Position, displacement, and average velocity – Problems and solutions
Example 1. 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.
a) What is your overall displacement from the beginning of your drive to your arrival at the station?
b) What is the time interval \( \Delta t \) from the beginning of your drive to your arrival at the station?
c) What is your average velocity \( {{\nu }_{avg}} \) from the beginning of your drive to your arrival at the station? Find it both numberically and graphically.
d) Suppose that to pump the gasoline, pay for it, and walk back to the truck takes you another 45 min. What is your average speed from the beginning of your drive to your return to the truck with the gasoline?
Solution:
a) Key idea
Assume, for convenience, that you move in the positive direction of an x axis, from a dirst position of x1 = 0 to a second position of x2 at the station. That second position must be at x2 = 8.4 km + 2.0 km = 10.4 km. Then your displacement \( \Delta x \) along the x axis is second position minus the first position.
Calculation: From Eq.2-1, we have
\( \Delta x={{x}_{2}}-{{x}_{1}}=10.4\text{ }km-0=10.4\text{ }km \) (Answer)
Thus, your overall displacement is 10.4 km in the positive direction of the x axis.
b) Key idea
We already know the walking time interval \( \Delta {{t}_{wlk}} \) (=0.50 h), but we lack the driving time interval \( \Delta {{t}_{dr}} \). However, we know that for the drive the displacement \( \Delta {{x}_{dr}} \) is 8,4 km and the average velocity \( {{\nu }_{avg,dr}} \) is 70 km/h. Thus, this average velocity is the ratio of the displacement for the drive to the time interval for the drive.
Calculations: We first write
\( {{\nu }_{avg,dr}}=\frac{\Delta {{x}_{dr}}}{\Delta {{t}_{dr}}} \)
Rearranging and substituting data then give us
\( \Delta {{t}_{dr}}=\frac{\Delta {{x}_{dr}}}{{{\nu }_{avg,dr}}}=\frac{8.4\text{ }km}{70\text{ }km/h}=0.12\text{ }h \)
So, \( \Delta t=\Delta {{t}_{dr}}+\Delta {{t}_{wlk}}=0.12\text{ }h+0.50\text{ }h=0.62\text{ }h \) (Answer)
c) Key idea
From Eq.2-2 we know that \( {{\nu }_{avg}} \) for the entire trip is the ratio of the displacement of 10.4 km for the entire trip to the time interval of 0.62 h for entire trip.
Calculation: Here we find
\( {{\nu }_{avg}}=\frac{\Delta x}{\Delta t}=\frac{10.4\text{ }km}{0.62\text{ }h}=16.8\text{ }km/h\approx 17\text{ }km/h \) (Answer)
To find \( {{\nu }_{avg}} \) graphically, first we graph the function x(t) as shown in Fig.2-5, where the beginning and arrival points on the graph are the origin and the point labeled as “Station”.Your average velocity is the slope of the straight line connecting those points; that is, \( {{\nu }_{avg}} \) is the ratio of the rise ( \( \Delta x=10.4\text{ }km \)) to the run (\Delta t=0.62 \text{ }h), which gives us \( {{\nu }_{avg}}=16.8\text{ }km/h \).
d) Key idea
Your average speed is the ratio of the total distance you move to the total time interval you take to make that move.
Calculation: The total distance is 8.4 km + 2.0 km + 2.0 km = 12.4 km. The total time interval is 0,12 h + 0,50 h + 0,75 h = 1.37 h.
Thus, Eq.2-3 gives us
\( {{s}_{avg}}=\frac{12.4\text{ }km}{1.37\text{ }h}=9.1\text{ }km/h \) (Answer)
Example 2. While driving a car at 90 km/h, how far do you move while your eyes shut for 0.50 s during a hard sneeze?
Solution:
The speed (assumed constant) is \(\nu =\frac{\left( 90\text{ }km/h \right)\left( 1000\text{ }km/h \right)}{3600\text{ }s/h}=25\text{ }m/s\)
Thus, in 0.5 s, the car travels a distance \( d=\nu t=\left( 25\text{ }m/s \right)\left( 0.50\text{ }s \right)\approx 13\text{ }m \)
Example 3. 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.
