Physics
Acceleration
Acceleration
• Acceleration is defined
as the rate at which
velocity changes.
• Acceleration is large
when there is a large
velocity change in a
small time interval.
• The ratio is called the
average acceleration
between the 2 times.
Units of acceleration
• Acceleration is
Relative.
• Velocity is measure in
meters per second,
m/s, so acceleration is
measured in m/s/s or
m/s2.
• Since position and
velocity are vectors,
so is acceleration.
• In straight-line motion,
acceleration can be
positive or negative.
Acceleration Practice Problems
1. The velocity of a car increases from 2.0
m/s at 1.0 s to 16 m/s at 4.5 s. What is
the car’s average acceleration?
2. A car goes faster and faster backwards
down a long driveway. We define forward
velocity as positive, so backward velocity
is negative. The car’s velocity changes
from -2.0 m/s to -9.0 m/s in a 2.0-s time
interval. Find its acceleration.
Answers
1. Given:
v1 = 2.0 m/s
v2 = 16 m/s
t1 = 1.0 s
t2 = 4.5 s
Solution: 14 m/s
3.5 s
Unknown: a
Equation:
=
4.0 m/s2
Answers
2. Given:
Unknown: a
V1 = -2.0 m/s
Equation:
V2 = -9.0 m/s
Dt = 2.0 s
Solution:
-9.0 m/s – (-2.0 m/s) = -7.0 m/s
A = -7.0 m/s = -3.5 m/s2
2.0 s
Negative Acceleration
• Negative acceleration can be 1 of 2 things.
• The object could be slowing down and
have a negative acceleration or the object
could be
speeding
up but moving
in a negative
direction.
Uniform Acceleration
• Acceleration that does not change in time is
uniform or constant acceleration.
• A velocity-time graph of constant
acceleration is a straight line.
• Initial velocity (v1) is the velocity at time =
0s. The slope of the line is the
acceleration, a, so the equation that
describes the curve is:
vf = vi + at
Final Velocity After Uniform Acceleration
• Ex: If a car with a velocity of 2.0 m/s at t =
0 accelerates at a rate of +4.0 m/s2 for 2.5
s, what is its velocity at time t = 2.5 s?
• Given: vi =2.0 m/s
Unknown: vf
a = 4.o m/s2
t = 2.5 s
Solution: vf = 2.0 m/s + (4.0 m/s2)(2.5 s)
= 12 m/s
Displacement When Velocity and Time
are Known
• To find the displacement if the object is
uniformly accelerating, the velocity is
replace by the average velocity.
d = ½ (vf + vi)t
• On a velocity-time graph for an object with
constant acceleration, the displacement if
the total area under the curve.
Displacement When Acceleration and
Time are Known
• If the initial velocity, acceleration, and time
interval are known, the displacement of the
object can be found by combining
equations already used.
d = vit + ½ at2
• Example: A car starting from rest
accelerates uniformly at 6.1 m/s2 for 7.0 s.
How far does the car move?
Solution
•
•
•
•
•
Given:
Unknown:
Vi = 0 m/s
d=?
a = 6.1 m/s2
t = 7.0 s
d = (0 m/s)(7.0 s) + ½ (6.1 m/s2)(7.0 s)2
• Answer: 150 m
Displacement When Velocity and
Acceleration are Known
• The equations for final velocity and
displacement can be combined to form an
equation relating initial and final velocities,
acceleration, and displacement in which
times does not appear.
• Vf2 = vi2 + 2ad
• Example: An airplane must reach a
velocity of 71 m/s for takeoff. If the runway
is 1.0 km long, what must the constant
acceleration be?
Solution
•
•
•
•
•
•
•
•
Given:
Unknown:
Vi = 0 m/s
a=?
Vf = 71 m/s
d = 1 km or 1000 m
Solution: (71m/s)2 = (0 m/s)2 + 2a(1000 m)
5041 m2/s2 = 2000(a) m
(5041 m2/s2 )/(2000m) = a
a = 2.52 m/s2
Acceleration Due to Gravity
• Galileo was the first to show that all objects
fall to Earth with a constant acceleration.
• No matter what the mass of the object is or
what height it is dropped from, the
acceleration due to gravity is always the
same.
• Acceleration due to gravity is called “g” and
is always negative (it pulls objects
downward).
The Value of “g”
• On the surface of Earth, a freely falling
object has an acceleration, g, or -9.8 m/s2.
• Actually, g varies slightly from -9.790 m/s2
in southern Florida to -9.810 m/s2 in
northern Maine.
• The value of g is also smaller at high
altitudes; g = -9.789 m/s2 on Pike’s Peak.
• The value of g would be higher at sea level.
Freely Falling Objects
• Ex: The time the Demon Drop ride at
Cedar Point, Ohio is freely falling in 1.5 s.
A. What is its velocity at the end of this
time? B. How far does it fall?
• Given:
Unknown:
• g = -9.8 m/s2
vf = ?
• vi = 0 m/s
• t = 1.5 s
Solution
• vf = 0 + (-9.8 m/s2)(1.5 s)
• vf = -15 m/s
• d = vit + ½ (-9.8 m/s2)(1.5 s)2
• d = 0 + (-11 m)
• d = -11 m
• Negative displacement indicates downward
motion.