Sec 2-1 Notes Displacement
and velocity
Motion – we will study 1D only first
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forward & back, left & right, or up and down
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takes place over time
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depends on the frame of reference
Frame of reference – a coordinate system for
specifying the precise location of object in space
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if an object is at rest, its position does not change
with respect to a frame of reference
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motion is relative
Displacement – the change in position of an
object
Dx = xf – xi final
position – initial position
- not always equal to the distance traveled
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can be positive or negative
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a vector quantity (magnitude and direction)
Velocity – how fast something moves
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a vector quantity
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not the same as speed (a scalar quantity)
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average speed= distance traveled/time traveled
Average velocity – the total displacement divided
by the time interval during which the displacement occurred
vavg
= Dx/Dt vavg
= displacement/time
ex:
d = 200 m west t = 40 s
Find the average
velocity.
Graphical interpretation of velocity
position (y – axis) vs. time (x-axis)
Slope of the line between any two points represents average
velocity during that time interval.
Slope = rise/run
= Dx/Dt
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what if the
graph is a line?
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what if the
graph is a curve?
instantaneous velocity – the
velocity of the object at some instant (or a specific point in its path)
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may not be the same as average velocity
to
graphically determine instantaneous velocity
-construct a tangent line to the graph at that instant
-the slope of the tangent line is the instantaneous velocity
at that point
Sec 2-2
Notes Acceleration
acceleration – the
rate of change of velocity
aavg
= Dv/Dt = (vf – vi)/(tf – ti)
units: m/s/s or m/s2
a vector quantity – has both
magnitude and direction
To analyze
acceleration – use a velocity-time graph
Velocity (m/s)
Slope = rise/run = Dv/Dt
average acceleration – line of best fit
instantaneous acceleration – use slope
of tangent line at that point (derivative)
displacement – the area under the curve
for the specified time interval (integral)
example: Find the acceleration of an amusement park
ride that falls from rest to a speed of 28m/s in 3.0s.
-You daily encounter non-uniform motion, where the
acceleration is not constant.
-We most often study uniform motion in class.
Uniform motion – motion in which the
acceleration is constant
Vavg
= Dx/Dt
There are
multiple forms of equations we can use to solve a variety of problems. Memorize
these equations.
What if we
want to find the final velocity?
recall:
a = (vf-vi)/Dt
solve for vf. aDt = vf – vi
aDt + vi
= vf
vf = vi + aDt
- use to find vf
of an object moving with uniform acceleration after it has accelerated at a
constant rate for any time interval
example: A plane starting at rest at one end of a
runway undergoes a uniform acceleration of 4.8m/s2 for 15 s before
takeoff. What is the speed at
takeoff? How long must the runway be for
the plane to be able to take off?
Working with UNIFORM
ACCELERATION, we can also calculate average velocity in another way. It is simply an average.
vavg = (vi +vf)/2
so… if we equate these two
expressions for vavg,
Dx/Dt = (vi + vf)/2
and rearranging…
Dx = ½ (vi + vf)Dt (some texts use d for Dx)
example: A bicyclist accelerates from 5.0 m/s to 16 m/s in 8s. Assuming uniform acceleration, what distance
does the bicyclist travel during this time interval?
What is
another way to find displacement if I don’t know the final velocity?
recall: vf = vi + aDt
so Dx = ½ (vi
+ vf)Dt becomes
Dx = ½ (vi + vi + aDt)Dt
Dx = ½ (2vi + aDt)Dt
Dx = viDt + ½ aDt2
Dx = vit + ½ at2
example: How long must the runway be for the plane to
be able to take off?
What if we
don’t know the time interval in which the uniform motion occurred?
recall:
Dx = ½(vi
+vf)Dt
solve for Dt Dt = (2Dx)/(vi
+ vf)
now use vf = vi + aDt
substitute and solve to yield
vf2
= vi2 + 2aDx or vf2
= vi2 + 2ad
example: An aircraft has a landing speed of
302km/h. The landing area of an aircraft
carrier is 195m long. What is the
minimum uniform acceleration required for a safe landing?
Sec 2-3
Notes Falling
objects
-freely falling objects undergo constant acceleration
free fall – motion
of an object falling with a constant acceleration
- neglect air
resistance
- near the surface of a planet
- freely falling objects always have the same downward acceleration
What if an object is tossed straight up?
- what happens to the velocity of the object?
- v at the top of the path is 0m/s
- what happens to the acceleration of the object?
- a is ALWAYS constant
Different planets and moons have different accelerations due
to gravity.
- for Earth (we also use g) a = 9.81
m/s2
- for Earth’s moon a
= 1.64 m/s2
- for Mars a = .377g = 3.70 m/s2
- for Jupiter a = 2.364g = 23.19 m/s2
example:
A ball is thrown straight up into the air at an initial velocity of 20.0
m/s. Create a table showing the ball’s
position, velocity, and acceleration for the first 5.00s of its motion. Find the ball’s time, position, velocity, and
acceleration at the top of its flight and at the bottom.
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Time (s) |
Position (m) |
Velocity (m/s) |
Acceleration (m/s2) |
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