Motion along a Straight Line/ Motion in One Dimension
What is motion?
A change of position over time.
The motion is along a straight line, then it is convenient to use x-axis of the
coordinate system (which is marked as a unit of lengths) to lie along the direction of
motion.
1. Position, Time, Speed, Displacement and Velocity
 A particle’s position (𝑥) is the location of the particle with respect to a chosen
reference point that we can consider to be the origin of a coordinate system (x =
0). The motion of a particle is completely known if the particle’s position in space
is known at all times.
 Distance (𝑑 ) is the length of a path followed by a particle.
 Time (𝑡 ) is the interval over which change occur. A progression of events from
the past to the present and into the future.
 Speed (𝑠) is the distance traveled per unit of time. It is a scalar quantity that
refers to how fast an object is moving.
𝑠=
𝑑
𝑡
 The displacement (∆𝑥) of a particle is defined as its change in position in some
time interval. As the particle moves from an initial position 𝑥1 to a final
position 𝑥2 , its displacement is given by ∆x = 𝑥2 − 𝑥1 .

 Velocity (𝑣) is equal to displacement divided by time. It means velocity is the
distance traveled per unit of time in a given direction. Velocity is a vector
quantity.
𝑥2 − 𝑥1 ∆𝑥
𝑣=
=
𝑡
𝑡
 Time Interval (∆𝑡 ) is the difference between occurrences of two events measured
in time.
∆𝑡 = 𝑡2 − 𝑡1
 The average speed (𝑠𝑎𝑣 ) of a particle, a scalar quantity, is defined as the total
distance traveled divided by the total time interval (∆𝑡 ).
𝑠𝑎𝑣
𝑑
𝑑
=
=
𝑡2 − 𝑡1 ∆𝑡
2. Average and Instantaneous Velocity
Average Velocity
Is defined as the particle’s displacement ∆𝑥 divided by the time interval ∆𝑡
during displacement occurs.
𝑣𝑎𝑣 =
𝑥2 − 𝑥1 ∆𝑥
=
𝑡2 − 𝑡1
∆𝑡
Instantaneous velocity
∆x
Is defined as the limit of the average velocity ( ∆t ) as the time
interval (∆t) approaches zero. By definition, this limit equals the derivative of x with
respect to t, or the time rate of change of the position:
𝑣𝑥 = lim
∆𝑡−0
∆x 𝑑𝑥
=
∆t 𝑑𝑡
The instantaneous speed of a particle is equal to the magnitude of its instantaneous
velocity.
EXAMPLE PROBLEM 1: Average and instantaneous velocities
A cheetah is crouched 20 m to the east of an observer. At time the cheetah begins to
run due east toward an antelope that is 50 m to the east of the observer. During the
first 2.0 s of the attack, the cheetah’s coordinate x varies with time according to the
equation:
𝒙 = 𝟐𝟎 𝒎 + (𝟓. 𝟎 𝒎/𝒔𝟐 )𝒕𝟐
At 𝒕𝟏 = 𝟏. 𝟎 𝒔 and 𝒕𝟐 = 𝟐. 𝟎 𝒔
