Ch 1 (Etkina) Notes Flashcards
When describing motion, we need to focus on two important aspects: the … whose motion we are describing (the …) and the … who is doing the describing (the …)
object; object of interest; person; observer
Motion is a change in an object’s …. relative to a …. during a certain change in ….
position; given observer; time
Without identifying the observer, it is impossible to say whether the object of interest
moved
physicists say motion is …., meaning that the motion of any object of interest depends on the …. of the observer
relative; point of view
people intuitively use … as the object of reference- the object with respect to which they …
earth; describe motion
what we view as the object of reference influences how we
describe motion
specifying the observer before describing the motion of an object of an interest is an extremely important part of constructing what physicists call a
reference frame
a reference frame includes an …, a …. with a … for measuring …, and a … to measure …
object of reference; coordinate system; scale; distances; clock; time
if the object of reference is large and cannot be considered a point-like object, it is important to specify where on the object of reference the … of the coordinate system is placed
origin
(reference frame includes) an object of reference with a specific …. on it
point of reference
(reference frame includes) a coordinate system, which includes one or more …., and an …. located at the … it also includes a … for specifying … along the xes
coordinate axes; origin; point of reference; scale; distances
(reference frame includes) a clock which includes an … called … and a unit of measurement for specifying … and ….
origin in time; t=0; time; time intervals
one-dimensional/linear motion is a model of motion that assumes that an object, considered as a …., moves along a …
point-like object; straight line
when the object travels at constant speed, the dots are
evenly spaced
when the object slows down, the dots get
closer
when the object moves faster and faster, the dots get
farther apart
we can represent motion in even more detail by adding … to each dot that indicate which … the object is moving and how … it is moving as it passes a particular position
velocity arrows; way; fast
dot diagrams + velocity arrows is
motion diagram
the longer the arrow, the … the motion
faster
vectors have … and …
direction; magnitude
velocity change arrow (Δv→) indicates change in … on …
velocity; motion diagram
a velocity change arrow is a … indication
qualitative
a velocity change arrow points in the same direction as the velocity arrows when object is …. it points in the opposite direction when the object is …
speeding up; slowing down
when velocity changes are the same for each time interval, all of the velocity change arrows will be the …, so we only need use … for the whole motion diagram
same length; one
time described in two ways:
time: the … clock reading
time interval (…): the … clock readings
actual; Δt; difference in
time and time interval are
scalar quantities
position: object’s … with respect to the …
location; coordinate system
displacement: … quantity that defines the change in … (…)
vector; position; Xf - Xi
distance: … of …
magnitude; displacement
path length: how … the object moved as it traveled from …
far; initial position to final position
the quantity determined through subtracting Xf-Xi is called …. or simply …., appreviated…
x-scalar component of the displacement; x-component of the displacement; dx
kinematics: description of
motion
the slope of an x-t graph equals … = ….
X2- X1 / t2 - t1 = Δx/Δt m/s
the slope of an x-t graph indicates how the object’s … changes with respect to …, tells … relative to ….
position; time; direction of motion; coordinate axis
the slope of an x-t graph is equal to
velocity
velocity = .. + …
speed; direction
V= Δx/Δt is only for
constant velocity linear motion
speed is the … and is always …
magnitude of velocity; positive
V=Δx/Δt can be arranged to devise the following equation:
X2 = X1 + Vx(t2-t1)
applying X2 = X1 + Vx(t2-t1) for time zero, when initial position is X0, the equation can be written as
X= X0 + Vx*t
Rearranging the equation for an object’s position (X= X0 + Vx*t ) yields …, this is … and is also the … on the v-t graph
x - x0 = vx*t;
displacement; area
for motion with constant velocity, the magnitude of the displacement of an object is the … between a v-t graph line and the …. between the time readings
area; time axis
displacement is the area with a + sign when velocity is …, and area with - sign when velocity is …
positive; negative
instantaneous velocity: velocity of an object at a
particular time
for accelerating objects, v = Δx/Δt can’t be used to determine …, but it can be used to find …
instantaneous velocity; average velocity
acceleration characterizes the rate at which
velocity of an object is changing
when object is moving along a straight line and the slope of the v-t graph is constant, acceleration follows this formula:
ax = Δvx/ Δt
average acceleration during a time interval: a =
Δv/ Δt
If Δt is very small, acceleration defined by the equation Δv/Δt is .., not …
instantaneous; average
units of acceleration:
m/s^2
it’s possible for an object to have … velocity and … acceleration: e.g. when an object starts ..
zero; nonzero; moving from rest
when object starts moving from rest, Vi is
0
an object can have .. velocity and … acceleration eg. if an object moves at …
nonzero; zero; constant velocity
If, for linear motion, t0=0, acceleration can be defined as:
ax = (vx - v0x)/ t - 0
ax = (vx - v0x)/ t - 0 can be arranged to get
Vx = V0x + ax*t
(displacement from constant a) for constant acceleration, noting that displacement is the area for the v-t graph, we can find this area by separating the graph into a
triangle and rectangle
(displacement from constant a) area of the triangle is
1/2bh
(displacement from constant a) base in this case is … which equals …, as initial time starts at … and is …
Δt; t - 0; origin; 0
(displacement from constant a) height in this case
Δv
(displacement from constant a) area of triangle is therefore
A = 1/2(t - 0)(vx - v0x)
(displacement from constant a) vx - vox = ax*t so area of triangle can be rewritten as
A = 1/2(t)(axt) = 1/2ax*t^2
(displacement from constant a) area of rectangle is width * height which therefore =
Arect = Vx0 * (t-0)
(displacement from constant a) thus, the area between curve and time axis (displacement) is:
x - x0 = arect + atriangle
x = x0 + V0xt + 1/2ax*t^2
alternate equation for linear motion with constant acceleration
2ax (x - x0) = (Vx)^2 - (V0x)^2
we can represent the motion of a falling ball mathematically using the equations of motion for constant acceleration with Ay =
9.8 m/s^2
For free fall (vertical motion), Vy =
V0y + ayt = V0y + (9.8 m/s^2)t
for free fall, displacement equation with acceleration becomes, y =
y0 + v0yt + 1/2(9.8 m/s^2)t^2
the magnitude of an object’s acceleration while falling without air resistance is given a special symbol, …, where it = …
g; 9.8 m/s^2
for an object that is at rest and remains at rest, the instant the object is not moving, its acceleration =
0 m/s^2
