acceleration physics montwood high school r. casao
TRANSCRIPT
Acceleration
Physics
Montwood High School
R. Casao
Acceleration Defined
• Acceleration is defined as the change in velocity per change in time.
• Velocity is speed and direction of motion.
• If speed changes, an acceleration is definitely occurring.
• Acceleration involves a:– change in speed.– change in direction.– change in speed and direction.
– Consider a car traveling at a constant 40 mph on a long straight road.
– The car is NOT acceleration because neither its speed or direction is changing. The car did accelerate to get its speed up to 40 mph, but did not accelerate after reaching 40 mph.
– Consider a car driving around a circular track at a constant speed.
– The car IS accelerating. The speed is not changing, but the direction of motion is changing. The velocity is changing and the car is accelerating.
Types of Acceleration
• Increasing speed– Example: Car speeds up at green light
• Decreasing speed– Example: Car slows down at stop light
• Changing Direction– Example: Car takes turn (can be at constant speed)
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Acceleration is a vector that points in the same direction as the change in velocity.
v0
v
v
v0 + v = v
Acceleration is a vector that points in the same direction as the change in velocity.
v0
v
v
v0 + v = v
• An object is accelerating when the velocity of the object changes over time.
• Average acceleration is the the change in velocity v divided by the change in time t.
– Average acceleration is the slope of the secant line between any two points on a velocity – time graph.
if
ifavg tt
vvΔtΔv
a
• The car moves from point A with velocity vi at time ti to point B with velocity vf at time tf.
• The slope of the line connecting A to B represents the average acceleration of the car between points A (vi) and B (vf). Acceleration is the derivative of the velocity.
• The slope of the tangent line passing through point B represents the instantaneous acceleration of the car at point B.
• Acceleration is the slope of a velocity – time graph (the derivative of the velocity).
• Acceleration has a magnitude and a direction and is a vector quantity.
• Units: m/s2 (most common); ft/s2, or km/hr2.• If you see the term “deceleration” in a problem, it
usually means that the acceleration is negative for that problem; generally, the object would be slowing down and its velocity is decreasing.
• On a velocity – time graph:– If a is positive, velocity is in the positive direction and
the speed of the object is increasing; acceleration is in the positive direction.
– The object is increasing its velocity in the positive direction; the object is speeding up and its displacement per unit time is increasing.
• On a velocity – time graph:– If a is positive, velocity is in the negative direction
and the speed of the object is decreasing; acceleration is in the positive direction.
– The object is decreasing its velocity in the negative direction; the object is slowing down and its displacement per unit time is decreasing.
• On a velocity – time graph:– If a is negative, velocity is in the positive direction and
the speed of the object is decreasing; the acceleration is in the negative direction.
– The object is slowing down in the positive direction; the object is slowing down and its displacement per unit time is decreasing.
• On a velocity – time graph:– If a is negative, velocity is in the negative direction
and the speed of the object is increasing; acceleration is in the negative direction.
– The object is speeding up in the negative direction and its displacement per unit time is increasing.
– If a is zero, velocity is constant (not changing); the object could be at rest. The slope of the velocity – time graph is zero (rise = 0).
– Three possibilities:– Object at rest:
– Constant positive velocity:
– Constant negative velocity:
• When an object’s velocity and acceleration are in the same direction, the object is speeding up.
– Two cases:1. Positive velocity with positive acceleration: object speeding
up in the positive direction.
2. Negative velocity with negative acceleration: object speeding up in the negative direction.
• When an object’s velocity and acceleration are in opposite directions, the object is slowing down.
– Two cases:1. Positive velocity with negative acceleration: object slowing
down in the positive direction.
2. Negative velocity with positive acceleration: object slowing down in the negative direction.
(a) instantaneous acceleration
- equal to slope of the tangent line through the point of interest
(b) average acceleration
- equal to the slope of the secant line between two points on the graph
(c) zero acceleration
- no change in velocity
Visualizing Motion with Constant Acceleration
Motion diagrams for three carts:
•An acceleration of 0 m/s2 tells you the velocity will remain the same each second of the motion.•Acceleration tells you the amount by which the velocity will increase (+a) or decrease (-a) each second of the motion.