(a) Find the cheetah’s displacement; (b) Find its average velocity during time
interval; (c) Find its instantaneous velocity 𝑉𝑥
SOLUTION:
(a) Using the equation 𝒙 = 𝟐𝟎 𝒎 + (𝟓. 𝟎 𝒎/𝒔𝟐 )𝒕𝟐 the cheetah’s positions 𝑥1 and 𝑥2
are:
At 𝑡1 = 1.0 𝑠
𝒙𝟏 = 20 𝑚 + (5.0 𝑚⁄𝑠 2 ) (1.0 𝑠)2 = 25 𝑚
At 𝑡2 = 2.0 𝑠
𝒙𝟐 = 20 𝑚 + (5.0 𝑚⁄𝑠 2 ) (2.0 𝑠)2 = 40 𝑚
And the displacement during this interval is
∆𝒙 = 𝑥2 − 𝑥1 = 40 𝑚 − 25 𝑚 = 𝟏𝟓 𝒎
(𝐴𝑛𝑠𝑤𝑒𝑟)
(b) The average velocity during this interval is
𝒗𝒂𝒗 =
𝑥2 − 𝑥1
15 𝑚
=
= 𝟏𝟓 𝒎⁄𝒔
𝑡2 − 𝑡1 2.0 𝑠 − 1.0 𝑠
(𝐴𝑛𝑠𝑤𝑒𝑟)
(c) Its instantaneous velocity 𝑉𝑥
𝒗𝒙 =
𝑑𝑥
𝑑𝑡
=
𝑑
𝑚
𝑚
[20 𝑚 + (5.0 2 ) 𝑡 2 ] = 0 + [(0)𝑡 2 + (5.0 2 ) 2𝑡] = (5.0 𝑚/𝑠 2 )2𝑡
𝑑𝑡
𝑠
𝑠
At 𝑡1 = 1.0 𝑠 is
(5.0 𝑚/𝑠 2 )(2 × 1.0 𝑠) = 𝟏𝟎. 𝟎 𝒎/𝒔 (𝐴𝑛𝑠𝑤𝑒𝑟)
And at 𝑡2 = 2.0 𝑠 is
(5.0 𝑚/𝑠 2 )(2 × 2.0 𝑠) = 𝟐𝟎. 𝟎 𝒎/𝒔 (𝐴𝑛𝑠𝑤𝑒𝑟)
EXAMPLE PROBLEM 2:
A particle moves along the x axis. Its position varies with time according to the
expression 𝒙 = −𝟒𝒕 + 𝟐𝒕𝟐 where x is in meters and t is in seconds. The position–
time graph for this motion is shown in the figure below. Notice that the particle
moves in the negative x direction for the first second of motion, is momentarily at
rest at the moment 𝒕 = 𝟏 𝒔, and moves in the positive x direction at times 𝒕 > 𝟏 𝒔.
(a) Determine the displacement of the particle in the time intervals:
𝐭 = 𝟎 to 𝐭 = 𝟏 𝐬
and
𝐭 = 𝟏 𝐬 to 𝐭 = 𝟑 𝐬
(b) Calculate the average velocity during these two time intervals.
(c) Find the instantaneous velocity of the particle at 𝒕 = 𝟐. 𝟓 𝒔.
SOLUTION:
(𝐚) Using the equation 𝒙 = −𝟒𝒕 + 𝟐𝒕𝟐 :
In the first time interval ( t = 0 to t = 1 s ) we get,
∆𝒙 = 𝑥2 − 𝑥1 = [−4(1) + 2(1)2 ] − [−4(0) + 2(0)2 ] = −𝟐 𝒎 (𝐴𝑛𝑠𝑤𝑒𝑟)
For the second time interval ( t = 1 s to t = 3 s ) we get,
∆𝒙 = 𝑥2 − 𝑥1 = [−4(3) + 2(3)2 ] − [−4(3) + 2(3)2 ] = 𝟖 𝒎 (𝐴𝑛𝑠𝑤𝑒𝑟)
(𝐛) The average velocity (𝑣𝑎𝑣 ) during the first time interval ( ∆𝑡 = 1 − 0 = 1 𝑠),
𝒗𝒂𝒗 =
𝑥2 − 𝑥1
−2 𝑚
−2 𝑚
=
=
= −𝟐 𝒎⁄𝒔
𝑡2 − 𝑡1 1 𝑠 − 0
1𝑠
(𝐴𝑛𝑠𝑤𝑒𝑟)
For the second time interval (∆𝑡 = 3 − 1 = 2 𝑠),
𝒗𝒂𝒗 =
𝑥2 − 𝑥1
8𝑚
8𝑚
=
=
= 𝟒 𝒎⁄𝒔
𝑡2 − 𝑡1 3 𝑠 − 1 𝑠
2𝑠
(𝐜) The instantaneous velocity (𝑉𝑥 ) of the particle is,
𝒗𝒙 =
𝑑𝑥
𝑑
= (−4𝑡 + 2𝑡 2 ) = −4 + 4𝑡
𝑑𝑡 𝑑𝑡
At 𝑡 = 2.5 𝑠,
𝑣𝑥 = −4 + 4(2.5 ) = 𝟔 𝒎⁄𝒔
(𝐴𝑛𝑠𝑤𝑒𝑟)
(𝐴𝑛𝑠𝑤𝑒𝑟)
3. Average and Instantaneous Acceleration
Acceleration describes the rate of change of velocity with time. Like velocity, acceleration is
a vector quantity. When the motion is along a straight line, its only nonzero component is
along that line. As we’ll see, acceleration in straight-line motion can refer to either
speeding up or slowing down.
