Velocity and Acceleration
Chapter 2, Sections 1 - 5,
Pages 28 - 43
Velocity
• Speed is a scalar. A number or measurement
with units only.
• Velocity is a vector. A number or measurement
with units and a direction.
• Can have positive and negative velocity.
Negative velocity occurs in relation to the point
of reference.
• Reference Point – zero location in a coordinate
system or frame of reference.
Velocity
Position – the separation between that object
and the reference point in a given direction.
Uses the symbol x. It is a vector. Also known as
displacement.
• Displacement is known as a “change in
Position.”
• Distance – has no frame of reference. It is the
separation between two points. It is a scalar.
Velocity
• Average velocity is figured by the change
in the position divided by the change in the
time.
x
v
, or change in postion in the amount it took to change postion.
t
• Usually measured in m/s.
Position – Time Graph
• Graph that shows the
position of an object vs
time. Time is the
independent variable, x
axis, position depends on
its time, not other way
around.
• Slope of the graph will
then represent the
velocity of the object. If
the object’s velocity is
constant, the slope is a
straight line.
Position-Time Graph
Answers
a. 400 m
b. 0 m
c. -200 m
Velocity of the line in Each case.
a. Slope a. = 10 m/s
b. Slope b. = 0 m/s
c. Slope c. = -20 m/s
Instantaneous velocity
• The velocity at a given instant in time.
• Figured from the slope of the tangent line to the
curve at that point if the velocity is not constant.
Acceleration
• The rate at which velocity changes over
time; an object accelerates if its speed,
direction, or both change.
Average Acceleration
• The change in the velocity divided by the
change in the time for it to occur.
Acceleration is a vector.
Δv
v
f  vi
a

t 2  t1 Δ t
This can be rewritten as
Acceleration
v f  vi  at
• Which is equation #1 under Newtonian
Mechanics of your equations sheet.
Average Acceleration
An Indy-500 race car’s velocity increases
from +4.0 m/s to +36 m/s over a 4.0 s
period. What is its average acceleration?
Answer
1.
8.0 m/s/s or 8.0 m/s2
Units
• Velocity is measured in m/s, time in s, so
units for acceleration are m/s/s or m/s x s
or m/s2.
• Can have negative acceleration – opposite
of positive. Usually means you are slowing
down or moving in the opposite direction
from the starting direction.
• The negative comes from negative
position from the velocity vector.
Slope of a velocity – time graph
• It is acceleration as long
as the acceleration is
constant.
• The area under the line
created by this graph is
the distance traveled.
• The displacement from its
original position to its
position at time t.
Slope of a velocity – time graph
Velocity-Time Graph
•
•
•
•
•
Answers
a. 75 m
b. 150 m
c. 125 m
d. 500 m
Velocity-Time Graph
Displacement with Constant Acceleration. Or
Finding the Displacement When Velocity and
Time are Known
Δx
average velocity v 
,
Δt
v  v
i
f
average velocity v 
.
2
Setting these two equal to each
other
v  v
Δx
i
f

,
Δt
2
Final Equation
Then solving for
displacement:
1

Δx   v  v  t
f
2 i
Sample Problem
1. A car accelerates uniformly from rest to a
speed of 6.6 m/s in 6.5 s. find the
distance the car travels during this time.
2. A driver in a car traveling at a speed of
21.8 m/s sees a cat 101 m away on the
road. How long will it take for the car to
accelerate uniformly to a stop in exactly
99 m?
Answers
1. 21 m
2. 9.1 s
Finding the displacement when
acceleration and time are known,
but the final velocity is not known.
• We know vf = vi + at from the
acceleration formula and
• x = ½ (vf + vi)t. from the equation
above.
• We then take the first equation for vf
and substitute it into the second
equation.
•
•
•
•
x = ½ ((vi + at) + vi)t,
simplifying this to get
x = ½(2vi + at)t,
simplifying further, and you get
Final Equation
1 2
x  v i t  at
2
Which is the 2nd equation on your equations
sheet.
Sample Question
1. A car starts from rest and travels from 5.0 s
with a constant acceleration of -1.5 m/s2.
a.
b.
What is the final velocity of the
car?
How far does the car travel in this
time interval?
Answer
1. a.
b.
-7.5 m/s,
19 m
Finding the displacement when velocity and
acceleration are known, but time is not known.
• Know x= ½ (vf + vi)t
• and vf = vi + at,
• Solving the second equation for t:
t
vf
v
a
i
• and substitute it into the first equation
½(v f  v i ) (v f  v i ) ½ (v 2f  v 2i )

x 
a
a
• simplifying this and you get something
easier to understand:
Final Equation
v  v  2ax
2
f
2
i
• Which is the 3rd equation on your equation
sheet.
Sample Question
1. A person pushing a stroller starts from
rest, uniformly accelerating at a rate of
0.500 m/s2. What is the velocity of the
stroller after it has traveled 4.75 m?
2. An aircraft has a liftoff speed of 33 m/s.
What minimum constant acceleration
does this require if the aircraft is to be
airborne after a take-off run of 240 m?
Answer
1. 2.18 m/s
2. 2.3 m/s2
Homework
• Ch 2 , pages53 – 55
• #1, 9,,18,20,24,25,27,28,30,and 31.
• Extra Credit #11 and 37.