chapter 2: kinematics - saint mary's university | astronomy &...

SMU PHYS1100.1, fall 2008, Prof. Clarke 1

Chapter 2: Kinematics

2.1 Motion in 1-D

Kinematics: the study of motion without regard to its cause.

- a car along a straight road

- a brick tossed vertically in the air

- a ball rolling down a hill

Recall the five representations:

verbal, pictorial, physical, graphical, mathematical

Here we introduce graphical representations, and converting from physical representations (motion diagrams) to graphicalrepresentations (graphs, plots).



1. Uniform motion: straight-line motion in which equal displacements occur during any successive equal time interval.

- velocity remains constant (magnitude and direction)- no curvature in particle path

We wish to represent uniform motion both as a motion diagram and as a position vs. time graph.

position can be represented by x, y, z, or generically, s.

First thing we need to create a graph is a coordinate system.

e.g., origin at front, left, bottom corner of room, +x point to your right along floor, +y points up.



Example 1: me walking across room from left to right at a constant speed.

a) draw the motion diagram.





b) draw the x vs. t graph.






c) draw the v vs. t graph.







v

t

Horizontal line

velocity is constant



Uniform motion is:

- equal spaced dots on a motion diagram;

- straight line on an x (or y or s) vs. t graph;

- horizontal line on a v vs. t graph.

The slope on a position vs. time graph is the velocity.

Slope is always the rise (∆∆∆∆s) over the run (∆∆∆∆t).

sf = si + vs∆∆∆∆t (uniform motion)

∆∆∆∆t∆∆∆∆s

∆∆∆∆tsf −−−− sivs = =



Clicker question 2.1

Which position-versus-time graph represents the motion shown in the motion diagram?




v points to left v negative x vs. t graph slopes down to the right (negative slope); and

x > 0 at t = 0.



2. Non-Uniform Motion: on a motion diagram, path needs not be straight, and displacements need not be equal between successive equal time intervals.

- expect varying slope on a position vs. time graph

- expect varying velocity on a v vs. t graph.

With non-uniform velocity, we need the concept of an instantaneous velocity: the rate at which position changes with respect to time at a particular moment in time.




Example 2: Suppose as I walk across the room from left to right, I speed up.







v

t

Line with positive slope

velocity is increasing



Instantaneous velocities and motion diagrams.

Motion diagrams give only average velocities. How do we measure the instantaneous velocity at x ?

As we “magnify” the region around x the time interval gets smaller and smaller and the velocity vectors become more and more “constant”.

This “constant” velocity is the instantaneous velocity we seek.



If we could magnify the motion diagram indefinitely so that ∆∆∆∆t 0, then the velocity vectors at x would truly be constant.

We write: (instantaneous velocity).vs = lim∆∆∆∆s

∆∆∆∆t

ds

dt∆∆∆∆t 0

=

ds/dt is the derivative of s(t)

with respect to t.

The average velocity is the slope of the secant that passes through points 2 and 3

instantaneous velocity is the slope of the tangent that “kisses” the graph at x.



So now for a little calculus…




…Don’t worry, Mr. Bill, it’s not that bad!



If s(t) = atn,

Until November, there’s only one derivative we’ll need:

ds(t)

dt

e.g., x = t2 (= 1t2): a = 1, n = 2, n−1 = 1; dx/dt = (1)(2)t1 = 2t

e.g., x = −3t4 : a = −3, n = 4, n−1 = 3; dx/dt = (−3)(4)t3 = −12t3

e.g., y = 3/t2 = 3t−2: a = 3, n = −2, n−1 = −3; dy/dt = (3)(−2)t−3 = −6t−3

e.g., s = 4t (= 4t1): a = 4, n = 1, n−1 = 0; ds/dt = (4)(1)t0 = 4

e.g., s = c (= ct0): a = c, n = 0, n−1 = −1; ds/dt = (0)(c)t−1 = 0

then = antn−−−−1 is its derivative.

ended here, 11/9/08

notes 2.1



s(t) = 2 t2 mRelating velocity and position graphs.

Position, s, as a function of time: s(t) = 2 t2 metres.