Average Acceleration
The change in the x-component of velocity, divided by the time interval ∆𝑡:
𝒂𝒂𝒗−𝒙 =
𝒗𝟐𝒙 − 𝒗𝟏𝒙 ∆𝒗𝒙
=
𝒕𝟐 − 𝒕𝟏
∆𝒕
Instantaneous Acceleration
The instantaneous acceleration is the limit of the average acceleration as the time interval
approaches zero. In the language of calculus, instantaneous acceleration equals the
derivative of velocity with time.
∆𝒗𝒙 𝒅𝒗𝒙
𝒂𝒙 = 𝐥𝐢𝐦
=
∆𝒕−𝟎 ∆𝒕
𝒅𝒕
EXAMPLE PROBLEM 3:
Suppose the instantaneous velocity 𝑣𝑥 of the car in the figure at any time t is given by the
equation
𝒗𝒙 = 𝟔𝟎 𝒎⁄𝒔 + (𝟎. 𝟓𝟎 𝒎⁄ 𝟑 ) 𝒕𝟐
𝒔
(a) Find the change in instantaneous velocity of the car in the time interval
𝒕𝟏 = 𝟏. 𝟎 𝒔
to
𝒕𝟐 = 𝟑. 𝟎 𝒔
(b) Find the average x-acceleration in this time interval.
(c) Find the instantaneous x-acceleration 𝒂𝒙 at time 𝒕𝟏 = 𝟏. 𝟎 𝒔 and 𝒕𝟐 = 𝟑. 𝟎 𝒔
SOLUTION:
(a) Before we can find the change in x-velocity (∆𝑣𝑥 ) of the car we must find the x-velocity
at each time from the given equation. At 𝒕𝟏 = 𝟏. 𝟎 𝒔 and 𝒕𝟐 = 𝟑. 𝟎 𝒔, the velocities are
𝑣1𝑥 = 60 𝑚⁄𝑠 + (0.50 𝑚⁄𝑠 3 ) (1)2 = 60.5 𝑚⁄𝑠
𝑣2𝑥 = 60 𝑚⁄𝑠 + (0.50 𝑚⁄𝑠 3 ) (3)2 = 64.5 𝑚⁄𝑠
The change in x-velocity (∆𝑣𝑥 ) at 𝒕𝟏 = 𝟏. 𝟎 𝒔 and 𝒕𝟐 = 𝟑. 𝟎 𝒔 is,
∆𝑣𝑥 = 𝑣2𝑥 − 𝑣1𝑥 = 64.5 𝑚⁄𝑠 − 60.5 𝑚⁄𝑠 = 𝟒. 𝟎 𝒎⁄𝒔
(𝐴𝑛𝑠𝑤𝑒𝑟)
(b) The average x-acceleration during this time interval (∆𝑡 ) of duration 𝒕𝟐 − 𝒕𝟏 = 𝟐 𝒔 is,
𝑎𝒂𝒗
∆𝑣𝑥 4.0 𝑚⁄𝑠
=
=
= 𝟐. 𝟎 𝒎⁄ 𝟐
𝒔
∆𝑡
2.0 𝑠
(𝐴𝑛𝑠𝑤𝑒𝑟)
(c) The instantaneous acceleration 𝒂𝒙 of the car is,
𝑎𝒙 =
𝑑𝑣𝑥
𝑑
=
[60 𝑚⁄𝑠 + (0.50 𝑚⁄𝑠 3 ) 𝑡 2 ]
𝑑𝑡
𝑑𝑡
= 0 + (0)𝑡 2 + (0.50 𝑚⁄𝑠 3 ) 2𝑡 = (𝟏. 𝟎 𝒎⁄ 𝟑 ) 𝒕
𝒔
When 𝒕 = 𝟏. 𝟎 𝒔,
𝒂𝒙 = (1.0 𝑚⁄𝑠 3 ) (1.0 𝑠) = 𝟏. 𝟎 𝒎⁄ 𝟐
𝒔
(𝐴𝑛𝑠𝑤𝑒𝑟)
𝒂𝒙 = (1.0 𝑚⁄𝑠 3 ) (3.0 𝑠) = 𝟑. 𝟎 𝒎⁄ 𝟐
𝒔
(𝐴𝑛𝑠𝑤𝑒𝑟)
When 𝒕 = 𝟑. 𝟎 𝒔,
EXAMPLE PROBLEM 4:
The velocity of a particle moving along the x axis varies according to the expression
𝒗𝒙 = 𝟒𝟎 − 𝟓𝒕𝟐 . Where 𝑣𝑥 is in meters per second and t is in seconds.
