Example 2-1 Average Velocity
You drive from Bismarck, North Dakota, to Fargo, North Dakota, on Interstate Highway 94, an approximately straight
road 315 km in length. You leave Bismarck and travel 210 km at constant velocity in 2.50 h, then stop for 0.50 h at
a rest area. You then drive the remaining 105 km to Fargo at constant velocity in 1.00 h. You then turn around immediately and drive back to Bismarck nonstop at constant velocity in 2.75 h. (a) Draw an x–t graph for the entire round trip.
Then calculate the average velocity for (b) the trip from Bismarck to the rest area, (c) the trip from the rest area to Fargo,
(d) the entire outbound trip from Bismarck to Fargo, (e) the return trip from Fargo to Bismarck, and (f) the round trip
from Bismarck to Fargo and back to Bismarck.
Set Up
This is an example of linear motion along
the straight highway between Bismarck and
Fargo. We set up our coordinates as shown,
with x = 0 at the starting position (Bismarck)
and the positive x direction from Bismarck
toward Fargo. We also choose t = 0 to be
when the car leaves Bismarck. Equation 2-1
will tell us the average velocity for each
portion of the trip.
Average velocity for linear motion:
vaverage, x =
x2 - x1
t2 - t1
Dx
=
Dt
t1 = 0
t5 = 6.75 h
t2 = 2.50 h
t3 = 3.00 h t4 = 4.00 h
210 km
(2-1)
x1, x5
Bismarck
105 km
x2 , x3
Rest area
x4
Fargo
Solve
(a) To draw the x–t graph for the car’s
motion, we first put a dot on the graph for
the car’s positions and the corresponding
times at the beginning and end of each leg
of the trip. The velocity is constant on each
leg, so we draw straight lines connecting
successive points. The slope of the graph is
positive (upward) for the legs in which the
car moves in the positive x direction
toward Fargo, and negative (downward)
in which the car moves in the negative
x direction back toward Bismarck. The
slope is zero (the line is flat) during the
time when the car is at the rest area and not
moving.
x ( km)
315
arrive
rest area
210
depart
rest area
105
0
depart Bismarck
Bismarck to rest area:
x2 - x1
Dx
=
vaverage, x =
Dt
t2 - t1
210 km - 0
=
= +84.0 km>h
2.50 h - 0
(c) Calculate the average velocity for the trip
from the rest area (position x3 = 210 km at
time t3 = 2.50 h + 0.50 h = 3.00 h) to Fargo
(position x4 = 315 km at time t4 =
3.00 h + 1.00 h = 4.00 h).
Rest area to Fargo:
x4 - x3
Dx
=
vaverage, x =
Dt
t4 - t3
315 km - 210 km
=
= +105 km>h
4.00 h - 3.00 h
(d) Calculate the average velocity for the
entire outbound trip from Bismarck (position
x1 = 0 at time t1 = 0) to Fargo (position
x4 = 315 km at time t4 = 4.00 h).
Bismarck to Fargo:
x4 - x1
Dx
=
vaverage, x =
Dt
t4 - t1
=
315 km - 0
= +78.8 km>h
4.00 h - 0
Fargo back to Bismarck:
x5 - x4
Dx
vaverage, x =
=
Dt
t5 - t4
=
arrive
Bismarck
0 1 2 3 4 5 6 7
(b) Calculate the average velocity for the
trip from Bismarck (position x1 = 0 at time
t1 = 0) to the rest area (position x2 = 210 km
at time t2 = 2.50 h).
(e) Calculate the average velocity for the
return trip from Fargo (position x4 = 315 km
at time t4 = 4.00 h) to ­Bismarck (position
x5 = 0 at time t5 = 4.00 h + 2.75 h = 6.75 h).
arrive Fargo,
depart Fargo
0 - 315 km
= -115 km>h
6.75 h - 4.00 h
t ( h)
(f) For the round trip from Bismarck
(position x1 = 0 at time t1 = 0) to Fargo
and back to Bismarck (position x5 = 0 at
time t5 = 6.75 h), the net displacement Dx =
x5 2 x1 is zero: The car ends up back where
it started. Hence the average velocity for the
round trip is also zero.
Reflect
Bismarck to Fargo and back to Fargo:
x5 - x1
Dx
=
vaverage, x =
Dt
t5 - t1
0 - 0
=
= 0
6.75 h - 0
The average velocity is positive for the trips from Bismarck toward Fargo (since the car moves in the positive x direction)
and negative for the trip from Fargo back to Bismarck (since the car moves in the negative x direction). The result
vaverage, x = 0 for the round trip means that the net displacement for the round trip was zero. The average speed for the
trip was definitely not zero, however: The car traveled a total distance of 2(315 km) = 630 km in a total elapsed time of
6.75 h, so the average speed was (630 km)>(6.75 h) = 93.3 km>h. Displacement and distance are not the same thing;
likewise, average velocity and average speed are not the same thing.