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Applied Natural Sciences
Leo Pel
e‐mail: [email protected]
http://tiny.cc/3NAB0
Content of the course 3NAB0 (see study guide)
17 November diagnostic test!
Week 1 : 14 November
Introduction, units (Ch1), Circuits (Ch25,26) Week 2 : 21 November
Heat (Ch17), Kinematics (Ch2‐3)
Week 3: 28 November
Newton, Energy (Ch4‐6) Week 4: 5 December Energy, Momentum (Ch7‐8)
8 December Intermediate assessment
Week 5: 12 December
Rotation, Elasticity, Fluid mechanics (Ch9‐12)
Week 6: 19 December
Harmonic oscillator and Waves (Ch14‐15)
Week 7: 9 January (2016)
Sound (Ch16) Light (Ch33)
25 January Final assessment
Chapter 2
Motion Along a
Straight Line
Application of calculus in Mechanics: 1D
PowerPoint® Lectures for
University Physics, Thirteenth Edition
– Hugh D. Young and Roger A. Freedman
Lectures by Wayne Anderson
Copyright © 2012 Pearson Education Inc.
LEARNING GOALS
• How to describe straight-line motion in terms of average
velocity, instantaneous velocity, average acceleration,
and instantaneous acceleration.
• How to interpret graphs of position versus time, velocity
versus time, and acceleration versus time for straightline motion.
• How to solve problems involving straight-line motion
with constant acceleration, including free-fall problems.
• How to analyze straight-line motion when the
acceleration is not constant.
Mechanics (‘the limitation’)
Each object is considered to be a point mass
mass M
mass M
“point”mass M
book: “particle”
Objects (therefore) do not rotate
Chapter 2: kinematics Kinematics: the description of motion
In 3 dimensions, this means determining the position, r, of an object at any instant of time, t. co-ordinates
unit vectors
We will limit ourselves to 1D/2D
Chapter 2: kinematics in 1D
Kinematics: the description of motion
Determination of motion of an object along a straight line (1 dimension), means determining the co‐ordinate, x, of an object at any instant of time, t. x co-ordinate
unit vector
Motion along a straight line
Which quantities of an object do you need to know to be able to describe its motion along a straight line?
Answer:
velocity and acceleration at any time
(as a vector, so including direction!)
What data is also needed for a unique determination of position of an object at any time?
Answer:
Position at t=0, and a definition of a co‐ordinate system (for determination of positive and negative direction)
Motion along a line
y-axis
Position, x, of an object at any time, t
(velocity and acceleration can be derived)
0
x1 at t1
 Choose co‐ordinate system
Choose x‐axis along the direction of motion  only x co‐ordinate
Choose direction ‘positive’ x axis
 Choose origin x=0 and t=0  Define reference point : which point represents the “position” of the object
x-axis
Position as a function of time x(t)
y-axis
position as a function of time : x(t)
x-axis
0
x1 = x(t1)
average
velocity
x2 = x(t2)
Average Velocity
This plot shows the average velocity being
measured over shorter and shorter intervals.
Average Velocity
Instantaneous Velocity
Definition:
This means that we evaluate the average
velocity over a shorter and shorter period of time;
as that time becomes infinitesimally small, we
have the instantaneous velocity.
instantaneous
velocity
Average velocity
Position function
average
velocity
slope (=steepness) of the connecting line
between x1(=x(t1)) en x2=x(t2)
Instantaneous velocity
position function
position function
position function
(instantaneous) velocity
Derivate of the position function with respect to time
slope (=steepness) tangent to x(t)
Motion along a line
The graph shows the position functions of two trains running along parallel tracks. Which statement is correct?
1. At t=tB both trains have the same velocity
2. Both trains accelerate all the time

3. Both trains have the same velocity at a
time instant t<tB
4. Both trains have the same acceleration
somewhere on the graph
Answer:
3. The slope of curve B is is at a certain time instant t<tB
equal to the slope of curve A
Velocity function v(t)
Velocity function v(t)
Acceleration function a(t)
Acceleration function a(t)
average acceleration
(instantaneous) acceleration
Acceleration function VS Decelartion
Acceleration (increasing speed) and deceleration
(decreasing speed) should not be confused with the
directions of velocity and acceleration:
Accelerating
Decelerating
Decelerating
Accelerating
Motion with Constant Acceleration
Velocity vs. time:
Average velocity:
Position as a function of time:
Velocity as a function of position:
Derivation of function v(x) for constant a
v v 0  at
v v 0
t
a
1
2
x  x 0  v 0 t  at
2
2
 v  v0  1  v  v0 
x  x0  v0 
  a

 a  2  a 
( x  x0 )2a  2v0 (v  v0 )  (v  v0 ) 2
( x  x0 )2a  v0  v 2
2
Motion with Constant Acceleration
The relationship between position and time
follows a characteristic curve.
Parabola
Example: Hit the Brakes!
A park ranger driving at 11.4
m/s in back country suddenly sees
a deer “frozen” in the headlights.
He applies the brakes and slows
with an acceleration of -3.80 m/s2.
(a) If the deer is 20.0 m from the
ranger’s car when the brakes
are applied, how close does
the ranger come to hitting
the deer?
(b) What is the stopping
time?
v 2  v02 (0) 2  (11.4 m/s) 2
x 