Solution:
a) Using the fact that time = distance/velocity while the velocity is constant, we find
\({{\nu }_{avg}}=\frac{73.2\text{ }m+73.2\text{ }m}{\frac{73.2\text{ }m}{1.22\text{ }m/s}+\frac{73.2\text{ }m}{3.05\text{ }m/s}}=1.74\text{ }m/s\)
b) Using the fact that distance = \( \nu t \) while the velocity \( \nu \) is constant, we find
\( {{\nu }_{avg}}=\frac{\left( 1.22\text{ }m/s \right)\left( 60\text{ }s \right)+\left( 3.05\text{ }m/s \right)\left( 60\text{ }s \right)}{120\text{ }s}=2.14\text{ }m/s \)
c) The graphs 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.
Example 4. 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
Solution:
Think. This one-dimensional kinematics problem consists of two parts, and we are asked to solve for the average velocity and average speed of the car.
Express. Since the trip consists of two parts, let the displacements during first and second parts of the motion be \( \Delta {{x}_{1}} \) and \( \Delta {{x}_{2}} \) and the corresponding time intervals be \( \Delta {{t}_{1}} \) and \( \Delta {{t}_{2}} \), respectively. Now, because the problem is one-dimensional and both displacements are in the same direction, the total displacement is simply \( \Delta x=\Delta {{x}_{1}}+\Delta {{x}_{2}} \), and the total time for trip is \( \Delta t=\Delta {{t}_{1}}+\Delta {{t}_{2}} \). Using the definition of average velocity given in Eq.2-2, we have
\( {{v}_{avg}}=\frac{\Delta x}{\Delta t}=\frac{\Delta {{x}_{1}}+\Delta {{x}_{2}}}{\Delta {{t}_{1}}+\Delta {{t}_{2}}} \)
To find the average speed, we note that during a time \( \Delta t \) 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 \( d=\left| \Delta x \right|=v\Delta t \)
Analyze
a) During the first part of the motion, the displacement is \( \Delta {{x}_{1}}=40\text{ }km \) and the time taken is
\( {{t}_{1}}=\frac{40\text{ }km}{30\text{ }km/h}=1.33\text{ }h \)
Similarly, during the sencond part of the trip the displacement is \( \Delta {{x}_{2}}=40\text{ }km \) and the time interval is
\( {{t}_{2}}=\frac{40\text{ }km}{60\text{ }km/h}=0.67\text{ }h \)
The total displacement is \( \Delta x=\Delta {{x}_{1}}+\Delta {{x}_{2}}=40\text{ }km+40\text{ }km=80\text{ }km \), and the total time elapsed is \( \Delta t=\Delta {{t}_{1}}+\Delta {{t}_{2}}=2.00\text{ }h \). Consequently, the average velocity is
\( {{v}_{avg}}=\frac{\Delta x}{\Delta t}=\frac{80\text{ }km}{2.0\text{ }h}=40\text{ }km/h \)
b) In this case, the average speed is the same as the magnitude of the average velocity: \( {{s}_{avg}}=40\text{ }km/h \)
c) The graph of the entire trip, shown below, consists of two contiguous line segments, the first having a slope of 30 km/h and connecting the origin to \( \left( \Delta {{t}_{1}},\Delta {{x}_{1}} \right)=\left( 1.33\text{ }h,40\text{ }km \right) \) and the second having a slope of 60 km/h and connecting \(\left( \Delta {{t}_{1}},\Delta {{x}_{1}} \right)\) to \(\left( \Delta t,\Delta x \right)=\left( 2.00\text{ }h,80\text{ }km \right)\).
From the graphical point of view, the slope of dashed line drawn from the origin to \(\left( \Delta t,\Delta x \right)\) represents the average velocity.
Learn. The average velocity is a vector quantity that depends only on the net displacement (also a vector) between the starting and ending points.
Example 5. A car moves uphill at 40 km/h and then back downhill at 60 km/h. What is the average speed for the round trip?
