Section 1: Displacement and Velocity
 Motion that takes place in one direction, such as a train, is one dimensional
motion.
 When looking at the motion of objects it is important to have a frame of
reference. This allows for the comparison of points.
 So… if an object is at rest, does it move from its frame of reference?
__________________
 Explain your reasoning.
_______________________________________________

When an object moves its change in position is known as displacement.


It can be calculated from final position compared to initial position.
It has an equation of: ∆x= xf-xi

The change in direction will correspond to the coordinate plane in which the
object moves: in the x direction for horizontal and y direction for vertical
movement.

*Remember: Keep displacement separate from distance! Displacement is the
comparison of where an object began and ended after movement.
 Yes or No: Is there displacement for a ball that rolls 2 feet from its
starting position?
 Yes or No: Is there displacement for a runner that begins and ends at
the same mark?

Displacement can also be described by positives and negatives, depending on
the direction of movement.
Y=positive
∆y=positive
x=negative
∆x=negative
y= negative
∆y=negative
x=positive
∆x=positive
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Velocity
 Motion can also be described by velocity, which states how fast an object is
moving in a particular direction.
 Average velocity described the average speed of an object’s motion covering
a period of time.
x  xi
x
 Average velocity: νavg = f

t f  ti
t
 Notice that velocity is also described by direction. Velocity can be a positive
or negative value based on its coordinates.
 The SI Units for velocity are: meters/second (m/s)
 Compare speed and velocity:
Speed:
expressed with
a quantity.
Similiarity: Both
measure how fast
an object moves.


Velocity:
Includes the
quantity and the
direction of
motion.
The velocity of an object can only be determined if the position is
known. Note the equation. It can be very helpful to graph the data.
Relate the slope of a line to the velocity of an object.


slope 
rise x

run t
Slope: +
Velocity: +
Slope: Velocity: -
Slope: 0
Velocity: 0
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


Curved lines are produced when velocity is not constant, or changes.
When a curved graph is seen, we can calculate instantaneous velocities.
Instantaneous Velocity: velocity of an object at a certain instant in time or
specific point on the object’s path.

Vin must be calculated with the tangent line corresponding to the time in
question.
Section 2- Acceleration
 We will not be able to relate changing velocity to the object’s period of
motion.
 Think of stopping at a stop light. You are driving at a constant velocity
and then decelerate to a stop, then accelerate again.

Acceleration is the rate of change in velocity over a time period.

v
t

Acceleration equation: a=


SI Units for acceleration: meters/second squared (m/s2)
Practice Problem B: As a shuttle bus comes to a sudden stop to avoid
hitting a dog, it accelerates uniformly at -4.1 m/s2 as it slows from 9.0
m/s to 0.0 m/s. Find the time interval of acceleration of the bus.
 Given: a=-4.1 m/s2
unknown: t=?
Vi= 9.0m/s
Vf= 0.0m/s
a
t

v f  vi
v f  vi
t f  ti
,t 
t f  ti
v f  vi
a
0.0m / s  9.0m / s
9.0m / s

 2.2s
2
4.1m / s
4.1m / s 2
Note that time cannot be negative. So, acceleration is determined by
velocity and the direction of motion.
 ∆v is +, a is +, speed increasing
 ∆v is -, a is -, speed is decreasing
 ∆v is 0, a is 0, speed is constant
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







Acceleration can also be graphed:
Positive slope increase in velocity, increase in acceleration
Negative slope decrease in velocity, decrease in acceleration
No slope constant velocity, constant acceleration
Constant acceleration is achieved when the object’s velocity increases at the
same interval each time interval.
 Think about this graphically. What do you get when velocity and tie
increase together? _____________________________________
For object’ s in free fall, the displacement increases as the time interval
increases causing the velocity of the object to increase in a chain reaction.
Constant acceleration: (vi+vf)/2
If the equations for acceleration and velocity were compared, we would get
an area graphically that shows the constant acceleration. From this deviation
of the equations, a new formula is calculated that gives the displacement in
terms of velocity and time.
∆x= ½ ((vi+vf)∆t
Practice Problem C: A car accelerates uniformly from rest to a speed of
6.6m/s in 6.5 s. Find the distance the car travels during this time.
 Given: vi= 0.0 m/s
unknown: x=?
Vf= 6.6m/s
t= 6.5s
∆x= ½ (vi+vf)∆t
½ (0.0m/s + 6.6m/s) (6.5s)=(3.3m/s)(6.5s)
∆x=21.45 m
This equation can also be used to solve for final velocity. Starting with
a

v
can go to v f  at  vi . Acceleration must be constant to use
t
this formula. Notice that velocity and time are the only variables that
change.
This new equation can also be used to calculate the displacement of an
object if vf is not known. (Substitute the equation in for vf in the ∆x eqn)
 x 

1
1
1
vi  v f  t  x   vi  vi  at  t  x  vi t  at 2

2
2
2
Practice Problem D: A car with an initial speed of 6.5 m/s accelerates at a
uniform rate of 0.92m/s2 for 3.6 s. Find the final speed and the
displacement of the car during this time.
 Given: vi=6.5 m/s
Unknown: vf=?
2
a= 0.92m/s
∆x=?
t= 3.6s
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v f  at  vi
v f   0.92m / s 2   3.6s    6.5m / s    3.312m / s    6.5m / s 
v f  9.8m / s
1
x  vi t  at 2
2
x   6.5m / s  3.6s  
1
2
0.92m / s 2   3.6s 

2
1
0.92m / s 2 12.96s 2   23.4m  5.96m

2
x  23.4m 
x  29.36m



If acceleration is uniform we can derive an expression that solves for time.
This can be useful for when time is not known and you are needing to solve
for a displacement.
t 
2x
vi  v f
v f 2  vi 2  2ax

If acceleration is known then a new formula can be used:

Practice Problem E:a car is traveling initially at 7 m/s accelerates uniformly
at the rate of 0.8 m/s2 for a distance of 245 m. What is the velocity at the
end of the acceleration?
 Given: vi=7m/s
Unknown: vf
2
a=0.8m/s
x=245m
v f 2  vi 2  2ax
v f  vi 2  2ax
vf 
 7m / s 
2
 2  0.8m / s 2   245m  
 49m / s   392m / s 
2
2
2
2
v f  441m2 / s 2
v f  21m / s
Section 3- Free Fall

Free fall refers to an object with constant acceleration as it falls. This
acceleration is caused by gravity and have a value of 9.8m/s2.
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
Even if the ball is at rest, it is still being pulled down.

If the velocity is positive and the acceleration is negative, the object will be
+ direction
-acceleration
- velocity
Anytime an object is affected by free fall, it has a negative acceleration
because gravity it pulling it downward.
-Direction
-Acceleration
+Velocity

slowing down.

If the velocity is positive and the acceleration is negative, the object will be
speeding up.

Sample Problem F: A robot probe drops a camera off the rim of a 239m
high cliff on Mars, where the free fall acceleration is -3.7m/s2. Find the
velocity with which the camera hits the ground. Find the time required for it
to hit the ground.
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