(a) Find the average acceleration in the time interval 𝒕𝟏 = 𝟎 𝒔
to
𝒕𝟐 = 𝟐. 𝟎 𝒔
(b) Determine the instantaneous acceleration at 𝒕𝟐 = 𝟐. 𝟎 𝒔
SOLUTION:
(a) The average acceleration in the time interval 𝒕𝟏 = 𝟎 𝒔
to
𝒕𝟐 = 𝟐. 𝟎 𝒔:
𝑣1𝑥 = 40 − 5(0)2 = 40.0 𝑚⁄𝑠
𝑣2𝑥 = 40 − 5(2)2 = 20.0 𝑚⁄𝑠
𝑎𝒂𝒗
𝑣2𝑥 − 𝑣1𝑥 20.0 𝑚⁄𝑠 − 40.0 𝑚⁄𝑠 −20.0 𝑚⁄𝑠
=
=
=
= −𝟏𝟎 𝒎⁄ 𝟐
𝒔
𝑡2 − 𝑡1
2.0 𝑠 − 0
2.0 𝑠 − 0
The negative acceleration means that the particle is slowing down.
(b) The instantaneous acceleration at 𝒕𝟐 = 𝟐. 𝟎 𝒔 is
𝑎𝒙 =
𝑑𝑣𝑥
𝑑
(40 − 5𝑡 2 )
=
𝑑𝑡
𝑑𝑡
= 0 + (0)𝑡 2 + (5)2𝑡 = (𝟏𝟎. 𝟎 𝒎⁄ 𝟑 ) 𝒕
𝒔
When 𝒕 = 𝟐. 𝟎 𝒔,
𝒂𝒙 = (10.0 𝑚⁄𝑠 3 ) (2.0 𝑠) = 𝟐𝟎. 𝟎 𝒎⁄ 𝟐
𝒔
(𝐴𝑛𝑠𝑤𝑒𝑟)
(𝐴𝑛𝑠𝑤𝑒𝑟)
4. Motion with Constant Acceleration
The simplest kind of accelerated motion is straight-line motion with constant acceleration.
In this case the velocity changes at the same rate throughout the motion.
If a particle moves in a straight
line with constant acceleration
𝒂𝒙 the velocity changes by
equal amounts in equal time
interval.
However, the position changes by different
amounts in equal time intervals because the
velocity is changing.
Equations of Motion with Constant Acceleration
Equation
Include Quantities
𝒗𝒙 = 𝒗𝟎𝒙 + 𝒂𝒙 𝒕
t
𝟏
𝒙 = 𝒙𝟎 + 𝒗𝟎𝒙 𝒕 + 𝒂𝒙 𝟐𝒕𝟐
𝟐
t
𝒗𝒙 𝟐 = 𝒗𝟎𝒙 𝟐 + 𝟐𝒂𝒙 (𝒙 − 𝒙𝟎 )
𝒙 − 𝒙𝟎 = (
𝒗𝟎𝒙 + 𝒗𝒙
)𝒕
𝟐
t
𝒗𝒙
x
𝒂𝒙
𝒂𝒙
x
𝒗𝒙
x
𝒗𝒙
𝒂𝒙
EXAMPLE PROBLEM 5:
A motorcyclist heading east through a small town accelerates at a constant 4.0 𝑚⁄𝑠 2 after
he leaves the city limits. At time 𝑡 = 0 he is 5.0 m east of the city-limits signpost, moving
east at 15 𝑚⁄𝑠 (a) Find his position and velocity at 𝑡 = 2.0 𝑠 (b) Where is he when his
velocity is 25 𝑚⁄𝑠?