Solution:
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 \( =\frac{D}{t} \). Thus, the average speed is
\(\frac{{{D}_{up}}+{{D}_{down}}}{{{t}_{up}}+{{t}_{down}}}=\frac{2D}{\frac{D}{{{v}_{up}}}+\frac{D}{{{v}_{down}}}}\)
Which, after canceling D and plugging in \( {{v}_{up}}=40\text{ }km/h \) and \( {{v}_{down}}=60\text{ }km/h \), yields 48 km/h for the average speed.
Example 6. The position of an object moving along an x axis is given by \( x=3t-4{{t}^{2}}+{{t}^{3}} \), 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
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 \( 0\le t\le 4\text{ }s \) and indicate how the answer for f) can be found on the graph.
Solution:
Think. In this one-dimensional kinematics problem, we’s given the position function x(t), and asked to calculate the position and velocity of the object at a later time.
Express. The position function is given as \(x(t)=(3\text{ }m/s)t-(4\text{ }m/{{s}^{2}}){{t}^{2}}+(1\text{ }m/{{s}^{3}}){{t}^{3}}\).
The position of the object at some instant tO is simply given by \( x({{t}_{0}}) \). For the time interval \( {{t}_{1}}\le t\le {{t}_{2}} \). The displacement is \( \Delta x=x({{t}_{2}})-x({{t}_{1}}) \). Similarly, using Eq.2-2, the average velocity for this time interval is
\({{v}_{avg}}=\frac{\Delta x}{\Delta t}=\frac{x({{t}_{2}})-x({{t}_{1}})}{{{t}_{2}}-{{t}_{1}}}\)
Analyze
a) Plugging in t = 1 s intro x(t) yields
\( x(1\text{ }s)=(3\text{ }m/s)(1\text{ }s)-(4\text{ }m/{{s}^{2}}){{(1\text{ }s)}^{2}}+(1\text{ }m/{{s}^{3}}){{(1\text{ }s)}^{3}}=0 \)
b) With t = 2 s we get \( x(2\text{ }s)=(3\text{ }m/s)(2\text{ }s)-(4\text{ }m/{{s}^{2}}){{(2\text{ }s)}^{2}}+(1\text{ }m/{{s}^{3}}){{(2\text{ }s)}^{3}}=-2\text{ }m \)
c) With t = 3 s we have \( x(3\text{ }s)=(3\text{ }m/s)(3\text{ }s)-(4\text{ }m/{{s}^{2}}){{(3\text{ }s)}^{2}}+(1\text{ }m/{{s}^{3}}){{(3\text{ }s)}^{3}}=0\text{ }m \)
d) Similarly, plugging in t = 4 s gives
\( x(4\text{ }s)=(3\text{ }m/s)(4\text{ }s)-(4\text{ }m/{{s}^{2}}){{(4\text{ }s)}^{2}}+(1\text{ }m/{{s}^{3}}){{(4\text{ }s)}^{3}}=12\text{ }m \)
e) The position at t = 0 is x = 0. Thus, the displacement between t = 0 and t = 4 s is \( \Delta x=x(4\text{ }s)-x(0)=12\text{ }m-0=12\text{ }m \).
f) The position at t = 2 s is subtracted from the position at t = 4 s to give the displacement: \( \Delta x=x(4\text{ }s)-x(2\text{ }s)=12\text{ }m-(-2\text{ }m)=14\text{ }m \).
Thus, the average velocity is \( {{v}_{avg}}=\frac{\Delta x}{\Delta t}=\frac{14\text{ }m}{2\text{ }s}=7\text{ }m/s \).
g) The position of the object for the interval \( 0\le t\le 4 \) is plotted bellow. The straight line drawn from the point at \( (t,x)=(2\text{ }s,-2\text{ }m) \) to \( (4\text{ }s,12\text{ }m) \) would represent the average velocity, answer for part (f).
Learn. Our graphical representation illustrates once again that the average velocity for a time interval depends only on the net displacement between the starting and ending points.
Example 7. The 1992 world speed record for a bicycle (human-powered vehicle) was set by Chris Huber. His time through the measured 200 m stretch was a sizzling 6.509 s, at which he commented, “Cogito ergo zoom!” (I think, therefore I go fast!). In 2001, Sam Whittinggham beat Huber’s record by 19.0 km/h. What was Whittingham’s time through the 200 m?
