Part 1 – Linear Dynamics
Unit 1 - Displacement & Velocity in 1D
Speed is a scalar quantity that refers to the rate at
which an object covers distance.
∆d
Speed= >0
∆t
Displacement is a vector quantity that refers to the
object’s overall change in position. It is given by:
x (t)
Average velocity is a vector quantity and refers to the
change in displacement ∆ x divided by the change in
time ∆ t .
∆ x x final−x initial
v= =
∆ t t final−t initial
If you make the time interval smaller and smaller, you get the
instantaneous velocity, which is essentially the
derivative of displacement with respect to time. One
way to calculate it is by determining the slope of the
displacement curve.
dx (t) dx
v ( t )= =
dt dt
Note that the displacement of an object can be
calculated by looking at the area underneath the
velocity curve of an object. In other words, we can
integrate velocity over a time interval to find
displacement.
dx
=v ( t )
dt
t2 t2 t2
dx
∫( dt )
dt=∫ v (t )dt → x ( t f )−x ( t i) =∫ v ( t ) dt
t1 t 1 t 1
t2
x ( t f )=x ( t i) +∫ v ( t ) dt → x=x 0 +v ∆ t (constant velocity)
t1
Similarly, we can define acceleration as the rate of change in velocity, or
more precisely the derivative of velocity with respect to time.
Acceleration can be calculated by determining the slope of the velocity
curve.
, dv ( t ) dv
a ( t )= =
dt dt
Note that the velocity of an object can be calculated by looking at the
area underneath the acceleration curve of an object. In other words, we
can integrate acceleration over a time interval to find velocity.
dv
=a ( t )
dt
t2 t2 t2
dv
∫( dt )
dt =∫ a(t ) dt → v ( t f )−v ( t i ) =∫ a ( t ) dt
t1 t 1 t 1
t2
v ( t f )=v ( t i ) +∫ a ( t ) dt
t1
Note that the concept of “slowing down” is not the same as having a
“negative acceleration.” (Do not use the word deceleration as it can be
ambiguous what it is referring to).
o If an object slows down in the positive direction, it has a negative
acceleration.
o If an object slows down in the negative direction, it has a positive
acceleration.
Deriving equations for motion at constant acceleration:
t2
v ( t f )=v ( t i ) +∫ a ( t ) dt → v=v 0+ a ∆ t
t1
t
x=x 0 +∫ v ( t ) dt
0
t
x=x 0 +∫ (v ¿¿ 0+a ∆ t) dt ¿
0
1 2
x=x 0 +v 0 t+ a t
2
1
∆ x=v 0 t+ a t 2
2
We can eliminate t, to derive a new equation:
v−v 0
v=v 0 + a ∆ t → ∆ t=
a
2
v −v 0 1 v−v 0
∆ x=v 0 ( a
+ a
2 a) ( )
v v 0−v20 1 v 2−2 v v 0+ v 20
∆ x=
a
+ a
2 a2 [ ]
Unit 1 - Displacement & Velocity in 1D
Speed is a scalar quantity that refers to the rate at
which an object covers distance.
∆d
Speed= >0
∆t
Displacement is a vector quantity that refers to the
object’s overall change in position. It is given by:
x (t)
Average velocity is a vector quantity and refers to the
change in displacement ∆ x divided by the change in
time ∆ t .
∆ x x final−x initial
v= =
∆ t t final−t initial
If you make the time interval smaller and smaller, you get the
instantaneous velocity, which is essentially the
derivative of displacement with respect to time. One
way to calculate it is by determining the slope of the
displacement curve.
dx (t) dx
v ( t )= =
dt dt
Note that the displacement of an object can be
calculated by looking at the area underneath the
velocity curve of an object. In other words, we can
integrate velocity over a time interval to find
displacement.
dx
=v ( t )
dt
t2 t2 t2
dx
∫( dt )
dt=∫ v (t )dt → x ( t f )−x ( t i) =∫ v ( t ) dt
t1 t 1 t 1
t2
x ( t f )=x ( t i) +∫ v ( t ) dt → x=x 0 +v ∆ t (constant velocity)
t1
Similarly, we can define acceleration as the rate of change in velocity, or
more precisely the derivative of velocity with respect to time.
Acceleration can be calculated by determining the slope of the velocity
curve.
, dv ( t ) dv
a ( t )= =
dt dt
Note that the velocity of an object can be calculated by looking at the
area underneath the acceleration curve of an object. In other words, we
can integrate acceleration over a time interval to find velocity.
dv
=a ( t )
dt
t2 t2 t2
dv
∫( dt )
dt =∫ a(t ) dt → v ( t f )−v ( t i ) =∫ a ( t ) dt
t1 t 1 t 1
t2
v ( t f )=v ( t i ) +∫ a ( t ) dt
t1
Note that the concept of “slowing down” is not the same as having a
“negative acceleration.” (Do not use the word deceleration as it can be
ambiguous what it is referring to).
o If an object slows down in the positive direction, it has a negative
acceleration.
o If an object slows down in the negative direction, it has a positive
acceleration.
Deriving equations for motion at constant acceleration:
t2
v ( t f )=v ( t i ) +∫ a ( t ) dt → v=v 0+ a ∆ t
t1
t
x=x 0 +∫ v ( t ) dt
0
t
x=x 0 +∫ (v ¿¿ 0+a ∆ t) dt ¿
0
1 2
x=x 0 +v 0 t+ a t
2
1
∆ x=v 0 t+ a t 2
2
We can eliminate t, to derive a new equation:
v−v 0
v=v 0 + a ∆ t → ∆ t=
a
2
v −v 0 1 v−v 0
∆ x=v 0 ( a
+ a
2 a) ( )
v v 0−v20 1 v 2−2 v v 0+ v 20
∆ x=
a
+ a
2 a2 [ ]