SOLUTION:
GIVEN:
Initial position: 𝒙𝟎 = 𝟓. 𝟎 𝐦
Velocity: 𝑽𝟎𝒙 = 𝟏𝟓 𝒎⁄𝒔
Acceleration: 𝒂𝒙 = 𝟒. 𝟎 𝒎⁄ 𝟐
𝒔
(a) Since we know the values of 𝑋0 , 𝑉0𝑥 and 𝑎𝑥 the table tells that we can find position
𝑥 at 𝒕 = 𝟐. 𝟎 𝒔 by using
1
𝑥 = 𝑥0 + 𝑉0𝑥 𝑡 + 𝑎𝑥 𝑡 2
2
1
= 5.0 m + (15 𝑚⁄𝑠)(2.0 𝑠) + (4.0 𝑚⁄𝑠 2 ) (2.0 𝑠)2
2
= 𝟒𝟑 𝒎 (𝐴𝑛𝑠𝑤𝑒𝑟)
And the x-velocity at this time by using
𝑉𝑥 = 𝑉0𝑥 + 𝑎𝑥 𝑡
= 15 𝑚⁄𝑠 + (4.0 𝑚⁄𝑠 2 ) (2.0 𝑠)
= 𝟐𝟑 𝒎⁄𝒔 (𝐴𝑛𝑠𝑤𝑒𝑟)
(b) To find the position of motorcyclist when his velocity is 25 𝑚⁄𝑠 we use
𝑉𝑥 2 = 𝑉0𝑥 2 + 2𝑎𝑥 (𝑥 − 𝑥0 )
𝑉𝑥 2 = 𝑉0𝑥 2 + 2𝑎𝑥 𝑥 − 2𝑎𝑥 𝑥0
2𝑎𝑥 𝑥 = 2𝑎𝑥 𝑥0 + 𝑉𝑥 2 − 𝑉0𝑥 2
𝑥 = 𝑥0 +
𝑉𝑥 2 −𝑉0𝑥2
s
(23 𝑚⁄𝑠)2 − (15 𝑚⁄𝑠 )2
= 5.0 m +
2 (4.0 𝑚⁄𝑠 2 )
= 𝟓𝟓 𝒎 (𝐴𝑛𝑠𝑤𝑒𝑟)
2𝑎𝑥
5. Freely Falling Bodies
Free fall is a case of motion with constant acceleration. The magnitude of the
acceleration due to gravity is a positive quantity, g. The acceleration of a body in free
fall is always downward.
Equation
Include Quantities
𝒗𝒚 = 𝒗𝟎𝒚 + 𝒂𝒚 𝒕
t
𝟏
𝒚 = 𝒚𝟎 + 𝒗𝟎𝒚 𝒕 + 𝒂𝒚 𝟐𝒕𝟐
𝟐
t
𝒗𝒚 𝟐 = 𝒗𝟎𝒚 𝟐 + 𝟐𝒂𝒚 (𝒚 − 𝒚𝟎 )
𝒚 − 𝒚𝟎 = (
𝒗𝟎𝒚 + 𝒗𝒚
)𝒕
𝟐
t
𝒗𝒚
y
𝒂𝒚
𝒂𝒚
y
𝒗𝒚
y
𝒗𝒚
𝒂𝒚
EXAMPLE PROBLEM 6:
A one-euro coin is dropped from the Leaning Tower of Pisa and falls freely from
rest. What are its position and velocity after 1.0 s, 2.0 s, and 3.0 s?
