Motion (Uniform And Accelerated) Flashcards
Velocity functions
A change in position over time
Position functions
The location of a given object at a point in time
Acceleration Functions
A change in velocity over time
Displacement
The net change in position
d=Δs
Displacement formula (given acceleration)
Δs=vit+(at^2)/2
Displacement formula (without acceleration)
Δs=t*(vi+vf)/2
Final position
Sf=v*t+si
Calculating the position function from the acceleration function
s(t)=∫∫a(t)
Average postion
Savg=(Sf+Si)/2
Average velocity (from position values)
Vavg=Δs/Δt
Average velocity (from velocity values)
Vavg=(vi+vf)/2 otherwise Vavg=(x-xi)/t
Final velocity (given time)
vf=a*t+vi
Final velocity (given displacement)
vf=√(2aΔs+vi^2)
Calculating the velocity function from the position function
v(t)=∂s(t)
Calculating the position function from the velocity function
s(t)=∫v(t)
Calculating the velocity function from the acceleration function
v(t)=∫a(t)
Average acceleration (given displacement)
Aavg=Δs/(Δt)^2
Average Acceleration (from velocity values)
Aavg=Δv/Δt
Calculating Acceleration (without time)
a=(vf^2-vi^2)/(2Δs)
Calculating acceleration (without final velocity)
a=(Δs-vi*t)/t^2
Acceleration function from the velocity function
a(t)=∂v(t)
Acceleration function from the position function
a(t)=∂∂s(t)
Velocity at a given position on a graph
v=[(s(t+h)-s(t)]/h
Where ‘h’ is an arbitrarily small number
Acceleration at a given velocity on a graph
a=[(v(t+h)-v(t)]/h
Where ‘h’ is an arbitrarily small number
Acceleration at a given position on a graph
a=(([(s(t+h)-s(t)]/h)-([(s(t)-s(t-h)]/h))/h
Difference of two velocity-from-position functions, that vary by ‘h’ in two different directions, divided by ‘h’
Time (from acceleration and displacement)
t=√(Δs/a)
Time (from acceleration and velocity)
t=Δv/a
Time (from final position and velocity)
t=(Sf-Si)/v
Time (from displacement and velocity)
t=Δs/v
Time (from final position and acceleration)
t=√((Sf-Si)/a)
Time (from final velocity and acceleration)
t=(Vf-Vi)/a
Maximum hight of vertical projectiles
Write it out
Ymax=(.5)(-9.81)(vi/9.81)^2+(vi^2)/9.81+yi
Time at which a vertical projectile reaches maximum hieght
t=vi/9.81
Time at which the vertical projectile returns to yi
t=2(vi)/9.81