Slope of tangent to graph is the

derivative, and is also the

instantaneous velocity.



Relating velocity and position graphs.

Position, s, as a function of time: s(t) = 2 t2 metres.

Slope of tangent to graph is the

derivative, and is also the

instantaneous velocity.

ds/dt = vs = 4 t m/s (units!)

We say that the position increases quadratically in time, whereas the velocity increases linearly in time.

s(t) = 2 t2 m

vs(t) = 4 t m/s



a) No, A and B never have the same speed.

b) Yes, at t = 0 s.c) Yes, at t = 2 s.d) Yes, when 0 s < t < 2 s.e) Yes, when 2 s < t < 4 s.





b) Yes, at t = 0 s.c) Yes, at t = 2 s.d) Yes, when 0 s < t < 2 s.e) Yes, when 2 s < t < 4 s.


A and B have the same position at t=2, but since their position graphs have different constant slopes, their speeds are always different.





b) Yes, at t = 1 s and 3 s.c) Yes, at t = 2 s.d) Yes, when 0 s < t < 1 s.e) Yes, when 1 s < t < 3 s.





b) Yes, at t = 1 s and 3 s.c) Yes, at t = 2 s.d) Yes, when 0 s < t < 1 s.e) Yes, when 1 s < t < 3 s.

A’s position graph has a varying slope, and only at t = 2 does its slope match the constant slope of B’s position graph. The instantaneous speeds are the same only where the slopes of the position graphs are the same.



To see this, let’s step back a bit to the average velocity before we let ∆∆∆∆t 0:

2.4 Finding position from velocity

vs = lim∆∆∆∆s

∆∆∆∆t

ds

dt∆∆∆∆t 0

= tells us how to find velocity from position. How do we do the reverse?

vavg = = ∆∆∆∆s

∆∆∆∆t

sf −−−− si

∆∆∆∆t

sf = si + vavg∆∆∆∆t = si + vavg(tf – ti)vavg

tfti

t

v

∆∆∆∆t

vavg

This is an “area”* on a v vs. t graph!

Solving for sf, we get:

*with units (m/s) s = m



We generalise this by saying:

∆s = sf – si is the area under the v vs. t graph.

What is ∆s at t = 1 s?

let ti = 0 and tf = t






(helpful hint: area of a triangle is ½ base x height)







½ (1 s)(4 ms−1) = 2 m







½ (1 s)(4 ms−1) = 2 m


½ (3 s)(12 ms−1) = 18 m







½ (1 s)(4 ms-1) = 2 m


½ (3 s)(12 ms-1) = 18 m

What is ∆s at any time t?







½ (1 s)(4 ms-1) = 2 m


½ (3 s)(12 ms-1) = 18 m

What is ∆s at any time t?

½ (t)(4t) = 2t2 mlet ti = 0 and tf = t



We do what any good physicist would do: We approximate!

We replace the curve with an “equivalent” set of rectangles, then add up the areas of the rectangles.

But this could mean lots of adding, especially if we use lots of rectangles to make our approximation a good one.

Straight-line v vs. t graphs are easy, so I’ll throw you a “curve”…

What about a curved v vs. t graph? Then what?



And so….more calculus!!!



We want to sum the areas of the

rectangles:

Calm down, Mr. Bill. It’s still not that bad!.




rectangles:

and, we want lots of rectangles to make our approximation a good one.





rectangles:


Then sf is just si plus this area:





rectangles:


Then sf is just si plus this area:


Trouble is, as ∆∆∆∆t 0, N , and none of us will live long enough to do all that adding! Now that is worth crying OHH NOOO!

88 88



And so thank goodness for Newton and Leibniz who taught us a very quick way to add an infinity of numbers: integrals!

Capital Greek letter “sigma”

(their S) which stands for “sum”.

The integral sign is a stretched Roman

letter “S” which also stands for “sum”.

∆∆∆∆t is a finite quantity, whereas dt is an infinitesimal. But vsdt is still an “area” (of an

infinitely thin rectangle), and the integral sign is still telling us to add them up!