SOLUTION:
At a time t after the coin is dropped, its position and y-velocity are
Position:
1
𝑦 = 𝑦0 + 𝑣0𝑦 𝑡 + 𝑎𝑦 𝑡 2
2
1
= 0 + 0 + (−𝑔)𝑡 2
2
1
= (−9.8 𝑚⁄𝑠 2 ) 𝑡 2
2
= (−4.9 𝑚⁄𝑠 2 ) 𝑡 2
𝐕𝐞𝐥𝐨𝐜𝐢𝐭𝐲
𝑣𝑦 = 𝑣0𝑦 + 𝑎𝑦 𝑡
= 0 + (−𝑔)𝑡
= (−9.8 𝑚⁄𝑠 2 ) 𝑡
When 𝐭 = 𝟏. 𝟎 𝐬
𝑦 = (−4.9 𝑚⁄𝑠 2 ) (1.0 s)2 = −𝟒. 𝟗 𝒎
𝑣𝑦 = (−9.8 𝑚⁄𝑠 2 ) (1.0 s) = −𝟗. 𝟖 𝒎⁄𝒔
When 𝐭 = 𝟐. 𝟎 𝐬
𝑦 = (−4.9 𝑚⁄𝑠 2 ) (2.0 s)2 = −𝟗. 𝟖 𝒎
𝑣𝑦 = (−9.8 𝑚⁄𝑠 2 ) (2.0 s) = −𝟏𝟗. 𝟔 𝒎⁄𝒔
When 𝐭 = 𝟑. 𝟎 𝐬
𝑦 = (−4.9 𝑚⁄𝑠 2 ) (3.0 s)2 = −𝟏𝟒. 𝟕 𝒎
𝑣𝑦 = (−9.8 𝑚⁄𝑠 2 ) (3.0 s) = −𝟐𝟗. 𝟒 𝒎⁄𝒔
Summary of Formulas:
Legends:
𝒕 = 𝑡𝑖𝑚𝑒
𝒙 = 𝑝𝑜𝑠𝑖𝑡𝑖𝑜𝑛
𝒚 = 𝑝𝑜𝑠𝑖𝑡𝑖𝑜𝑛 (𝒇𝒓𝒆𝒆 𝒇𝒂𝒍𝒍)
𝒗𝒙 = 𝑖𝑛𝑠𝑡𝑎𝑛𝑡𝑎𝑛𝑒𝑜𝑢𝑠 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦
𝒗𝒂𝒗 = 𝑎𝑣𝑒𝑟𝑎𝑔𝑒 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦
𝒂𝒙 = 𝑖𝑛𝑠𝑡𝑎𝑛𝑡𝑎𝑛𝑒𝑜𝑢𝑠 𝑎𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛
𝒂𝒂𝒗 = 𝑎𝑣𝑒𝑟𝑎𝑔𝑒 𝑎𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛
𝒗𝒚 = 𝑖𝑛𝑠𝑡𝑎𝑛𝑡𝑎𝑛𝑒𝑜𝑢𝑠 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 (𝒇𝒓𝒆𝒆 𝒇𝒂𝒍𝒍)
