Average Velocity and Average Speed Problems and Solutions

 Problem#1

During a hard sneeze, your eyes might shut for 0.50 s. If you are driving a car at 90 km/h during such a sneeze, how far does the car move during that time?

Answer:
The speed (assumed constant) is
v = (90 km/h)(1000 m/km)/(3600 s/h) = 25 m/s.
Thus, in 0.50 s, the car travels a distance is
d = vt = (25 m/s)(0.50 s) ≈ 13 m.

Problem#2
Compute your average velocity in the following two cases: (a) You walk 73.2 m at a speed of 1.22 m/s and then run 73.2 m at a speed of 3.05 m/s along a straight track. (b) You walk for 1.00 min at a speed of 1.22 m/s and then run for 1.00 min at 3.05 m/s along a straight track. (c) Graph x versus t for both cases and indicate how the average velocity is found on the graph.

Answer:
(a) Using the fact that time = distance/velocity while the velocity is constant, we find
v_avg = (73.2 m + 73.2 m)/[(73.2 m/1.22 m/s) + (73.2 m/3.05 m/s)]
v_avg = 1.74 m/s

(b) Using the fact that distance = vt while the velocity v is constant, we find
v_avg = [(1.22 m/s)(60 s) + (3.05 m/s)(60 s)]/120 s
v_avg = 2.14 m/s

(c) The graphs (Fig.1a) and (Fig.1b) are shown below (with meters and seconds understood). The first consists of two (solid) line segments, the first having a slope of 1.22 and the second having a slope of 3.05. The slope of the dashed line represents the average velocity (in both graphs). The second graph also consists of two (solid) line segments, having the same slopes as before — the main difference (compared to the first graph) being that the stage involving higher-speed motion lasts much longer.

Fig.1

Problem#3
An automobile travels on a straight road for 40 km at 30 km/h. It then continues in the same direction for another 40 km at 60 km/h. (a) What is the average velocity of the car during the full 80 km trip? (Assume that it moves in the positive x direction.) (b) What is the average speed? (c) Graph x versus t and indicate how the average velocity is found on the graph.

Answer:
Since the trip consists of two parts, let the displacements during first and second parts of the motion be ∆x₁ and ∆x₂, and the corresponding time intervals be ∆t₁ and ∆t₂, respectively. Now, because the problem is one-dimensional and both displacements are in the same direction, the total displacement is ∆x = ∆x₁ + ∆x₂, and the total time for the trip is ∆t= ∆t₁ + ∆t₂. Using the definition of average velocity given in Eq. 2-2, we have

v_avg = ∆x/∆t
v_avg = (∆x₁ + ∆x₂)/(∆t₁ + ∆t₂)

To find the average speed, we note that during a time ∆t if the velocity remains a positive constant, then the speed is equal to the magnitude of velocity, and the distance is equal to the magnitude of displacement, with d = |∆x| = v∆t .

(a) During the first part of the motion, the displacement is ∆x₁ = 40 km and the time interval is
t₁ = ∆x₁/v₁ = 40 km/(30 km/h) = 1.33 h

Similarly, during the second part the displacement is ∆x₂ = 40 km and the time interval is
t₂ = ∆x₂/v₂ = 40 km/(60 km/h) = 0.67 h

The total displacement is ∆x = ∆x₁ + ∆x₂ = 40 km + 40 km = 80 km, and the total time elapsed is ∆t = ∆t₁ + ∆t₂ = 2.00 h. Consequently, the average velocity is

v_avg = ∆x/∆t = 80 km/2 h = 40 km/h

(b) In this case, the average speed is the same as the magnitude of the average velocity:
s_avg = 40 km/h.

(c) The graph (Fig.2) of the entire trip is shown below; it consists of two contiguous line segments, the first having a slope of 30 km/h and connecting the origin to (∆t₁, ∆x₁) = (1.33 h, 40 km) and the second having a slope of 60 km/h and connecting (∆t₁, ∆x₁) to (∆t, ∆x) = (2.00 h, 80 km).

Fig.2

Problem#4
A car travels up a hill at a constant speed of 40 km/h and returns down the hill at a constant speed of 60 km/h. Calculate the average speed for the round trip.

Answer:
Average speed, as opposed to average velocity, relates to the total distance, as opposed to the net displacement. The distance D up the hill is, of course, the same as the distance down the hill, and since the speed is constant (during each stage of the motion) we have speed = D/t.
Thus, the average speed is

S_avg = (D_up + D_down)/(t_up + t_down)

= (2D)/(D/v_up + D/v_down)

S_avg = 2v_upv_down/( v_up + v_down)

= 2 x 40 km/h x 60 km/h/(40 km/h + 60 km/h)
S_avg = 48 km/h

Problem#5
The position of an object moving along an x axis is given by x = 3t⁴ – 4t² – t³, where x is in meters and t in seconds. Find the position of the object at the following values of t: (a) 1 s, (b) 2 s, (c) 3 s,and (d) 4 s. (e) What is the object’s displacement between t = 0 and t = 4 s? (f) What is its average velocity for the time interval from t = 2 s to t = 4 s? (g)Graph x versus t for 0 ≤ t ≤ 4 s and indicate how the answer for (f) can be found on the graph.

Answer:
Using x = 3t – 4t² + t³ with SI units understood is efficient (and is the approach we will use), but if we wished to make the units explicit we would write
x = (3 m/s)t – (4 m/s²)t² + (1 m/s³)t³

We will quote our answers to one or two significant figures, and not try to follow the significant figure rules rigorously.

(a) Plugging in t = 1 s yields x = 3 – 4 + 1 = 0.

(b) With t = 2 s we get x = 3(2) – 4(2)²+(2)³ = –2 m.

(c) With t = 3 s we have x = 0 m.

(d) Plugging in t = 4 s gives x = 12 m.

For later reference, we also note that the position at t = 0 is x = 0.

(e) The position at t = 0 is subtracted from the position at t = 4 s to find the
displacement ∆x = 12 m.

(f) The position at t = 2 s is subtracted from the position at t = 4 s to give the
displacement ∆x = 14 m. Then,

v_avg = ∆x/∆t = 14 m/2 s = 7 m/s

(g) The position of the object for the interval 0 ≤ t ≤ 4 is plotted below (Fig.3). The straight line drawn from the point at (t, x) = (2 s , –2 m) to (4 s, 12 m) would represent the average velocity, answer for part (f).

Fig.3