 17.1 m d  20.0 m  17.1 m  2.9 m
2
2a
2(3.80 m/s )
v
(11.4 m/s)
v  v0  at  0  t   0  
 3.00 s
2
a
(3.80 m/s )
Free fall
A strobe light begins to fire as the apple is dropped. Notice how the space between images increases as the bal’s velocity grows.
Free fall
g = 9.8 m/s2
~105
31 km/h
Free fall
Aristotle thought that heavier bodies would fall faster.
Galileo is said to have dropped two objects, one light and one heavy, from the top of the Leaning Tower of Pisa to test his assertion that all bodies fall at the same rate.
Apollo 15 1971
Motion along a line
The graph shows the position function of a train that moves along a straight track. Which statement is correct?
1. The train moves with constant velocity

2. The train decelerated all the time
3. The train accelerates the all the time
4. The train accelerates and after that it
decelerates
Answer:
2. The slope of the curve indicates the velocity; the
slope, and hence the velocity, decreases all the time =>
the train decelerates.
Speed of a Lava Bomb
A volcano shoots out blobs of molten lava
(lava bombs) from its summit. A geologist
observing the eruption uses a stopwatch to
time the flight of a particular lava bomb that
is projected straight upward.
If the time for it to rise and fall back to its
launch height is 4.75 s, what is its initial
speed and how high did it go?
(Use g = 9.81 m/s2.)
x  x0  v0t  12 gt 2
 x  0  t (v0  12 gt )
Either t  0 or v0  12 gt  0
v0  12 gt  12 (9.81 m/s 2 )(4.75 s)  23.3 m/s
At maximum height, v 2  0  v02  2 g x
v02
(23.3 m/s) 2
x 

 27.7 m
2
2 g 2(9.81 m/s )
Summary
From the position function, x(t), one can obtain the velocity function, v(t), and the acceleration function, a(t), via differentiation with respect to time
2
dx
dv d x
x(t )  v(t ) 
 a(t ) 
 2
dt
dt dt
If a(t) is known, can v(t) and x(t) be determined?
Suppose a(t) is known!
dv
a (t ) 
dt
v(t )  Cv   a (t )dt
dx
v(t ) 
dt
x(t )  C x   v(t )dt
If the acceleration function, a(t), is known, then the velocity function, v(t), can be obtained by integration and from another integration step the position function, x(t),
can be determined if Cv en Cx are known
Determine v(t) and x(t) if a(t) is known
Initial values of v(t) and x(t) have to be known
t
v(t )  v(t0 )   a(t ) dt
t0
t
x(t )  x(t0 )   v(t ) dt
t0
Apply to: constant acceleration, i.e. a(t)=a0=constant
a(t), v(t) en x(t) van een valbeweging
a (t )  g  9.8(1) m/s 2
v(t  0)  v0 ; x(t  0)  x0
Initial values
t1
t1
0
0
t1
t1
0
0
v(t1 )  v(0)   a(t ) dt  v0   g dt  v0  g t1
x(t1 )  x(0)   v(t ) dt  x0   (v0  g t ) dt  x0  v0t1 
0
x
Constant acceleration
0
Area under
the curve
1 2
g t1
2
0
9.8 m/s2
Constant acceleration
Motion for which the acceleration is constant
a(t) = a0
v(t) = v0 + a0 t
v0 initial velocity
x(t) = x0 + v0 t + ½ a0 t2
x0 initial position
Summary
One‐dimensional motion object: x(t), v(t), a(t)
dx
dv
x(t )  v(t ) 
 a (t ) 
dt
dt
differentiation
t
v(t )  v(t0 )   a(t ) dt
t0
t
x(t )  x(t0 )   v(t ) dt
Integration
Initial values needed
t0
Motion with constant acceleration: a = constant
Acceleration: free fall (a=g)
Free fall
A ball is thrown straight upwards from a height, h, with a velocity, v0, and hits the ground at velocity v1. What is the velocity of the ball when hitting the ground when the ball is thrown from the same height, h, but now straight downwards with velocity v0, (ignore air resistence). 1. The velocity is larger than v1
 2. The velocity is equal to v1
3. The velocity is smaller than v1
4. Not enough data to know
Answer:
2. In the former situation, when coming back at height
h, the ball will have a velocity equal to the velocity with
which is was thrown upwards, but now pointing
downwards
v(t) = v0 - a0 t
Vo
v0 initial velocity
x(t) = 0 + v0 t - ½ a0 t2
initial position
Back at initial position as:
h
Vo
V?
v0 t - ½ a0 t2=0
t( v0 - ½ a0 t)=0
t=0 or t=2v0/ a0
v(t) = v0 - a0 2v0/ a0 =-v0
So same final velocity
Summary
Summary