And the fundamental theorem of the calculus gives us the power…

If vs(t) = a tn

vs(t) dt = a tn dt = a tn+1 = a (tfn+1 – ti

n+1) (n = −1)tf

ti

tf

ti

1n +1

tf

ti

1n +1



A worked example (2.10)

Suppose the particle starts at x = 30 m at t = 0 s.

a) Draw a motion diagram from the v vs. t graph.






40 (m)3020100

each rectangle has an “area” of (5 ms−−−−1)(1 s) = 5 m

each triangle is half a rectangle = 2.5 m

7.5m






40 (m)3020100



7.5m

2.5m

“turning point”






40 (m)3020100



7.5m

2.5m

“turning point”

−−−−2.5m

−−−−7.5m






40 (m)3020100



7.5m

2.5m

−−−−2.5m

−−−−7.5m

−12.5m −17.5m

“turning point”vx




Alternative way to compute distances: use the integral formula.

First need vx(t), a straight line

∆∆∆∆t = 2 s

∆∆∆∆vx = −10 m/s

vx(t) = m t + b = −5 t + 10

slope m = ∆∆∆∆vx/ ∆∆∆∆t = −−−−5 ms−−−−2






∆∆∆∆t = 2 s

intercept b = 10 ms−−−−1

vx(t) = m t + b = −5 t + 10

∆∆∆∆vx = −10 m/s

slope m = ∆∆∆∆vx/ ∆∆∆∆t = −−−−5 ms−−−−2






∆∆∆∆t = 2 s


slope m = ∆∆∆∆vx/ ∆∆∆∆t = −−−−5ms−−−−2

∆∆∆∆vx = −10 m/s

vx(t) = m t + b = −5 t + 10

xf = xi + vx(t) dt = 30 + −5t dt + 10 dtt

0

5

2= 30 − t2 + 10 t

t

0

tf

ti






∆∆∆∆t = 2 s


t 0 1 2 3 4 5 6

xf 30.0 37.5 40.0 37.5 30.0 17.5 0.0vx 10.0 5.0 0.0 −5.0 −10.0 −15.0 −20.0

“turning point”

slope m = ∆∆∆∆vx/ ∆∆∆∆t = −−−−5ms−−−−2

∆∆∆∆vx = −10 m/s

vx(t) = m t + b = −5 t + 10

xf = xi + vx(t) dt = 30 + −5t dt + 10 dtt

0

5

2= 30 − t2 + 10 t

t

0

tf

ti



Which of the blue x vs. tgraphs goes with the green v vs. t graph?

Clicker question 2.4 Let the particle’s position at ti = 0 s be xi = –10 m .



Clicker question 2.4 Let the particle’s position at ti = 0 s be xi = –10 m .

Which of the blue x vs. tgraphs goes with the green v vs. t graph?



vfs = vis + as∆∆∆∆t

as = =∆∆∆∆t∆∆∆∆vs

∆∆∆∆tvfs−−−− vis

2.5 Motion with constant acceleration

If a particle is undergoing constant acceleration, its v vs. tgraph is a straight line. Then, we can write:

sf = si + vis∆∆∆∆t + ½ as ∆∆∆∆t2

and since displacement is the area under the v vs. t

graph, we have:

(m/s2)



Three kinematical equations for constant acceleration

vfs = vis + as∆∆∆∆t

sf = si + vis∆∆∆∆t + ½ as∆∆∆∆t2

vfs2 = vis

2 + 2as∆∆∆∆s (derive on board)

∆∆∆∆s = sf −−−− s

i

For “free-fall”problems, as = −g, where g = +9.80 m/s2

vfy = viy −−−− g∆∆∆∆t

yf = yi + viy∆∆∆∆t − ½ g∆∆∆∆t2

vfy2 = viy

2 −−−− 2g∆∆∆∆y

g is never negative; it is the magnitude of the acceleration vector. The direction is given by the negative sign: −.

is the acceleration of gravity



Examples for the board…

1. A rocket sled accelerates at 50 ms−2 for 5.0 s, coasts for 3.0 s, then deploys a parachute and accelerates at −2.5 ms−2 until stopping.

a) What is the maximum speed of the rocket sled?

b) What is the total distance traveled?

out of fuel deploy parachute

notes 2.2

ended here, 16/9/08




2. A cannon ball is shot straight up with an initial speed of 100 ms-1.

a) How high does it go?

b) How long does it take to reach its maximum height?

notes 2.3




3. A ball falling by a window 2 m high can be seen from inside the room for 0.25 s. From what height above the window sill was the ball dropped if it was dropped from rest?

