Physics 101 Assignment #1 Solutions
Text Problems:
Chapter
2
Conceptual
Questions
10, 13, 15
Multiple Choice
Questions
16, 17, 18, 21,
22, 26
Problems
7, 17, 23, 47
Conceptual Questions
10. Figure Q2.10 shows the position graph for an object moving along the horizontal axis.
x
t
Figure Q2.10
a. Write a realistic motion story for an object that would have this position graph.
The position graph presents an object that is stationary some distance away from the origin for
two time units and then rapidly approaches and crosses the origin at a constant negative velocity
for 1 time unit arriving at a point equal distance from the origin as its starting distance. The
object then remains stationary for 3 time units and then slowly moves back towards the origin at
a constant positive velocity where it stops 3 time units later. The magnitude of the negative
velocity is 6 times larger than the magnitude of the positive velocity. Any story that captured
these points will be correct.
b. Draw the corresponding velocity graph.
Velocity Graph
v
t
13. Figure Q2.13 Shows a position versus time graph for a moving object. At which lettered points
a. Is the object moving the fastest?
Position D is the fastest position because the slope of x vs t is the greatest.
b. Is the object moving to the left?
Position C, D, & E represent motion to the left (negative x) because the slope of x vs t is
negative (e.g. velocity is negative)
c. Is the object speeding up?
The object is speeding up between points B and C since the slope (Velocity) is increasing
in magnitude in this interval.
d. Is the object slowing down?
The object is slowing down between A and B and also between E and F because the slope
of x vs t is decreasing in magnitude in these intervals.
e. Is the object turning around?
The object is turning around at point B because the velocity (slope of x vs t) is changing
sign (from positive to negative) at this point.
x
B
A
C
t
D
Figure Q2.13
E
F
15. A car traveling at velocity v takes distance d to stop after the brakes are applied. What is the
stopping distance if the car is initially traveling at velocity 2v? Assume that the acceleration due
to braking is the same in both cases.
The relationship between velocity, displacement and acceleration in a constant acceleration
scenario is
v 2f  vi2  2a  x
The final velocity is zero (rest), the displacement is given as
v 2f  vi2  2a  x
0  vi2  2a  x
vi2
2a
 x  vi2
x  
The negative sign will disappear because the acceleration is negative (a deceleration). So if a is
constant, the displacement x is proportional to the initial velocity squared. So if the initial
velocity is twice as large in the second circumstance of the problem, the stopping distance (x)
will be 22 or 4 times larger. Doubling your speed will quadruple your stopping time – something
we were taught in Driver Education classes or DWI school.
This result could be derived more explicitly using a ratio technique that is a very simple yet
powerful way to solve problems like this. For example, let x1 be the stopping distance when the
initial velocity is v and x2 be the stopping distance when the initial velocity is 2 v. Take the ratio
of the two stopping distances and the velocity and acceleration terms cancel. See below.
2v 
v2
x1  
and x 2  
2a
2a
2
2v 

2
x 2

2v 
2 a 4v 2 2 a 4
2
a







 4
x1
2a
2a v 2 1
v2
v2

2a
2
Explicitly we see that the stopping distance is four times larger at twice the initial velocity.
Multiple Choice Questions
16. Figure Q2.16 shows the position graph of a car traveling on a straight road. At which labeled
instant is the speed of the car greatest?
x
E
D
C
A
B
t
Figure Q2.16
The speed is greatest at point C because that is where the slope of x vs t is greatest.
17. Figure Q2.17 shows the position graph of a car traveling on a straight road. The velocity at
instant 1 is ____________________ and the velocity at instant 2 is ________________.
x
1
2
t
Figure Q2.17
The velocity at instant 1 and 2 are both negative since the slope of x vs t is negative at both
points.
A. positive, negative
B. positive, positive
C. negative, negative
D. negative, zero
E. positive, zero
18. Figure Q2.18 shows an object’s position-versus-time graph. What is the velocity of the object
a t = 6 s?
The slope needs to be calculated from the x vs t graph to yield the velocity.
Slope 
x x f  xi 38 m  5 m
m


