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ap c mech physics work paper
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Motion along a straight line (one dimension) is described in terms of
the position x(t) of the object at time t with respect to the origin, the
d.
instantaneous velocity " ~ a? and the instantaneous acceleration
do
a dt * Speed is the absolute value of the velocity. Average velocity
and acceleration are defined in terms of the net change of the
displacement and velocity, respectively. When the acceleration is
constant (UAM), the average and instantaneous accelerations are
equal, and the position and instantaneous velocity of the object can be
found in terms of the acceleration, and the initial position and velocity.
An important physical example of UAM is “free-fall,” the motion of an
object under the influence of gravity alone near the surface of the
Earth.
PRACTICE EXERCISES
Multiple-Choice Questions
1. The position of an object is given by the equation x(f) = 2 + 4t-
#2, where position is measured in meters and time in seconds.
What is the particle’s average acceleration from f= 0 to t= 2?
(A) -4 mis@
(B) -2 m/s?
(C) 0 m/s2
(D) 2 m/s2
(E) 4 mis?
2. A. 400-kg car can accelerate from rest to a final speed v over a
distance d. To what speed can the car accelerate within a distance
of 2d (again, starting from rest)? Assume the same value of
acceleration in both cases. i
(a) V 2u
(B) av
wa
(©) 2V2v
(D) 4v
(E) 8v
3. Figure 2.4 compares the v(f) curves of two objects. The area
under both curves from ¢ = 0 to t= tfis the same. Which of the
following quantities is the same from t= 0 to t= t/?
(A) average position
(B) average velocity
(C) average acceleration
(D) total displacement
(E) both (B) and (D) are correct
Object 2
Object 1
Figure 2.4
4. In the situation introduced in question 3, if both objects start at x =
0 at t= 0, which of the following statements is correct? (A) Object 1 will be ahead of object 2 for the entire interval, 0 mw>rm p>
ANSWERS EXPLAINED
Multiple-Choice
G=(u 10) [
1. (B) The formula for average acceleration is a= (vj ~ ro) /At.
We can obtain the velocity at f = 0 and t = 2 by differentiating the
position equation to obtain v(t) = 4 - 2t. Therefore, v(t = 0) = 4 m/s
and v(t = 2) = 0 m/s, such that the average velocity is
we can differentiate the position function twice to find
that a(t) = Because the acceleration is constant, this value
must equal the average acceleration between any two times. 2 — a2
, _ w=w~t2aAz. |.
(A) Consider the equation / 0 In this case,
2 2 = QaAz.
% = 9) 56 that YF ~ “°9*: Doubling the displacement
i /2
increases the speed by a factor of V +:
. (E) The area under the v(t) curve is equal to the net displacement.
Therefore, if both curves have the same enclosed area at t7, both
particles will have experienced the same net displacement. But in
that case their average velocities are also the same
. (A) Object 2 begins with a smaller velocity than object 1, so
initially object 1 must be in the lead. However, eventually object 2
begins moving at a speed greater than object 1, so it begins to
catch up. Because both objects have the same displacement at ty,
object 2 catches up exactly at tz. Therefore, throughout the entire
interval, object 1 is in the lead.
. (B) From calculus, the average value of a function is given by the
formula
Using this formula here yields
f (443? )at
7_a _
f= —=3
(In the language of kinematics, this is equivalent to integrating v(t)
to get the displacement, Ax, and dividing the displacement by the
time increment to obtain the average velocity.) 6. (E) For constant acceleration, the position as a function of time is
= + 1442
given by ” (t) = 20 + vot + pat * Substituting x = 2 at f= 1,
we find that the acceleration is 4 m/s?. The velocity as a function
of time is then given by the equation v(f) = vg + at = 0 + (4)t
Substituting into this equation, we find that the object’s velocity at t
=2sis 8 mis.
7. (A) The velocity is the derivative of the position function.
Therefore, if the velocity approaches a constant positive value, the
position will approach a constant positive slope. This leaves
choices (A) and (B). The slope of the v(t) curve is initially positive,
meaning that the acceleration is positive and thus the second
derivative of the position is positive. Because the second
derivative is positive, the position curve will initially be concave
upward, which is true only of choice (A).
