16.102 labs – from web page
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Recommended problems, chapter 1 – from web page
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Chapter 2: Kinematics in One Dimension
Will cover motion in a straight line with constant acceleration:
• Displacement
• Speed, velocity, acceleration
• Equations of motion in one dimension
• Free fall under gravity
• Graphical representation
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Displacement, average speed, velocity
Car starts at xo at time to,
reaches x at time t
Distance travelled = x - xo
Displacement, D!x =!x −!x0
Average speed =
x − x0
Distance
=
Elapsed time t − t0
Average velocity =
D!x
Displacement
=
Elapsed time t − t0
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Displacement and distance not necessarily the same
Example: Car travels 50 km to east, then 20 km to
west in 1 hour.
Distance travelled = 50 + 20 = 70 km
Average speed = 70 km/h
Displacement =!x f inal −!xinitial = 30 km to east
Average velocity = 30 km/h to east
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Example: A car makes a trip due north for 3/4 of the time and due south for
1/4 of the time. The average northward velocity has a magnitude of 27 m/s.
The average southward velocity has a magnitude of 17 m/s.
What is the average velocity for the entire trip?
Put T = time for the entire trip.
x1 = (3T/4) x (27 m/s)
x2 = (T/4) x (17 m/s)
Average velocity = Displacement/Time
= (x1 – x2)/T, to the north
= 3 × 27/4 – 17/4 m/s
= 16 m/s, to the north
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Instantaneous Velocity
The velocity measured during a vanishingly small time interval.
D!x
!v = lim
Dt→0 Dt
This differs from the average velocity because the average is
measured over an extended time during which the object may
be accelerating.
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Acceleration
Change in velocity !v −!v0
Average acceleration =
=
Elapsed time
t − t0
D!v
Instantaneous acceleration = lim
Dt→0 Dt
Any change of velocity, including slowing down, is an “acceleration”.
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Average acceleration
9−0
= 9 km/h per second
1−0
9000
=
= 2.5 m/s2
3600
ā =
18 − 9
ā =
= 9 km/h per second
2−1
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Equations of Motion
Consider an object that has speed vo at time t = 0.
It is accelerated in a straight line at a constant rate to speed v at time t.
Acceleration:
v − v0
a=
, so v = v0 + at
t
(1)
Average speed:
v̄ =
v + v0 v0 + at + v0
x − x0
=
=
2
2
t
1
x − x0 = v0t + at 2
2
(2)
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From previous page:
v̄
x − x0
t
v
v0
2
v0
at
2
v0
1
x − x0 = (v + v0)t
2
And:
v − v0 = at
v + v0 x − x0
=
2
t
(3)
(v − v0) ×
(v + v0)
(x − x0)
= at ×
2
t
v2 − v20 = 2a(x − x0)
(4)
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The famous four formulae
v = v0 + at
(1)
1 2
x − x0 = v0t + at
2
(2)
1
x − x0 = (v + v0)t
2
(3)
v2 − v20 = 2a(x − x0)
(4)
You will definitely need to know these!
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Example: A runner accelerates to a velocity of 5.36 m/s due west in 3
seconds. His average acceleration is 0.640 m/s2, also directed due west. What
was his velocity when he began accelerating?
Take quantities pointing to the east (right) as positive.
v0 = ?
v = −5.36 m/s
a = −0.640 m/s2
t=3s
(1)
v − v0 = at
So:
v0 = v − at = −5.36 − (−0.640) × 3 = −3.44 m/s
Answer: 3.44 m/s due west.
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Example: A car accelerates from rest to a final speed in two stages. Each
stage takes the same time T.
In stage 1, the car’s acceleration is a = 3.0 m/s2 and ends at speed v1.
At the end of stage 2, the car is travelling 2.5 times as fast as at the end of
stage 1. The acceleration is a'.
What is the acceleration during stage 2?
Speed at the end of stage 1 is v1 = aT
(v0 = 0)
Speed at the end of stage 2 is v2 = v1 + a!T = 2.5v1
a! = ?
So, a!T = (2.5 − 1)v1 = 1.5aT
Therefore, a! = 1.5a = 4.5 m/s2
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