LGHS Physics
Linear Motion
MR. Manildi
Room 107
Physics Equation Summary
Equation
When to use the equation
Example
∆x = xi - xf
To find the distance from the
starting point
A person begins at the 5 meter
mark and walks backward to the
one meter mark, the
displacement is -4 meters.
1. A runner covers 200 m in
20 seconds. The running
speed is 10 m/s.
Displacement
Average velocity
To find the velocity when
there is no acceleration or to
find the average velocity given
the total displacement and time
∆v = at
To find the change in velocity
Instantaneous velocity if
acceleration
when there is acceleration
Where ∆v
To find the velocity that an
object will have after a given
time in free fall. This works
when there is no air resistance
here at the surface of the earth.
A ball falls for 3 seconds. It will
be going nearly 30 m/s after 3
seconds.
To find the displacement of
an object (how far) when there
is acceleration for a given
amount of time.
A car can go from zero to 30 m/s
in 3 seconds. So the acceleration
is 10 m/s2 and the distance is 45
meters.
Vave = ∆x/t
= vf - vi
When starting from rest, vi = 0
** for gravity where a = -9.8m/s2
∆x = vot + ½ at2
Displacement if acceleration
When starting from rest, vo = 0
** for gravity where a = -9.8m/s2
ΣF = ma
Newton’s Second Law
Weight = -9.8N/kg (mass)
1 N/kg = 1 m/s2
Centripetal acceleration
ac =
v2/r
for a given time.
To find the height that an object
will free fall. This works when
there is no air resistance here at
the surface of the earth.
To find the acceleration of an
object you need the NET FORCE
F and the mass.
Calculate weight from mass.
-9.8 N/kg is only for the surface
of Earth at Sea Level.
To find the acceleration of an
object moving with a constant
speed in a circular path.
A car can go from zero to 30 m/s
in 3 seconds. So the acceleration
is 10 m/s2
A ball falls for 3 seconds. It will
be going 30 m/s after 3 seconds
and fall 45 meters.
What is the net force on a
skydiver that reaches terminal
velocity? F(net) = 0, so there is
no acceleration
What is the Net force on a
Rocket that accelerates at 5
m/s/s and has a mass of 10 kg?
50 N. What is the weight of the
rocket? 100 N. What is the
Thrust of the Rocket? 150 N
What is your weight if your mass
is 65 kg? -637 N
A car going 20 m/s over a hill
with radius of 50 meters: 8m/s2
LGHS Physics
MR. Manildi
Room 107
To find the centripetal force of
an object moving with a
constant speed in a circular
path.
To find the gravitational force of
attraction between any two
objects at a distance “r”
A car 1000 kg car going 20 m/s
over a hill with radius of 50
meters: 8000 Newtons
G = 6.67 x 10-11 Nm2/kg2
Universal Gravitation Constant
Measured by Cavendish
Acceleration due to gravity
on the surface of Earth
We usually round this to
10 m/s/s for Earth.
This depends on the Mass of
Earth and the distance from the
earth but is accepted as constant
on the surface of earth.
ΣFc = m v2/r
FG = G
𝒎𝟏 𝒎𝟐
𝒓𝟐
Law of Universal Gravitation
The distance between them is
‘r’
If the force between the two
objects is F, and you double both
masses and ½ the distance, what
is the new force? 16 F
CONSTANTS
g = -9.8 m/s2