Physics 2210 Fall 2015

smartPhysics Unit 01 Kinematics in 1D

08/26/2015

Physics Assessment Test Results

Average 7 Median 7 Standard Deviation 2

4 or less: Strongly Recommend Transfer to PHYS 1500 5 or 6: Should consider attending PHYS 1500 to see if it’s better fit

Strongly Recommend Transfer to PHYS 1500

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• There should be 4 polls opened for you to answer

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Math Assessment Test Results

Average 13 Median 13 Standard Deviation 3

• 8 or less: Almost certainly not going to pass PHYS 2210

• Will need to review math

Almost certainly not going to pas PHYS 2210

Will need review

Unit 01

Unit 01

Kinematics in 1D • Description of motion

along a straight line • Motion: Changing

“position” occurring with time.

• Time: we all agree on a common experience of time passing

• Position: the specification of where an object is located at some time

• Mathematical Description requires a “coordinate system”: – Position is given by a

coordinate x – A chosen origin (O):

where we set x=0 – A reference positive

direction (in which the value of x increases)

– A reference “length”: usually called “unit” in elementary physics we use “meter” (m)

• Also choose origin and unit for time (but not direction?)

Example

East—west addresses numbering on 800S in Salt Lake City • Origin: at Main Street

(State Street is actually 100 E)

• Reference direction (east is +: towards U!!!) – Convention is suffix of E

for + and W for − ) • Reference Unit:

centiblocks – Metric prefix: “centi”

means 1/100.

Description of Motion • We think of the position

x of an object as a function of time t

• The function 𝑥 𝑡 : 𝑡 → 𝑥

maps each value of t into a unique value of 𝑥.

• You can have the same value of x for two or more different times t

• But you cannot have two or more values of x at the same time t

t

x

t

x Cannot be at THREE places at the same time

Graphical Representation

Closed Form Math. Representation

𝑥 𝑡 = �𝑥0 + 𝑣0𝑡 0 < 𝑡 < 𝑡1

𝑥1 + 𝑣0 𝑡 − 𝑡1 + 12𝑎(𝑡 − 𝑡1)2 𝑡 > 𝑡1

t

x

t1

𝑥0

This is an example of a piece-wise function definition

𝑣0>0, a<0

Definitions Displacement ∆𝑥 over times interval ∆𝑡 that takes an object from position 𝑥𝑖 at time 𝑡𝑖 to position 𝑥𝑓 at time 𝑡𝑓: ∆𝑥 = 𝑥𝑓 − 𝑥𝑖 ∆𝑡 = 𝑡𝑓 − 𝑡𝑖 Average velocity �̅� or 𝑣 :

�̅� ≡∆𝑥∆𝑡

=𝑥𝑓 − 𝑥𝑖𝑡𝑓 − 𝑡𝑖

Instantaneous velocity

𝑣 ≡ lim∆𝑡→0

∆𝑥∆𝑡

≡𝑑𝑥𝑑𝑡

≡ �̇�

Average acceleration 𝑎� or 𝑎 :

𝑎� ≡∆𝑣∆𝑡

=𝑣𝑓 − 𝑣𝑖𝑡𝑓 − 𝑡𝑖

Instantaneous acceleration

𝑎 ≡ lim∆𝑡→0

∆𝑣∆𝑡

≡𝑑𝑣𝑑𝑡

≡ �̇�

≡𝑑2𝑥𝑑𝑡2

≡ �̈�

We also have that the object has instantaneous velocity 𝑣𝑖 at time 𝑡𝑖 that changes to 𝑣𝑓 at time 𝑡𝑓:

Poll 08-26-01

(A)

(B)

Which plot best represents the acceleration curve associated with the displacement and velocity curves shown to the left? (A) or (B)

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Simplest Cases Three simplest cases of motion: i. Stationary: constant

position: 𝑥 = 𝑥0 for all times

ii. Constant velocity: �̇� = 𝑑𝑑

𝑑𝑡= 𝑣0for all

times. iii. Constant Acceleration:

�̈� = �̇� = 𝑑𝑑𝑑𝑡

= 𝑎0for all times

• We study these special cases because they represent very good approximations to what actually happens in real the real world

• Example: airliners cruise at constant speed

• Objects slide at constant speed on frictionless surfaces

• Free fall at surface of Earth occurs at constant acceleration

Constant Velocity Motion • To solve problems we

want to express position as a function of time: �̇� = 𝑣0.

