Important Formulas
• Displacement = final position – initial position
– Δx = xf – xi
– Δy = yf – yi
• Average Velocity = displacement/change in time
– vavg = Δx/Δt = xf – xi/tf – ti
Sec. 2–2: Acceleration
Coach Kelsoe
Physics
Pages 48–59
Objectives
• Describe motion in terms of changing
velocity.
• Compare graphical representations of
accelerated and nonaccelerated motions.
• Apply kinematic equations to calculate
distance, time, or velocity under conditions
of constant acceleration.
Changes in Velocity
• In most real-life
situations, we rarely
have constant velocity.
There is usually a
speeding up or slowing
down period in motion.
• This change in velocity
is what we call
acceleration.
Changes In Velocity
• Acceleration is the rate at which velocity
changes over time.
aavg
v f  vi
v


t
t f  ti
average acceleration =
change in velocity
time required for change
• An object accelerates if its speed,
direction, or both change.
• Acceleration has direction and magnitude.
Thus, acceleration is a vector quantity.
Acceleration
• The units for acceleration are meters per
second squared (m/s2).
• We generally use the term “acceleration” to
mean “speeding up,” and “deceleration” to
mean slowing down.
• A negative acceleration indicates
deceleration. Remember that accelerations
can be positive or negative.
Sample Problem B
• A shuttle bus slows down with an average
acceleration of -1.8 m/s2. How long does it
take the bus to slow from 9.0 m/s to a
complete stop?
• Givens:
– vi = 9.0 m/s
– vf = 0 m/s
– aavg = -1.8 m/s2
• Unknown:
– Δt = ?
Sample Problem B
• We don’t have a formula for time, but we can
rearrange our acceleration formula to solve
for time:
– aavg = Δv/Δt

Δt = Δv/aavg
– Δt = vf – vi /aavg = 0 m/s – 9.0 m/s/-1.8 m/s2
– Δt = 5.0 s
• Look for statements such as “starts at rest”
(vi = 0 m/s) and “comes to rest” (vf = 0 m/s)
as keys to your given information.
I’ll Take the Shuttle
• A shuttle bus is at rest and increases its
speed to 10 m/s in 5.0 seconds. What is it’s
acceleration?
– 10 m/s – 0 m/s/5 s = 2.0 m/s2
• The shuttle bus is cruising along at 15 m/s
when a dog runs in front of it. It slams on
the brakes and comes to a complete stop in
1.5 seconds. What is its acceleration?
– 0 m/s – 15 m/s/1.5 s = -10 m/s2 (deceleration)
Acceleration
• Imagine a high-speed train leaving a station
moving to the right so that both displacement
and velocity is positive.
• The velocity increases in magnitude as the
train picks up speed. As the velocity
increases, so does the acceleration.
• As long as Δv is positive, acceleration will be
positive.
Acceleration
• Imagine the same train slowing down as it
approaches the next station.
• The velocity is still positive because it is
traveling in the same direction.
• In this case, the initial velocity is higher than
the final velocity, so Δv will be negative and
acceleration will be negative.
• Can there still be motion with no
acceleration?
Changes In Velocity
• Consider a train moving to the right, so that the
displacement and the velocity are positive.
• The slope of the velocity-time graph is the average
acceleration.
– When the velocity in the positive
direction is increasing, the
acceleration is positive, as at
A.
– When the velocity is constant,
there is no acceleration, as at B.
– When the velocity in the positive
direction is decreasing, the
acceleration is negative, as at
C.
Velocity and
Acceleration
Motion with Constant
Acceleration
• On page 51 of your textbook, there is a strobe
photo of a billiard ball falling. There were 10
images taken within one second.
• As the ball’s velocity increases, the ball travels a
greater distance during each time interval.
• The velocity increases by exactly the same amount
during each time interval. Thus the acceleration is
constant.
• The relationships between displacement, velocity,
and constant acceleration are expressed by
equations that apply to any object moving with
constant acceleration.
Motion with Constant
Acceleration
• Displacement depends on acceleration, initial
velocity, and time.
• As we learned in Section 2–1 , average velocity is
equal to displacement divided by time:
– average velocity = displacement/time
– vavg = (Δx/Δt)
• For an object moving with constant acceleration,
the average velocity is equal to the average of the
initial velocity and the final velocity:
– average velocity = initial velocity + final velocity/2
– vavg = (vi + vf/2)
Motion with Constant
Velocity
• To find an expression for the displacement in terms
of the initial and final velocity, we can set the
expressions for average velocity equal to each
other:
– Δx/Δt = (vi + vf)/2
• Multiplying both sides of the equation by time (Δt)
gives us an expression for the displacement as a
function of time. This equation can be used to find
the displacement of any object moving with
constant acceleration.
– Δx = ½(vi + vf)Δt
– displacement = ½(initial + final velocity)(time interval)
Sample Problem
• A race car reaches a speed of 42 m/s. It then
begins a uniform negative acceleration, using its
parachute and braking system, and comes to rest
5.5 s later. Find the distance that the car travels
during braking.
Sample Problem
Solution
• Identify givens and unknowns:
– vi = 42 m/s
– Δt = 5.5 s
– vf = 0 m/s
– Δx = ?
• Choose correct equation:
– The variables we know are initial velocity, final
velocity, and time. We don’t know displacement.