One-Dimensional Motion with Constant Acceleration
• Constant or uniform acceleration means that the velocity increases or decreases at the same rate throughout the motion.– The rate of change in the velocity is always the same.
• Variables to consider:– xi = initial displacement
– xf = final displacement
– vi = initial velocity
– vf = final velocity
– a = acceleration
– t = time
• Velocity as a function of time:
• Displacement as a function of time:
tavv if
2i
fi
fiif
ta0.5tvΔx2
tvvΔx
2tvv
xx
• Velocity as a function of displacement:
• For motion in which a = 0 m/s2 (no change in velocity):vi = vf
Use the constant velocity equation:
• Often best to choose xi as 0 m at t = 0 s so that the displacement x = xf (the final position).
Δxa2vv2
i
2
f
tΔx
v
• For an object at rest: vi = 0 m/s.
• For an object that stops: vf = 0 m/s.
• The term “deceleration” indicates that the acceleration is negative.
• Examine the known variables before deciding which equation to use to solve the problem.
• Don’t be too surprised if you have to use one equation to find a variable and then use a second equation to get the desired answer!
Distance in the nth second• For problems involving the acceleration of an
object from an initial velocity without any changes in acceleration, determine the distance traveled during the nth second by using the distance equation twice.
• Example: determine the distance traveled during the 2nd second. You are being asked to find the distance the object travels from t = 1 s to t = 2 s.
• Find x for t = 2 s. Find x for t = 1 s. Subtract to get x during the 2nd second.
2Δx (vi t) (0.5 a t )
• For multi-part problems in which the object has changed its acceleration, you may have to determine the initial velocity during the time period in which you are asked to find the distance traveled in the nth second.
• Ex. Object has constant acceleration for 5 s, changes acceleration for the next 5 s. Determine the distance traveled during the 8th second.– You must find the distance traveled from t = 7 s to
t = 8 s.– Determine the final velocity at the end of the 5 s; this
becomes the initial velocity for the next 5 s.
– Find x for t = 3 s (8s – 5 s, the second interval begins at 5 s with the final velocity you just found becoming the initial velocity vi for this interval of time).
– Find x for t = 2 s (7 s – 5 s). Use the same initial velocity vi that you used to find x for t = 3 s.
• Subtract to get x during the 8th second.
Reaction Time• Reaction time is the time it takes for a person to
recognize a situation (stimulus) and react to that situation with a response.
• Reaction time problems usually involve a car moving with an initial velocity toward an object a measured distance away.
• Reaction time is the time it takes for the driver to take their foot off of the gas pedal and press on the brake pedal.
• During the reaction time, the car does not accelerate and maintains a constant speed; therefore, the only equation you should use is d
vt
Reaction Time• If you solve for distance d, this is the distance the car
will travel before the brakes are applied and the car begins to decelerate (negative acceleration).
• Subtract this reaction distance from the given distance to determine the total distance the car has to travel prior to arriving at the designated final point:
Total distance = reaction distance + acceleration distance• Examine the variables to determine the appropriate
acceleration equation to use to determine the unknown.
Graphing Acceleration:Speed vs. Time Graphs
0
2
4
6
8
10
12
14
0 1 2 3 4 5 6
Time (s)
Sp
ee
d (
m/s
)
1)Speed is increasing with time = accelerating2)Line is straight = acceleration is constant
Graphing Acceleration:Speed vs. Time Graphs
0
2
4
6
8
10
12
14
0 1 2 3 4 5 6
Time (s)
Sp
ee
d (
m/s
)
In Speed vs. Time graphs:Acceleration = Rise/Run
Acceleration= 4 m/s ÷ 2 s = 2 m/s2
Run = 2 s
Rise = 4 m/s
Graphing Acceleration:Distance vs. Time Graphs
0
5
10
15
20
25
30
35
0 1 2 3 4 5
Time (s)
Dis
tan
ce
(m
)
1)On Distance vs. Time graphs a curved line means the object is accelerating.
2)Curved line also means your speed is increasing. Remember: slope = speed.
Graphical Look at Motion: Velocity – Time curve
• The slope gives the acceleration.