8. (D) v= dx/dt. This is really a calculus problem: How many of the
derivatives correctly match their functions? In choice (A), the
position is a positive sine function, so the derivative is a positive
cosine function with the same period. In choice (B), the position
function’s slope increases from zero to a constant value and then
remains at the constant value. In choice (C), the position function
is a concave-upward parabola whose slope is initially zero.
Because the position function is quadratic, the velocity is linear
(beginning at zero and increasing at a constant rate). In choice
(D), the position function is a concave-downward parabola whose
slope is initially positive. Because the position function is
quadratic, the velocity is linear (beginning at a positive value and
decreasing at a constant rate). In choice (E), the initial value of
vit) is negative, but the slope of x(f) is initially zero (rather than
being negative, as it should be). Therefore, this pair does not
match.
Free-Response
1. (a) Let's call the balloon's direction of motion y, set the +y
direction pointing upward, and let the ground lie at y = 0. In this case, we have yo = 10 m, vo = 2 mis, and a= -9.8 m/s2. We can
plug this into the x(f) equation for UAM:
= yo + vot + dat? = 10m + (2m/s)t — (4.9m/s*) e
We can solve for the time when the balloon will reach the height of
the pedestrian’s head by setting y(f) = 1.2 m:
1.2m = 10 m+ (2 mis)t- (4.9 m/s2)t2
Solving this equation with the quadratic formula yields t = 1.56 s.
Because this value is greater than 1 s, the Physics C student
misses the pedestrian. Note that when we use the quadratic
formula, we also obtain a negative solution, which we discard
(clearly this solution is meaningless)
Note: What happens if we choose y = 0 at the top of the
building? This is equally valid so long as we are consistent
about our conventions. For example, if we set y = 0 at the top of
the building, in order to calculate the time when the balloon will
be at pedestrian height, we must solve for y(f) = -10.0+1.2 =
8.8m
(b) This is a very common problem. The key is to realize that
when the ball is at its maximum height, the y-component of the
instantaneous velocity is zero. Therefore, we can solve for the
time when the ball is its maximum height by solving for when v = 0
(since there is no x-component of velocity in this problem):
V=vo + at=2 mis - (9.8 m/s2)t= 0
This yields t= 0.20 s.
(c) What is the acceleration at this point? The velocity is
instantaneously zero, but the acceleration is always -9.8 mis2,
even at the peak of the trajectory. 2. (a) Imagine a car undergoing each of these scenarios: (A)
accelerating to the right (speed increases), (B) moving to the right
and braking (speed decreases), (C) accelerating to the left (speed
increases), (D) moving to the left and braking (speed decreases),
(E) moving to the right with constant speed, (F) moving to the left
with constant speed, (G) accelerating to the right from rest (speed
increases), (H) accelerating to the left from rest (speed increases).
Note: If the velocity is parallel to the acceleration, the speed
increases; if the velocity is antiparallel to the acceleration, the
speed decreases. In Chapter 5 we will see that this can be
understood in terms of power.
(b) This is much simpler than the last example: Velocity increases
whenever acceleration is positive, decreases whenever
acceleration is negative, and stays the same when acceleration is
zero. Thus, velocity increases for (A), (D), and (G), decreases for
(B), (C), and (H), and stays the same for (E) and (F)
3. (A) and (H), (B) and (E), (C) and (G), (D) and (F). In graph (A) the
object starts with zero velocity and then moves to the right. The
slope of the graph is constant (e.g., the object's acceleration is
uniform), so that x(f) is a parabola with zero slope at t = 0 [graph
(H)]. In graph (B) the object moves first with decreasing speed to
the right and then switches direction (where the curve crosses the
axis) and moves to the left. This motion corresponds to what is
shown in graph (E) (a parabola whose slope is initially positive at t
= 0 and then becomes negative). Graph (C) looks similar to graph
(A), with the exception that the initial velocity is nonzero. Thus, it
corresponds to graph (G), where the parabola has a positive slope
at t= 0. Finally, in graph (D), the velocity is constant, so that the
x(t) graph is a straight line [graph (F)]
4. As a general approach, define two equations for x(t), one for Bond
and another for Gump, and solve for the time when they are equal
(the time when Forrest catches up with Bond). Substituting this
time into the derivative of each of the position equations yields the
velocity of each runner at that time, and then subtraction gives the
difference in their speeds. Let f= 0 when the race begins. Then, the x(f) equation for Bond is
®pond = @o + Vot + Fat? = 4 (15 m/s?) e
Let t = 0 when Gump begins to run. Then, x(t’) for Gump is
Zcump = Zo + vot! + $at!? = 4(3 m/s”) t?