We break this problem into two parts. First we deal with the window, then with the fall before the window.

v1i v2f

v2i = 0

∆∆∆∆s2

∆∆∆∆s1∆∆∆∆t

−−−−g

−−−−g

notes 2.4



2.10 Motion down an inclined plane

Consider an object sliding freely (no friction) down an inclined plane. The object cannot pass through the plane, and so the object must accelerate only with the component of the free fall acceleration parallel to the plane. Thus,

as = a|| = ± g sinθ

where the sign will depend on which way the plane is inclined.




A horse pulls a sleigh along the snowy street at a constant speed of 2.0 m/s.

A kid on a toy sleigh at the top of a 10.0 m long frictionless ramp inclined at 10° with the horizontal starts sliding down the ramp the instant the horse passes him by.

H

Will the kid overtake the horse and, if so, how far along the ramp?k

10°10 m

notes 2.5



The ball rolls up the ramp, then back down.

Which is the correct acceleration graph?




Which of the green v vs. tgraphs goes with the orange a vs. t graph?


vx

t0

(a)

vx

t0

(b)

vx

t0

(c)

vx

t0

(d)



If the v vs. t graph is not a straight line, the acceleration is not constant, and we must define an instantaneous acceleration, as.

as = lim =∆∆∆∆vs

∆∆∆∆t

dvs

dt∆∆∆∆t 0

(c) example of non-uniform acceleration

as

t0

vs

tvis

t

s

si

cubic

parabola



Determining v from an a vs. t graph

Just like ∆∆∆∆s is the area under a v vs. t graph, ∆∆∆∆v is the area under the a vs. t graph:

vfs = vis + lim (as)k ∆∆∆∆t = vis + as dtΣΣΣΣk=1

N

∆∆∆∆t 0

tf

ti

tfti

t (s)

as (ms−2)

The area of a rectangle on an as vs. t graph has units ms−2

(height) x s (width) = ms−1

(units of velocity). The sum of the areas of all the green rectangles is the net change in velocity from ti to tf.

0




Suppose a(t) = 6t − 12 (ms−2).

a) If v(0) = 9, find v(t).

b) If x(0) = 0, find x(t).

c) What are the position, velocity, and acceleration at t = 2? What can you say about the velocity at this time?

d) When is (are) the turning point(s), and what is the acceleration and position there?

e) Graph a(t), v(t), and x(t).

notes 2.6

ended here, 18/9/08




A sprinter accelerates at 2.5 m/s2 until reaching her top speed of 10 m/s. She then continues to run at top speed. How long does it take her to run the 100 m dash?

(answer: 12 s)

v(t)

ttfti tm

10

0

∆∆∆∆t

∆∆∆∆v

notes 2.7




(EOC 31) Three particles, A, B, and C, move along the x-axis each with vx(0) = 10 ms−1. Using the graphs only (and not the kinematical equations), find the velocity of each at t = 7 s.

(answers: vA(7) = −10 ms−1; vB(7) = −20 ms−1; vC(7) = 75 ms−1.)

notes 2.8



A) If the particle starts from rest at t = 0 s, what is the particle’s speed at t = 1 s?

a) 2 ms-1 b) 5 m/sc) 6 m/s d) 8 ms-1


0 1 2 3 40

2

4

6

8

t (s)

a (ms-2)




a) 2 ms-1 b) 5 m/sc) 6 m/s d) 8 ms-1


0 1 2 3 40

2

4

6

8

t (s)

a (ms-2)




a) 2 ms-1 b) 5 m/sc) 6 m/s d) 8 ms-1


0 1 2 3 40

2

4

6

8

t (s)

a (ms-2)

B) At what time does the particle first reach its maximum speed?

a) 1 s b) 2 s c) 3 s d) 4 s

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