Solution:
Huber’s speed is: \({{v}_{0}}=\frac{200\text{ }m}{6.509\text{ }s}=30.72\text{ }m/s=110.6\text{ }km/h\)
Where we have used the conversion factor 1 m/s = 3.6 km/h. Since Whittingham beat Huber by 19.0 km/h, his speed is \( {{v}_{1}}=110.6\text{ }km/h+19.0\text{ }km/h=129.6\text{ }km/h \), or 36 m/s (1 km/h = 0,2778 m/s). Thus, using Eq.2-2, the time through a distance of 200 m for Whittingham is:
\( \Delta t=\frac{\Delta x}{{{v}_{1}}}=\frac{200\text{ }m}{36\text{ }m/s}=5.554\text{ }s \)
Example 8. Two trains, each having a speed of 30 km/h, are headed at each on the same straight track. A bird that can fly 60 km/h flies off the front of one train when they are 60 km apart and heads directly for the other train. On reaching the other train, the (crazy) bird flies directly back to the first train, and so forth. What is the total distance the bird travels before the trains collide?
Solution:
Recognizing that the gap between the trains is closing at a constant rate of 60 km/h, the total time that elapses thay crash is \( t=\frac{60\text{ }km}{60\text{ }km/h}=1.0\text{ }h \).
During this time, the bird travels a distance of \( x=vt=(60\text{ }km/h)(1.0\text{ }h)=60\text{ }km \).
Example 9. Panic escape. Figure 2-24 shows a general situation in which a stream of people attempt to escape through an exit door that turns out to be locked. The people move toward the door at speed \( {{\nu }_{s}}=3.50\text{ }m/s \), are each d = 0.25 m in depth, and are separated by L = 1.75 m. The arrangement in Fig.2-24 occurs at time t = 0.
a) At what average rate does the layer of people at the door increase?
b) At what time does the layer’s depth reach 5.0 m? (The answers reveal how quickly such a situation becomes dangerous.)
Solution:
The amount of time it takes for each person to move a distance L with speed vs is \( \Delta t=\frac{L}{{{v}_{s}}} \). With each additional person, the depth increases by one body depth d
a) The rate of increase of the layer of people is
\( R=\frac{d}{\Delta t}=\frac{d}{\frac{L}{{{v}_{s}}}}=\frac{{{v}_{s}}d}{L}=\frac{(3.50\text{ }m/s)(0.25\text{ }m)}{1.75\text{ }m}=0,50\text{ }m/s \)
b) The amount of time required to reach a depth of D = 5.0 m is
\( t=\frac{D}{R}=\frac{5.0\text{ }m}{0.50\text{ }m/s}=10\text{ }s \)
Example 10. In 1 km races, runner 1 on track 1 (with time 2 min, 27.95 s) appears to be faster than runner 2 on track 2 (2 min, 28.15 s). However, length L2 of track 2 might be slightly greater than length L1 of track 1. How large can L2 – L1 be for us still to conclude that runner 1 is faster?
Solution:
Converting to seconds, the running times are t1 = 147.95 s and t2 = 148.15 s, respectively. If the runners were equally fast, then
\({{s}_{avg1}}={{s}_{avg2}}\Rightarrow \frac{{{L}_{1}}}{{{t}_{1}}}=\frac{{{L}_{2}}}{{{t}_{2}}}\)
From this we obtain
\( {{L}_{2}}-{{L}_{1}}=\left( \frac{{{t}_{2}}}{{{t}_{1}}}-1 \right){{L}_{1}}=\left( \frac{148.15}{147.95}-1 \right){{L}_{1}}=0.00135{{L}_{1}}\approx 1.4\text{ }m \)
Where we set \( {{L}_{1}}\approx 1000\text{ }m \) in the last step.
Thus, if L1 and L2 are no different than about 1.4 m, then runner 1 is indeed faster than runner 2. However, if L1 is shorter than L2 by more than 1.4 m, then runner 2 would actually be faster.