• The straight line indicates a constant acceleration.
• The zero slope indicates a constant acceleration.
Graphical Look at Motion: Acceleration – Time curve
Graphical Motion with Constant Acceleration
• A change in the acceleration affects the velocity and position.
• Note especially the graphs when a = 0.
Acceleration – Time GraphsThe area under an acceleration-time graph is equal to the change in velocity of the object.
• The area under the curve is the distance; the rectangle represents vo·t and the triangle represents 0.5·a·t·t.
• The total distance represented under the curve is:
20.5ox v t a t
• If you know the velocity of an object, can you tell me it’s position?– No. An object’s position is independent of the velocity.
– You would need to be told the position of the object.
• If you know the acceleration of an object, can you tell me it’s velocity?– No. An object’s velocity is independent of the acceleration.
– You would need to be told the velocity of the object.
• If you know only the velocity of an object, can you tell me it’s acceleration?– No. An object’s acceleration is the rate of change in velocity,
which is independent of the actual velocity at a given instant (the instantaneous velocity).
– You would need to know the initial and final velocities of the object to determine the acceleration.
Acceleration – Time Graphs Summarized
• the y-coordinate at any time gives the acceleration of the object.
• horizontal graph segments indicate that the object has constant acceleration.
• a horizontal graph segment on the x-axis indicates that the object has constant velocity (no acceleration).
• graph segments above the x-axis imply increasing velocities.
• graph segments below the x-axis imply decreasing velocities.
• no changes in direction may be inferred from these graphs.
We typically only deal with constant acceleration situations, so acceleration graphs generally consist of horizontal segments only.
Question
0
2
4
6
8
10
12
14
0 1 2 3 4 5 6
Time (s)
Sp
ee
d (
m/s
)
Above is a graph showing the speed of a car over time.1) How is the speed of the car changing (speeding up,
Slowing down, or staying the same)?2) What is this car’s acceleration?
1) The car is slowing down2) Acceleration = rise/run = -6m/s ÷3s = -2 m/s2
Run = 3 s
Rise = -6 m/s
Question:
0
5
10
15
20
25
30
35
0 1 2 3 4 5
Time (s)
Dis
tan
ce
(m
)
1)Which line represents an object that is accelerating?
The black and red lines represent a objects that are accelerating. Black is going a greater distance each second, so it must be speeding up. Red is going less each second, so must be slowing down
Remember: in distance vs. time graphs: curved line = accelerating, flat line = constant speed
Example: The Aircraft Carrier
• Jet needs to be traveling 62 m/s for take-off• Catapult can accelerate jet up to 31 m/s2
• How long does the carrier have to be?
Example: The Aircraft Carrier
What do we know?
vi = 0 m/svf = +62 m/sa = +31 m/s2
What do we want to know?x = ?
fiv v a t 1
( )2 i fx v v t
212ix v t a t
2 2 2fiv v a x
Implied data!!
Example: The Aircraft Carrier
2 2
2 2
2 2
2 2
2
2
2
2(62 m/ s) (0 m/ s)
62 meters(2 31 m/ s )
fi
fi
fi
v v a x
a x v v
v vx
a
x
Approach to Problem Solving
1. Write down what you know in mathematical terms. (e.g. v0=0 m/s) Is there any implied data?
2. Write down what we are trying to figure out (e.g. x=?).
3. Examine the relevant equations and choose the one that will allow us to solve for the quantity in 2) above.
4. Do the algebra!!5. Plug in the numbers.
What is its final velocity after travelling 215 km?
Example:
A spacecraft, traveling at +3250 m/s fires its reverse thrusters to slow down. The thrusters slow the craft down by 10 m/s every second.
Example:
What do we know?
vi = +3250 m/sa = -10 m/s2
x = +215 km = +215,000 m
What do we want to know?v = ?
fiv v a t
0
1( )
2 fx v v t
212ix v t a t
2 2 2fiv v a x
2 2 2fiv v a x
Example:
2 2 2
2 2
(3250 m/ s) 2( 10 m/ s )(215,000 m)
(3250 m/ s) 2( 10 m/ s )(215,000 m)
2500 m/ s
f
f
f
v
v
v