We have to define xBong and XGump in terms of the same time
parameter. This can be done using the relation f = t' + 3s because
Gump starts running 3 s after Bond. Substituting this into xGump
yields
ZGump = 3 (3m/s*) t= 4 (3m/s*) (oo 3s)”
Looking ahead, we don't bother expanding, but immediately solve
for xBond = XGump:
(sms?) (t— 38)? =apona = 5 (15 m/s?) e
Gump =
Taking the square root yields a linear equation that can be quickly
solved to yield the time when Gump passes Bond, fpass = 10.2 s.
Bond's velocity at this time is
Bond = i =+ [3 (1.5m/s*) 2] = (1.5m/s*) ¢
Upond (t = 10.28) = 15.3m/s
Gump’s velocity at this time is Youn = “St = 4 [2 (3m/s*) (t-39)"] = (8m/s?) (8)
UGump (t = 10.28) = 21.6m/s
Therefore, Gump is running faster than Bond by 21.6 m/s - 15.3
m/s = 6.3 m/s. Intuitively, it makes sense that Gump is running
faster (otherwise, how could he possibly catch up?)
CHALLENGE
5. (a) In time interval tgown; the rock falls and hits the water. In time
interval tup, the sound travels back up the well. We need to
express both fqown and typ in terms of the depth of the well, set
their sum equal to 20 s, and solve for the depth of the well
Calculating tgqwn (time from dropping the rock until it hits the
water): Setting y = 0 at the top of the well and +y pointing down,
4 (9.8m/s”) e
Calculating tup: Sound moves at constant speed, so we can
simply divide length by velocity to calculate the time
(mathematically, if a = 0, then Ax = vAf):
y =o + vot + dt
Yowell = (4.9m/s’) ? town 2 town =
Yaeell
340 m/s
tup =
Summing and setting the equation equal to 20 s, Yevell
20s = tdown + tup = 340 m/s
We can get rid of the square root by isolating it and squaring:
2
2
(o0s- sts)" = ( 5)
400s? + ‘all wet __ Yell
115,600 m2 /s? Ww 49 m/s
We can solve with a quadratic equation, obtaining Ywel| = 1287 m
and 35,904 m. By squaring, we have introduced an extra
mathematical solution. Plugging these two solutions into the
equation before squaring indicates that only the solution 1287 m
works. From the result we find tgown = 16.2 s.
(b) The velocity is given by v= vg + aldown, which yields |v| = 0 +
(9.8)(16.2) = 159 m/s.
(c) The acceleration of an object in free-fall is always g = 9.8
mis. Two-Dimensional Kinematics 3
> INSTANTANEOUS VELOCITY, SPEED, AND ACCELERATION IN TWO.
DIMENSIONS
> UNIFORMLY ACCELERATED MOTION (UAM) INCLUDING PROJECTILE
MOTION
~ RELATIVE POSITION, VELOCITY, AND ACCELERATION
> UNIFORM CIRCULAR MOTION (UCM)
Velocity and Acceleration Are Really Vectors As discussed at the
beginning of the last chapter, velocity and acceleration are actually vectors
(although in one dimension, there is no need for formal vector notation, so
this is easily confused). In two dimensions, more formal vector notation is
required.
Defining the Position Vector A natural way to describe the position of an
object in more than one dimension is to define a position vector that points
from the origin to the location of the object. Its magnitude gives the
distance from the origin to the object's location. Mathematically, in two
dimensions,
Position vector =
=2i+yj
Complete Definitions of Velocity, Speed, and Acceleration Again, we'll
start by defining velocity, speed, and acceleration
Instantaneous velocity:
VG) = fim = 2 oy ar fa
[row
Instantaneous spe