– Δx = ½(vi + vf)Δt
Sample Problem
Solution
• Plug our values into the equation:
– Δx = ½(vi + vf)Δt
– Δx = ½(42 m/s + 0 m/s)(5.5 s)
– Δx = ½(42 m/s)(5.5 s)
– Δx = 120 m
• Notice how our units cancel out correctly to
give us the units we expect to get.
Motion with Constant
Acceleration
• What happens if the final velocity of an
object is not known but we still want to
calculate the displacement?
• If we know the initial velocity, the
acceleration, and the elapsed time, we can
find the final velocity. We can then use this
value for the final velocity to find the total
displacement of the object.
– a = Δv/Δt = (vf - vi)/Δt
– aΔt = vf – vi  aΔt + vi = vf
Motion with Constant
Acceleration
• You can use this equation to find the final
velocity of an object after it has accelerated
at a constant rate for any time interval.
– vf = vi + aΔt
• If you want to know the displacement of an
object moving with constant acceleration
over some certain time interval, you can
obtain another useful expression for
displacement by substituting the expression
for vf into the expression for Δx.
Motion with Constant
Acceleration
• By substituting our vf equation into our Δx
equation, we derive a new equation.
–
–
–
–
Δx = ½(vi + vf)Δt
Δx = ½(vi + vi + aΔt)Δt
Δx = ½[2viΔt + a(Δt)2]
Δx = viΔt + ½a(Δt)2
• This equation is useful not only for finding the
displacement of an object moving with constant
acceleration but also for finding the displacement
required for an object to reach a certain speed or
come to a stop.
Sample Problem
• A plane starting at rest at one end of a runway
undergoes a uniform acceleration of 4.8 m/s2 for
15 s before takeoff. What is its speed at takeoff?
How long must the runway be for the plane to be
able to take off?
Sample Problem
Solution
• Identify givens and unknowns:
– vi = 0 m/s
– Δt = 15 s
– Δx = ?
– a = 4.8 m/s2
– vf = ?
• Choose the correct equations:
– vf = vi + aΔt
– Δx = viΔt + ½a(Δt)2
Sample Problem
Solution
• Plug our values into the equations.
– vf = vi + aΔt
• vf = 0 m/s + (4.8 m/s2)(15 s)
• vf = 72 m/s
– Δx = viΔt + ½a(Δt)2
• Δx = (0 m/s)(15 s) + ½(4.8 m/s2)(15 s)2
• Δx = 0 m + 540 m
• Δx = 540 m
Motion with Constant
Acceleration
• To this point, all of the equations for motion under
constant acceleration have required knowing the
time interval.
• We can also obtain an expression that relates
displacement, velocity, and acceleration without
using the time interval.
• This method involves rearranging one equation to
solve for Δt and substituting that expression in
another equation, making it possible to find the
final velocity of a uniformly accelerated object
without knowing how long it has been accelerating.
Motion with Constant
Acceleration
• We’ll start with one of the equations for
displacement:
– Δx = ½(vi + vf)Δt  Multiply by 2
– 2Δx = (vi + vf)Δt  Divide by (vi + vf)
– 2Δx/(vi + vf) = Δt  Now use vf = vi + aΔt
– vf = vi + a[2Δx/(vi + vf)]  Consolidate the vi
– vf - vi = a[2Δx/(vi + vf)]  Multiply by (vi + vf)
– (vf - vi)(vf + vi) = 2aΔx  Consolidate left side
– vf2 - vi2 = 2aΔx  Move the vi2 term to the left
– vf2 = vi2 + 2aΔx
Motion with Constant
Acceleration
• When using this equation, you must take the
square root of the right side of the equation
to find the final velocity. Keep in mind that
the square root may be either positive or
negative.
• If you have been consistent in your use of
the sign convention, you will be able to
determine which value is the right answer by
reasoning based on the direction of motion.
Sample Problem
Final Velocity After Any Displacement
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?
Sample Problem
Solution
• Define
– Given:
• vi = 0 m/s
• a = 0.500 m/s2
• Δx = 4.75 m
– Unknown:
• vf = ?
– Diagram:
• Choose a coordinate system. The most convenient
one has an origin at the initial location of the stroller.
The positive direction is to the right.
Sample Problem
Solution
• Plan
– Choose an equation or situation: Because the
initial velocity, acceleration, and displacement
are known, the final velocity can be found using
the following equation:
vf2 = vi2 + 2aΔx
– Rearrange the equation to isolate the
unknown: Take the square root of both sides to
isolate vf.
vf = ±√(vi)2 + 2aΔx
Sample Problem
Solution
• Calculate
– Substitute the values into the equation and
solve:
vf = ±√(0 m/s)2 + 2(0.500 m/s2)(4.75 m)
vf = ±√0 m2/s2 + 4.75 m2/s2
vf = ±√4.75 m2/s2
vf = +2.18 m/s
• Evaluate
– The stroller’s velocity after accelerating for 4.75
m is 2.18 m/s to the right.
Vocabulary
• Acceleration
Important Formulas
• Average acceleration
– a = Δv/Δt
• Displacement with Constant Acceleration
– Δx = ½(vi + vf)Δt
– Δx = viΔt + ½a(Δt)2
• Velocity with Constant Acceleration
– vf = vi + aΔt
• Final Velocity After Any Acceleration
– vf2 = vi2 + 2aΔx