• Reversing the differentiation ̇ ≡ 𝑑

𝑑𝑡 we

integrate �̇� = 𝑣0 from 𝑡 = 𝑡𝑖 to 𝑡 = 𝑡𝑓 (a definite integral):

*** The object starts at position 𝑥𝑖 at time 𝑡𝑖 and moves to position 𝑥𝑓 at time 𝑡𝑓

𝑥𝑓 − 𝑥𝑖 = � 𝑣0𝑑𝑡𝑡𝑓

𝑡𝑖

= 𝑣0 𝑡𝑓 − 𝑡𝑖

𝑥𝑓 = 𝑥𝑖 + 𝑣0(𝑡𝑓 − 𝑡𝑖)

OR more generally replace 𝑡𝑓 by 𝑡 : valid for 𝑡 > 𝑡𝑖 to as long as velocity remains constant:

𝑥 = 𝑥𝑖 + 𝑣0(𝑡 − 𝑡𝑖) (1)

If we take the particular case 𝑡𝑖 = 0, and 𝑥𝑖 = 𝑥0 at that time, then

𝑥 = 𝑥0 + 𝑣0𝑡 (2)

Example 1 (1/3) A submarine can use sonar (sound traveling through water) to determine its distance from other objects. The time between the emission of a sound pulse (a "ping") and the detection of its echo can be used to determine such distances. Alternatively, by measuring the time between successive echo receptions of a regularly timed set of pings, the submarine's speed may be determined by comparing the time between pings. Assume you are the sonar operator in a submarine traveling at a constant velocity underwater. Your boat is in the eastern Mediterranean Sea, where the speed of sound is 1522 m/s. If you sent out pings every 4.6 s, and your apparatus receives echoes reflected from an undersea cliff every 4.58 s, how fast is your submarine approaching the cliff?

Example 1 (2/3) Speed of sound is 1522 m/s. If you sent out pings every 4.600 s, and you receive echoes every 4.580 s, how fast is your submarine approaching the cliff? /* This problem takes advantage of the fact that both the emitted and reflected pings are equally spaced with distance S, and traveling at speed vsound = 1522 m/s. We will first determine the spacing of the sub. At time t=0, the submarine emits the first ping from the origin, the ping travels to the right with speed vsound, so at time t=T=4.6 s, it is at position */

xp1: 0 + vsound*T;

(%o1) vsound T

(%i2) /* at time t=T, the submarie is now at location */

xs1: 0 + vsub*T;

(%o2) vsub T

(%i3) /* and since the sub now emits a second ping, at t=T, ping 2 is at */

xp2: xs1;

(%o3) vsub T

(%i5) /* The pings are traveling spaced by distance S given by */

S: xp1 - xp2;

(%o5) vsound T - vsub T

Example 1 (3/3) Speed of sound is 1522 m/s. If you sent out pings every 4.600 s, and you receive echoes every 4.580 s, how fast is your submarine approaching the cliff? (%i6) /* now we look at the situation after reflection. The pings are traveling in the -x direction while the sub continues to travel in the +x direction. At time t=0, ping 1 reaches the sub, and clearly ping 2 is now at x=+S. At time t=t2=4.58s, the sub meets ping2. Asssuming the position of the sub at t=0 to be the origin, then the position of the sub at t=t2 is given by */

xs: 0 + vsub*t2;

(%o6) t2 vsub

(%i7) /* ping 2, however, started at S, and travels in the -x direction with speed vsound, so its osition at t=t2 is */

xp: S - vsound * t2;