 3.3
t
t
10 s
s
The velocity is 3.3 m/s.
A. 0.67 m/s
B. 0.83 m/s
C. 3.3 m/s
D. 4.2 m/s
E. 25 m/s
21. Velocity-versus-time graphs for three drag racers are shown in Figure Q2.21. At t = 5.0 s,
which car has traveled the furthest?
Betty will have the traveled the farthest because (1) she sustained the highest velocity the longest
period of time, and (2) the area under her v vs t curve is the greatest (Recall: Area under the v vs
t curve equals the displacement.
A. Andy
B. Betty
C. Carl
D. All have traveled the same distance.
22. Which of the three drag racers in Question 21 had the greatest acceleration at t = 0 s?
Again, Betty had the greatest acceleration “off the line” at t = 0 s because the slope of the v vs t
curve is greatest at t = 0 s for Betty compared to the other drivers. It is slightly non-intuitive that
acceleration can be large even when the initial velocity is zero.
A. Andy
B. Betty
C. Carl
D. All have the same acceleration.
26. Figure Q2.26 shows a motion diagram with the clock reading (in seconds) shown at each
position. From t = 9 s to t = 15 s the object is in the same position. After that, it returns along the
same track. The positions of the dots for t>15 s are offset for clarity. Which graph best represents
the object’s velocity?
The motion diagram of this problem indicates that the velocity stars large and then decreases
from t = 0 s to t = 9 seconds (judging from the decreasing displacement between dots). The
particle is then at rest (v = 0) for 6 seconds and then reverses sign (heading back toward the
starting point at a constant velocity (judging by the equal displacements) between t = 15 s and t =
23 s.
The only graph that fits this interpretation is graph C
Problems
7. Larry leaves home at 9:05 a.m. and runs at constant speed to the lamppost shown in Figure
P2.7. He reaches the lamppost at 9:01 a.m., immediately turns, and runs to the tree, again at
constant speed. Larry arrives at the tree at 9:10 a.m. What is Larry’s velocity during each of
these two intervals?
17. Figure p2.17 shows the velocity graph of a bicycle moving along the x-axis. Its initial position
is x1 = 2.0 m at t1 = 0.0 s. At t = 2.0 s, what are the bicycle’s (a) position, (b) velocity, and (c)
acceleration?
The secret here is too determine the bicycle’s acceleration from the v vs t graph. All things are a
is known.
The acceleration is best determined by finding the slope of the v vs t graph.
m
v
m
a
 s 2 2
t
3s
s
6
The acceleration is 2 m/s2.
Therefore at t= 2.0 s, the velocity and position are given by
v f  vi  ai  t  0m / s  2.0m / s 2  2.0 s  4m / s
1
2
 2.0m / s 2  2.0s   6m
2
The velocity is m m/s and the position is 6 m after 2 s
x f  2.0m  0m / s  2.0 s 
p.s. You could have read velocity off the chart.;
23. A car travels with constant acceleration along a straight road. It was at the origin and at rest
at t1 = 0 s; its position is x2 = 144 m at t2=6.00 s.
a. What was the car’s position at t = 3.00 s?
b. What will its position be at t = 9.00 s?
This problem requires the use of one of the constant acceleration equations that are summarized
on page 60 of the text under APPLICATIONS. The three equations are
v f  vi  ai  t
x f  xi  vi  t 
1
a  t 2
2
v 2f  vi2  2a  x
In this problem we are given
ti = 0 s
xi = 0 m
vi = 0 m/s
tf = 6.00 s
xf = 144 m
1
a  t 2 ) is the one
2
we want to use, but we were not given the acceleration. If there was a way to determine the acceleration
a, then the problem is easy because the constant acceleration equation, with xi and vi equal to zero,
becomes simply
and we are asked to find a later position. The middle equation ( x f  xi  vi  t 
xf 
1
a  t 2 .
2
So the question becomes how we find the acceleration with what we are given. The secret to finding the
acceleration in this problem is to remember that, for constant acceleration as specified in this problem,
the velocity is linear with time. With a linear velocity increase we can write that the average velocity
over a time interval is just one-half the sum of the initial and final velocities over the interval. And the
average speed is also given by the displacement over the interval divided by the time. Mathematically,
v AVERAGE 


1
vi  v f .
2
and
v AVERAGE 
x
t
We know that the car starts from rest and accelerates for 6.00 s, reaching a position of 144 m. The
average speed over is interval then 144m divided by 6.00 seconds or 24 m/s. We place this into the
equation
1
v AVERAGE  vi  v f . .
2
This yields
m 1 m

24  0  v f . .
s 2 s



Solving for vf gives vf = 48 m/s.
So the final velocity after the first 6 seconds is 48 m/s. This means the acceleration must be
m
48
v
s 8 m
a

t
6s
s2
Now we just use x f 
1
a  t 2 , plugging in the appropriate values for time to find the final position.
2
1 m
1 m
2
2
x f   8 2  3.0 s  and x f   8 2  9.0 s 
2 s
2 s
x f  36 m
and
x f  324 m
The position of the car after 3.0 seconds of travel is 36 m and the position of the car after 9.0 seconds is
324 m.
47. The takeoff speed for an Airbus A320 jetliner is 80 m/s. Velocity data measured during
takeoff are as follows:
t (s)
v (m/s)
0
0
10
23
20
46
30
69
a. What is the takeoff speed in miles per hour?
The inside cover of our text gives some conversion factors. The metric speed of 80 m/s is
equivalent to (80 x 2.24 = ) 179.2 miles per hour.
b. What is the jetliner’s acceleration during takeoff?
Examining the takeoff velocity table indicates that the velocity changes by 23 m/s every
10 seconds. This then yields an acceleration of
m
23
v
s  2 .3 m
a

t
10 s
s2
The jetliner accelerates at 2.3 m/s2 during takeoff.
c. At what time do the wheels leave the ground?
`
We need to solve for the time when the velocity of the jetliner is 80 m/s (takeoff speed).
v
The simplest way to do this is to rearrange the definition of acceleration a 
to
t
v
v
read t 
. (Notice that the units of
will work out to be seconds. Try it.). Use this
a
a
relation to find time given that v = 80 m/s.
v
t 

a
m
s  34.8 s
m
2.3 2
s
80
The jetliner will need 34.8 seconds to reach takeoff speed and have the wheels leave the
ground.
d. For safety reasons, in case of an aborted takeoff, the runway must be three times the
take off distance. Can an A320 takeoff safely on a 2.5 mile long runway?
This question s is really asking us to first find the length of runway needed to achieve
takeoff speed. Once we find this minimum length, we’ll multiple it by three and compare it to a
2.5 mile runway. The simplest way to solve for this minimum length is to use the answers to the
last questions. We’ll plug acceleration we found and the minimum time required into
xf 
1
1
m
2
a  t 2   2.3 2  34.8 s   1,393 m  1.39 km
2
2
s
And we find the minimum runway length is 1.39 km or 0.863 miles (using the conversions in the
inside cover of the text). The safe runway length is then 2.59 miles. So the 2.5 mile long
runway in the question is too short, by a smidgen, to launch this jetliner safely.