Example 11. To set a speed record in a measured (straght-line) distance d, a race car must be driven first in one direction (in time t1) and then in the opposite direction (in time t2).
a) To eliminate the effects of the wind and obtain the car’s speed vc in a windless situation, should we find the average of \( \frac{d}{{{t}_{1}}} \) and \( \frac{d}{{{t}_{2}}} \) (method 1) or should we divide d by the average of t1 and t2?
b) What is the fractional difference in the two methods when a steady wind blows along the car’s route and the ratio of the wind speed vw to the car’s speed vc is 0.0240?
Solution:
Let \( {{v}_{w}} \) be the speed of the wind and {{v}_{c}} be the speed of the car.
a) Suppose during time interval t1, the car moves in the same direction as the wind. Then the effective speed of the car is given by \( {{v}_{eff,1}}={{v}_{c}}+{{v}_{w}} \), and the distance traveled is \( d={{v}_{eff,1}}{{t}_{1}}=\left( {{v}_{c}}+{{v}_{w}} \right){{t}_{1}} \).
On the other hand, for return trip during time interval t2, the car moves in the opposite direction of the wind and the effective speed would be \( {{v}_{eff,1}}={{v}_{c}}+{{v}_{w}} \). The distance traveled is \( d={{v}_{eff,2}}{{t}_{2}}=\left( {{v}_{c}}-{{v}_{w}} \right){{t}_{2}} \).
The two expressions can be rewritten as
\( {{v}_{c}}+{{v}_{w}}=\frac{d}{{{t}_{1}}} \) and \( {{v}_{c}}-{{v}_{w}}=\frac{d}{{{t}_{2}}} \)
Adding the two equantions and dividing by two, we obtain \( {{v}_{c}}=\frac{1}{2}\left( \frac{d}{{{t}_{1}}}+\frac{d}{{{t}_{2}}} \right) \).
Thus, method 1 gives the car’s speed vc a in windless situation.
b) If method 2 is used, the result would be
\( {{{v}’}_{c}}=\frac{d}{\frac{{{t}_{1}}+{{t}_{2}}}{2}}=\frac{2d}{{{t}_{1}}+{{t}_{2}}}=\frac{2d}{\frac{d}{{{v}_{c}}+{{v}_{w}}}+\frac{d}{{{v}_{c}}-{{v}_{w}}}} \) \( =\frac{v_{c}^{2}-v_{w}^{2}}{{{v}_{c}}}={{v}_{c}}\left[ 1-{{\left( \frac{{{v}_{w}}}{{{v}_{c}}} \right)}^{2}} \right] \)
The fractional difference is
\( \frac{{{v}_{c}}-{{{{v}’}}_{c}}}{{{v}_{c}}}={{\left( \frac{{{v}_{w}}}{{{v}_{c}}} \right)}^{2}}={{(0.0240)}^{2}}=5.76\times {{10}^{-4}} \)
Example 12. You are to drive 300 km to interview. The interview is at 11:15 am. You plan to drive at 100 km/h, so you leave at 8:00 am to allow some extra time. You drive at speed for the first 100 km, but then construction work forces you to slow to 40 km/h for 40 km. What would be the least speed needed for the rest of the trip to arrive in time for the interview?
Solution:
The values used in the problem statement make it easy to see that the first part of the trip (at 100 km/h) takes 1 hour, and the second part (at 40 km/h) also takes 1 hour. Expressed in decimal form, the time left is 1.25 hour, and the distance that remains is 160 km.
Thus, a speed \( v=\frac{160\text{ }km}{1.25\text{ }h}=128\text{ }km/h \) is needed.
Example 13. Traffic shock ware. An abrupt slowdown in concentrated traffic can travel as a pulse, termed a shock wave, along the line of cars, either downstream (in the traffic direction) or upstream, or it can be stationary. Figure 2-25 shows a uniformly spaced line of cars moving at speed \( \nu =25.0\text{ }m/s \) toward a uniformly spaced line of slow cars moving at speed \( {{v}_{s}}=5.00\text{ }m/s \). Assume that each faster car adds length L = 12.0 m (car length plus buffer zone) to the line of slow cars when it joins the line, and assume it slows abruptly at the last instant. (a) For what separation distance d between the faster cars does the shock wave remain stationary? If the separationis twice that amount, what are the (b) speed and (c) direction (upstream or downstream) of the shock wave?
