula

Distance = Velocity * Time
or
x = V * t

so
88km/h * (3.2 h) = 281.6 km
75km/h * (2.8 h) = 210.0 km

and remember that the bus stopped for 20 minutes but it obviously didnt travel anywhere.

so add up your total distances
281.6km + 210.0km = 491.6km

Now use the velocity equation

Velocity = Distance / Time
or
V = x / t

V = 491.6 km / 6.33333 hours

Vaverage = 77.62 km/h

Answer 3

The distance is 3.2*88+2.8*75 = 491.6 km(directions don´t matter)
The total time involved is 6h+1/3h = 19/3 h
dist = vel.*time so velocity = 491.6*3/19=77.62 km/h
If you need the separation between the start and end point
=sqrt[(3.2*88)^2+2.8*75)^2]