(%o7) - vsub T + vsound T - t2 vsound

(%i8) /* The two meet at t=t2, so xp = xs, and we solve for vsub */

vsub2: rhs(solve(xp=xs, vsub)[1]);

vsound T - t2 vsound

(%o8) --------------------

T + t2

(%i9) /* substittue in the given values */

vsub2, t2=4.58, T=4.60, vsound=1522;

(%o9) 3.315904139433507

Answer: 3.32 m/s

Constant acceleration Motion Constant �̈� = �̇� = 𝑎. integrate �̇� = 𝑎 from 𝑡 = 𝑡𝑖 to 𝑡 = 𝑡𝑓 (a definite integral):

𝑣𝑓 − 𝑣𝑖 = � 𝑎𝑑𝑡𝑡𝑓

𝑡𝑖

= 𝑎 𝑡𝑓 − 𝑡𝑖

𝑣𝑓 = 𝑣𝑖 + 𝑣0(𝑡𝑓 − 𝑡𝑖) 𝑣 𝑡 = 𝑣𝑖 + 𝑎(𝑡 − 𝑡𝑖) (3)

*** The object starts with velocity 𝑣𝑖 at time 𝑡𝑖 and accelerates to 𝑣𝑓 at time 𝑡𝑓

𝑣 𝑡 = 𝑣0 + 𝑎𝑡 (4)

𝑥𝑓 − 𝑥𝑖 = � 𝑣𝑖 + 𝑎(𝑡 − 𝑡𝑖)𝑑𝑡𝑡𝑓

𝑡𝑖

= 𝑣𝑖 𝑡𝑓 − 𝑡𝑖 + 12𝑎 𝑡𝑓 − 𝑡𝑖

2

𝑥𝑓=𝑥𝑖 + 𝑣𝑖 𝑡𝑓 − 𝑡𝑖 + 12𝑎 𝑡𝑓 − 𝑡𝑖

2

*** The object starts at position 𝑥𝑖 at time 𝑡𝑖 and moves to position 𝑥𝑓 at time 𝑡𝑓

𝑥(𝑡)=𝑥𝑖 + 𝑣𝑖 𝑡 − 𝑡𝑖 + 12𝑎 𝑡 − 𝑡𝑖 2

(5)

If we take the particular case 𝑡𝑖 = 0, and 𝑥𝑖 = 𝑥0 and 𝑣𝑖 = 𝑣0 at that time, then

𝑥(𝑡)=𝑥0 + 𝑣0𝑡 + 12𝑎𝑡

2 (6)

Poll 08-26-02

(A) 3.00 foot (B) 1.00 feet (C) 6.00 feet (D) 4.00 feet (E) 2.00 feet

At t=0 a ball, initially at rest, starts to roll down a ramp with constant acceleration. You notice it moves 1.00 foot between t=0 seconds and t = 1.00 second.

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Example 2 (1/3) Allison runs north along a very long, straight track at a constant speed of 5.6 m/s

Brad was standing at rest at the 50m mark, At 4.0 s after Allison passes the −10m line, he starts to move southward at a constant acceleration with magnitude of 1.6 m/s2.

(a) How long after passing the -10m mark will Allison collide with Brad?

(b) At what position will Allison collide with Brad?

(%i1) /* Allison: at xA0=-10m at t=0, north +

constant velocity vA=+5.6m/s^2 */

xA: xA0 + vA*t;

(%o1) xA0 + t vA

(%i2) /* Brad: at xB1=50m/s at t=t1=4.0s, starting at rest:

vB1=0 constant acceleration(southward is negative) aB=-1.6

xB = xB1 + vB1*(t-t1) + aB/2*(t-t1)^2 */

xB: xB1 +aB/2*(t-t1)^2;