𝒂𝒚 = 𝒈 = 𝑔𝑟𝑎𝑣𝑖𝑡𝑦 = 𝑖𝑛𝑠𝑡𝑎𝑛𝑡𝑎𝑛𝑒𝑜𝑢𝑠 𝑎𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛 = 𝟗. 𝟖 𝒎⁄𝒔𝟐
𝐕𝐞𝐥𝐨𝐜𝐢𝐭𝐲 𝐨𝐫 𝐈𝐧𝐬𝐭𝐚𝐧𝐭𝐚𝐧𝐞𝐨𝐮𝐬 𝐯𝐞𝐥𝐨𝐜𝐢𝐭𝐲:
𝒗𝒙 =
𝒅𝒙
𝒅𝒕
⟹
𝒗=
𝒅𝒙
(𝑆𝑖𝑟 𝑅𝑢𝑏𝑒𝑛 𝑛𝑜𝑡𝑒𝑠)
𝒅𝒕
𝐀𝐯𝐞𝐫𝐚𝐠𝐞 𝐯𝐞𝐥𝐨𝐜𝐢𝐭𝐲:
𝒗𝒂𝒗−𝒙 =
𝒙𝟐 − 𝒙𝟏 ∆𝒙
=
𝒕𝟐 − 𝒕𝟏
∆𝒕
⟹
𝒗𝒂𝒗 =
𝒙𝟐 − 𝒙𝟏 ∆𝒙
(𝑆𝑖𝑟 𝑅𝑢𝑏𝑒𝑛 𝑛𝑜𝑡𝑒𝑠)
=
𝒕𝟐 − 𝒕𝟏
∆𝒕
𝐀𝐜𝐜𝐞𝐥𝐞𝐫𝐚𝐭𝐢𝐨𝐧 𝐨𝐫 𝐈𝐧𝐬𝐭𝐚𝐧𝐭𝐚𝐧𝐞𝐨𝐮𝐬 𝐚𝐜𝐜𝐞𝐥𝐞𝐫𝐚𝐭𝐢𝐨𝐧:
𝒅𝒗𝒙
𝒂𝒙 =
𝒅𝒕
⟹
𝒂=
𝒅𝒙
𝒅𝒕
(𝑆𝑖𝑟 𝑅𝑢𝑏𝑒𝑛 𝑛𝑜𝑡𝑒𝑠)
𝐀𝐯𝐞𝐫𝐚𝐠𝐞 𝐚𝐜𝐜𝐞𝐥𝐞𝐫𝐚𝐭𝐢𝐨𝐧:
𝒂𝒂𝒗−𝒙
𝒗𝟐𝒙 − 𝒗𝟏𝒙
=
𝒕𝟐 − 𝒕𝟏
⟹
𝒂𝒂𝒗
𝒗𝟐𝒙 − 𝒗𝟏𝒙
(𝑆𝑖𝑟 𝑅𝑢𝑏𝑒𝑛 𝑛𝑜𝑡𝑒𝑠)
=
𝒕𝟐 − 𝒕𝟏
Motion with constant acceleration:
𝟏.
𝟐.
𝟑.
𝟒.
𝒗𝒙 = 𝒗𝟎𝒙 + 𝒂𝒙 𝒕
⟹
𝟏
𝒙 = 𝒙𝟎 + 𝒗𝟎𝒙 𝒕 + 𝒂𝒙 𝟐𝒕𝟐
⟹
𝟐
𝒗𝒙 𝟐 = 𝒗𝟎𝒙 𝟐 + 𝟐𝒂𝒙 (𝒙 − 𝒙𝟎 ) ⟹
𝒗𝟎𝒙 + 𝒗𝒙
𝒙 − 𝒙𝟎 = (
)𝒕
⟹
𝟐
𝒗 = 𝒗𝟎 + 𝒂𝒕 (𝑆𝑖𝑟 𝑅𝑢𝑏𝑒𝑛 𝑛𝑜𝑡𝑒𝑠)
𝟏
𝒙 = 𝒙𝟎 + 𝒗𝟎 𝒕 + 𝒂𝟐𝒕𝟐
𝟐
𝒗𝟐 = 𝒗𝟎 𝟐 + 𝟐𝒂(𝒙 − 𝒙𝟎 )
𝒗𝟎 + 𝒗
𝒙 − 𝒙𝟎 = (
)𝒕
𝟐
𝐅𝐫𝐞𝐞 𝐟𝐚𝐥𝐥𝐢𝐧𝐠 𝐛𝐨𝐝𝐢𝐞𝐬 (gravity = constant acceleration) ∶
𝟏.
𝟐.
𝟑.
𝟒.
𝒗𝒚 = 𝒗𝟎𝒚 + 𝒂𝒚 𝒕
⟹
𝟏
𝒚 = 𝒚𝟎 + 𝒗𝟎𝒚 𝒕 + 𝒂𝒚 𝟐𝒕𝟐
⟹
𝟐
𝒗𝒚𝟐 = 𝒗𝟎𝒚𝟐 + 𝟐𝒂𝒚 (𝒚 − 𝒚𝟎 ) ⟹
𝒗𝟎𝒚 + 𝒗𝒚
𝒚 − 𝒚𝟎 = (
)𝒕
⟹
𝟐
𝒗 = 𝒗𝟎 + 𝒂𝒕 (𝑆𝑖𝑟 𝑅𝑢𝑏𝑒𝑛 𝑛𝑜𝑡𝑒𝑠)
𝟏
𝒚 = 𝒚𝟎 + 𝒗𝟎 𝒕 + 𝒂𝟐𝒕𝟐
𝟐
𝟐
𝟐
𝒗 = 𝒗𝟎 + 𝟐𝒂(𝒙 − 𝒙𝟎 )
𝒗𝟎 + 𝒗
𝒚 − 𝒚𝟎 = (
)𝒕
𝟐