Untitled Flashcard Set

LESSON 1 — VELOCITY

Average velocity
v_avg = Δx / Δt

v_avg = average velocity
Δx = displacement
Δt = change in time

Displacement
Δx = x_f - x_i

x_f = final position
x_i = initial position

Change in time
Δt = t_f - t_i

t_f = final time
t_i = initial time

Speed
speed = distance / time

Speed and velocity
speed = |v|

Speed is the magnitude (absolute value) of velocity.

Slope of a position-vs-time graph
v = Δx / Δt

The slope of a position-versus-time graph represents velocity.

Sources: Lesson 1 defines displacement as the change in position and average velocity as displacement divided by elapsed time.

LESSON 2 — ACCELERATION

Average acceleration
a_avg = Δv / Δt

a_avg = average acceleration
Δv = change in velocity
Δt = change in time

Change in velocity
Δv = v_f - v_i

v_f = final velocity
v_i = initial velocity

Velocity with constant acceleration
v_f = v_i + at

v_f = final velocity
v_i = initial velocity
a = acceleration
t = time

Change in velocity
Δv = at

Displacement during constant acceleration from rest
Δx = 1/2 at²

Δx = displacement
a = constant acceleration
t = time

Average velocity during constant acceleration
v_avg = (v_i + v_f) / 2

Displacement using average velocity
Δx = v_avg t

Slope of a velocity-vs-time graph
a = Δv / Δt

The slope of a velocity-versus-time graph represents acceleration.

Area under a velocity-vs-time graph
Δx = area under the v-t graph

The signed area between the velocity graph and the time axis represents displacement.

Area under an acceleration-vs-time graph
Δv = area under the a-t graph

The signed area between the acceleration graph and the time axis represents change in velocity.

Special elevator formula from the lesson
t_tot = 2√(H / a₀)

t_tot = total trip time
H = total height
a₀ = constant acceleration during the accelerating/decelerating portions

The lesson explicitly derives this special relationship for the idealized constant-acceleration elevator model.

LESSON 3 — PROJECTILE MOTION

Horizontal position
x = x₀ + v_x t

x = horizontal position
x₀ = initial horizontal position
v_x = horizontal velocity
t = time

Horizontal velocity
v_x = constant

For ideal projectile motion, horizontal acceleration is zero.

Vertical position
y = y₀ + v_y₀t + 1/2 a_y t²

y = vertical position
y₀ = initial vertical position
v_y₀ = initial vertical velocity
a_y = vertical acceleration
t = time

Vertical velocity
v_y = v_y₀ + a_y t

v_y = final vertical velocity
v_y₀ = initial vertical velocity
a_y = vertical acceleration
t = time

For projectile motion near Earth
a_y = -g

g = magnitude of gravitational acceleration

Free-fall distance from rest
Δy = 1/2 gt²

or, when using positive downward distance:

d = 1/2 gt²

d = distance fallen
g = gravitational acceleration
t = time

Time to fall from rest
t = √(2d / g)

d = vertical distance fallen
g = gravitational acceleration

Speed from horizontal and vertical velocity components
v = √(v_x² + v_y²)

v = total speed
v_x = horizontal velocity
v_y = vertical velocity

Distance between two points
d = √[(x₂ - x₁)² + (y₂ - y₁)²]

The lesson uses the Pythagorean theorem / equivalent distance formula when comparing projectile positions.

Key projectile relationships:
Horizontal motion → constant velocity
Vertical motion → constant acceleration
Both components use the same time t.

The lesson's free-fall derivation gives the distance-from-rest equation and uses it to calculate fall time.

LESSON 4 — ENERGY CONSERVATION

Conserved kinematics quantity
v² + 2gy = constant

v = speed
g = gravitational acceleration
y = height

Equivalent form
v² + 2gy = c

c = constant

This is the main conservation relationship developed before introducing energy.

Solving for speed
v = √(c - 2gy)

Solving for height
y = (c - v²) / (2g)

Energy
E = 1/2 mv² + mgy

E = total mechanical energy
m = mass
v = speed
g = gravitational acceleration
y = height

Kinetic energy
K = 1/2 mv²

K = kinetic energy
m = mass
v = speed

Gravitational potential energy
U = mgy

U = gravitational potential energy
m = mass
g = gravitational acceleration
y = height

Total mechanical energy
E = K + U

Therefore:
E = 1/2 mv² + mgy

Energy conservation
E_initial = E_final

or

1/2 mv_i² + mgy_i = 1/2 mv_f² + mgy_f

Mass-weighted conserved quantity for multiple objects
m(v_m² + 2gy_m) + M(v_M² + 2gy_M) = constant

m, M = masses of the two objects
v_m, v_M = their speeds
y_m, y_M = their heights

Atwood-machine speed relation derived from the conservation law
v² = 2gH(M - m) / (M + m)

v = final speed
g = gravitational acceleration
H = distance the masses move
M = larger mass
m = smaller mass

Atwood-machine acceleration
a = g(M - m) / (M + m)

a = acceleration
g = gravitational acceleration
M = larger mass
m = smaller mass

The lesson modifies v² + 2gy by multiplying by each mass to analyze the Atwood machine, then connects this directly to energy.