2

(t - t1) aB

(%o2) xB1 + ------------

2

Example 2 (2/3) (%i3) soln1: solve(xA=xB, t); 2 sqrt(- 2 aB xB1 + 2 aB xA0 + vA + 2 t1 aB vA) - vA - t1 aB (%o3) [t = - -----------------------------------------------------------, aB 2 sqrt(- 2 aB xB1 + 2 aB xA0 + vA + 2 t1 aB vA) + vA + t1 aB t = -----------------------------------------------------------] aB (%i4) /* note we have two solutions from the quadratic pick positive solution [2] */ t2: rhs(soln1[2]),numer; 2 0.5 (- 2 aB xB1 + 2 aB xA0 + vA + 2 t1 aB vA) + vA + t1 aB (%o4) ---------------------------------------------------------- aB (%i5) t2, aB=-1.6, xB1=50, xA0=-10, vA=5.6, t1=4.0; (%o5) - 7.197402159170326 /* turns out to be negative, try the other solution */ t2: rhs(soln1[1]),numer; 2 0.5 (- 2 aB xB1 + 2 aB xA0 + vA + 2 t1 aB vA) - vA - t1 aB (%o6) - ---------------------------------------------------------- aB

Example 2 (3/3) (%i7) t2, aB=-1.6, xB1=50, xA0=-10, vA=5.6, t1=4.0;

(%o7) 8.197402159170327

Answer: (a) They collide at time t=8.2 s after Allison passes -10m mark

(%i9) xA2: xA, t=t2;

2 0.5

vA ((- 2 aB xB1 + 2 aB xA0 + vA + 2 t1 aB vA) - vA - t1 aB)

(%o9) xA0 - ---------------------------------------------------------------

aB

(%i10) xA2, aB=-1.6, xB1=50, xA0=-10, vA=5.6, t1=4.0;

(%o10) 35.90545209135383

(%i11) xB2: xB, t=t2;

2 0.5

(- 2 aB xB1 + 2 aB xA0 + vA + 2 t1 aB vA) - vA - t1 aB 2

aB (- ---------------------------------------------------------- - t1)

aB

(%o11) -----------------------------------------------------------------------

2

+ xB1

(%i12) xB2, aB=-1.6, xB1=50, xA0=-10, vA=5.6, t1=4.0;

(%o12) 35.90545209135382

Answer: (b) They collide at x = 35.9 m

Extra Example 3 (1/2) Allison runs north along a very long, straight track at a constant speed of 5.6 m/s

At 11 s after Allison passes the −15m line, Brad was at the 70m mark, running north also but with a speed of 4.8m/s.

(a) Will Allison ever catch Brad?

(b) How long after passing the -15m mark will Allison catch Brad?

(c) At what position will Allison catch Brad?

(%i1) /* Allison passes xA0=-15m at t=0, with velocity vA=5.6m */ xA: xA0 + vA*t; (%o1) xA0 + t vA (%i2) /* Brad passes xB1=70m at time t=t1=11s, with constant velocity vB=4.8m/s */ xB: xB1 + vB*(t-t1); (%o2) xB1 + (t - t1) vB (%i3) /* solve for time t2 at which they meet */ soln1: solve(xA=xB, t); xB1 - xA0 - t1 vB (%o3) [t = - -----------------] vB - vA

Extra Example 3 (2/2) (%i4) t2: rhs(soln1[1]); xB1 - xA0 - t1 vB (%o4) - ----------------- vB - vA (%i5) /* substitute in the given values */ t2, xB1=70, xA0=-15, t1=11, vA=5.6, vB=4.8; (%o5) 40.25000000000001

Answers: (a) Yes she catches him, (b) at 40.25s after passing -15m (%i6) xA: xA, t=t2; vA (xB1 - xA0 - t1 vB) (%o6) xA0 - ---------------------- vB - vA (%i7) xA, xB1=70, xA0=-15, t1=11, vA=5.6, vB=4.8; (%o7) 210.4 (%i8) xB: xB, t=t2; xB1 - xA0 - t1 vB (%o8) vB (- ----------------- - t1) + xB1 vB - vA (%i9) xB, xB1=70, xA0=-15, t1=11, vA=5.6, vB=4.8; (%o9) 210.4

Answer (c) at position x=210.4m