Solution:
a) Let the fast and the slow cars be separated by a distance d at t = 0.
If during the time interval \( t=\frac{L}{{{v}_{s}}}=\frac{12.0\text{ }m}{5.0\text{ }m/s}=2.40\text{ }s \) in which the slow car hs moved a distance of L = 12.0 m, the fast car moves a distance of \( vt=d+L \) to join the line of slow cars, then the shock wave would remain stationary.
The condition implies a separation of
\( d=vt-L=(25\text{ }m/s)(2.4\text{ }s)-12.0\text{ }m=48.0\text{ }m \)
b) Let the initial separation at t = 0 be d = 96.0 m.
At a later time t, the slow and the fast cars have traveled \( x={{v}_{s}}t \) and the fast car joins the line by moving a distance \( d+x \).
From \( t=\frac{x}{{{v}_{s}}}=\frac{d+x}{v} \)
We get \( x=\frac{{{v}_{s}}}{v-{{v}_{s}}}d=\frac{5.00\text{ }m/s}{25.0\text{ }m/s-5.00\text{ }m/s}(96.0\text{ }m)=24.0\text{ }m \)
Which in turn gives \( t=\frac{24.0\text{ }m}{5.00\text{ }m/s}=4.80\text{ }s \). Since the rear of the slow-car pack has moved a distance of \( \Delta x=x-L=24.0\text{ }m-12.0\text{ }m=12.0\text{ }m \) downstream, the speed of the rear of slow-car pack, or equivalently, the speed of shock wave, is
\( {{v}_{shock}}=\frac{\Delta x}{t}=\frac{12.0\text{ }m}{4.80\text{ }s}=2.50\text{ }m/s \)
c) Since \( x > L \), the direction of the shock wave is downstream.
Example 14. You drive on Interstate 10 from San Antonio to Houston, half the time at 55 km/h and the other half at 90 km/h. On the way back you travel half the distance at 55 km/h and the other half at 90 km/h. What is your average speed (a) from San Antonio to Houston, (b) from Houston back to San Antonio, and (c) for the entire trip? (d) What is your average velocity for the entire trip? (e) Sketch x versus t for (a), assuming the motion is all in the positive x direction. Indicate how the average velocity can be found on the sketch.
Solution:
a) Denoting the travel time and distance from San Antonio to Houston as T and D, respectively, the average speed is
\( {{s}_{avg}}=\frac{D}{T}=\frac{(55\text{ }km/h)(T/2)+(90\text{ }km/h)(T/2)}{T}=72.5\text{ }km/h \)
Which should be rounded to 73 km/h.
b) Using the fact that time = distance/speed while the speed is constant, we find
\( {{s}_{avg2}}=\frac{D}{T}=\frac{D}{\frac{D/2}{55\text{ }km/h}+\frac{D/2}{90\text{ }km/h}}=68.3\text{ }km/h \)
Which should be rounded to 68 km/h.
c) The total distance traveled (2D) must not be confused with the net displacement (zero). We obtain for the two-way trip
\( {{s}_{avg}}=\frac{2D}{\frac{D}{72.5\text{ }km/h}+\frac{D}{68.3\text{ }km/h}}=70\text{ }km/h \)
d) Since the net displacement vanishes, the average velocity for the trip in its entirety is zero.
e) In asking for a sketch, the problem is allowing the student to arbitraly set the distance D (the intent is not to make the student go to an atlas to look it up); the student can just as easily arbitrarily set T instead of D, as will be clear in the following discussion. We briefly describe the graph (with kilometers-per-hour understood for the slopes): two contiguous line segments, the first having a slope of 55 and connecting the origin to \( ({{t}_{1}},{{x}_{1}})=\left( \frac{T}{2},\frac{55T}{2} \right) \) and the second having a slope of 90 and connecting (t1, x2) to (T, D) where \( D=\frac{(55+90)T}{2} \).
The average velocity, from the graphical point of view, is the slope of a line drawn from the origin to (T,D). The graph (not drawn to scale) is depicted below:
