Introduction to Mechanics Practice using the Kinematics ...nebula2.deanza.edu/~lanasheridan/P50/Phys50-Lecture10.pdfIntroduction to Mechanics Practice using the Kinematics Equations

Introduction to MechanicsPractice using the Kinematics Equations

Lana Sheridan

De Anza College

Jan 24, 2018

Last time

• finished deriving the kinematics equations

• some problem solving practice

Overview

• using kinematics equations

• more problem solving practice

• free fall (?)

The Kinematics Equations

For constant acceleration:

v = v0 + at

∆x = v0t +1

2at2

∆x = vt −1

2at2

∆x =v0 + v

2t

v2 = v20 + 2 a∆x

For zero acceleration:

∆x = vt

Using the Kinematics Equations to solve problems

Process:

1 Identify which quantity we need to find and which ones we aregiven.

2 Is there a quantity that we are not given and are not askedfor?

1 If so, use the equation that does not include that quantity.

2 If there is not, more that one kinematics equation may berequired or there may be several equivalent approaches.

3 Input known quantities and solve.

Example 4

A car driver sees a speed trap ahead, while driving at 30.0 m/s(∼ 67 mph). Assume the car brakes with constant deceleration of2.30 m/s2. It slows to 20.0 m/s, what was the time taken to slow?

Sketch:

28 CHAPTER 2 ONE-DIMENSIONAL KINEMATICS

continued from previous page

InsightNote that the average acceleration for these six seconds is not simply the average of the individual accelerations;

and The reason is that different amounts of time are spent with each acceleration. In addition, theaverage acceleration can be found graphically, as indicated in the v-versus-t sketch on the previous page. Specifically, the graphshows that is 2.5 m/s for the time interval from to

Practice ProblemWhat is the average acceleration of the train between and [Answer:

]

Some related homework problems: Problem 32, Problem 34

16.0 s - 2.0 s2 = -0.38 m/s2aav = ¢v>¢t = (3.0 m/s - 4.5 m/s)>t = 6.0 s?t = 2.0 s

t = 6.0 s.t = 0¢v

-1.5 m/s2.2.0 m/s2, 0 m/s2,

In one dimension, nonzero velocities and accelerations are either positive ornegative, depending on whether they point in the positive or negative direction ofthe coordinate system chosen. Thus, the velocity and acceleration of an object mayhave the same or opposite signs. (Of course, in two or three dimensions the rela-tionship between velocity and acceleration can be much more varied, as we shallsee in the next several chapters.) This leads to the following two possibilities:

• When the velocity and acceleration of an object have the same sign, thespeed of the object increases. In this case, the velocity and accelerationpoint in the same direction.

• When the velocity and acceleration of an object have opposite signs, thespeed of the object decreases. In this case, the velocity and accelerationpoint in opposite directions.

These two possibilities are illustrated in Figure 2–11. Notice that when a particle’sspeed increases, it means either that its velocity becomes more positive, as inFigure 2–11 (a), or more negative, as in Figure 2–11 (d). In either case, it is themagnitude of the velocity—the speed—that increases.

v

a

(a)

x

v

a

(b)

x

v

a

(d)

x

v

a

x

(c)

FIGURE 2–11 Cars accelerating ordeceleratingA car’s speed increases when its velocityand acceleration point in the same direc-tion, as in cases (a) and (d). When thevelocity and acceleration point in oppo-site directions, as in cases (b) and (c), thecar’s speed decreases.

▲

The winner of this race was traveling ata speed of 313.91 mph at the end of thequarter-mile course. Since the winningtime was just 4.607 s, the average accelera-tion during this race was approximatelythree times the acceleration of gravity(Section 2–7).

▲

WALKMC02_0131536311.QXD 12/9/05 3:26 Page 28

Hypothesis: He’s slowing by 10 m/s at about 2 m/s2, perhapsaround 10 ÷ 2 = 5 s.

Given: v0 = +30.0 m/s i, v = +20.0 m/s i, a = −2.30 m/s2 iAsked for: t

Example 4


Sketch:






]



t = 6.0 s.t = 0¢v

-1.5 m/s2.2.0 m/s2, 0 m/s2,





v

a

(a)

x

v

a

(b)

x

v

a

(d)

x

v

a

x

(c)


▲


▲

WALKMC02_0131536311.QXD 12/9/05 3:26 Page 28



Example 4


Sketch:






]



t = 6.0 s.t = 0¢v

-1.5 m/s2.2.0 m/s2, 0 m/s2,





v

a

(a)

x

v

a

(b)

x

v

a

(d)

x

v

a

x

(c)


▲


▲

WALKMC02_0131536311.QXD 12/9/05 3:26 Page 28



Example 4


Sketch:






]



t = 6.0 s.t = 0¢v

-1.5 m/s2.2.0 m/s2, 0 m/s2,





v

a

(a)

x

v

a

(b)

x

v

a

(d)

x

v

a

x

(c)


▲


▲

WALKMC02_0131536311.QXD 12/9/05 3:26 Page 28



Example 4

Strategy: Usev = v0 + at

Rearrange:

t =v − v0

a

=20.0 m/s − 30.0 m/s

−2.30 m/s2

= 4.35 s

Reasonable / Agrees with hypothesis: yes, just a bit smaller. Thatmakes sense because the acceleration had magnitude 2.30 m/s2

rather than 2 m/s2. It is a reasonable amount of time to need toslow for a speed trap.

Example 4

Strategy: Usev = v0 + at

Rearrange:

t =v − v0

a

=20.0 m/s − 30.0 m/s

−2.30 m/s2

= 4.35 s

Reasonable / Agrees with hypothesis: yes, just a bit smaller. Thatmakes sense because the acceleration had magnitude 2.30 m/s2

rather than 2 m/s2. It is a reasonable amount of time to need toslow for a speed trap.

Example 5A car driver sees an obstacle in the road and applies the brakes. Ittakes him 4.33 s to stop the car over a distance of 55.0 m.Assuming the car brakes with constant acceleration, what was thecar’s deceleration?

Sketch:






]



t = 6.0 s.t = 0¢v

-1.5 m/s2.2.0 m/s2, 0 m/s2,





v

a

(a)

x

v

a

(b)

x

v

a

(d)

x

v

a

x

(c)


▲


▲

WALKMC02_0131536311.QXD 12/9/05 3:26 Page 28

Hypothesis: 55 m is not very far and 4.33 s is not a long time: thedeceleration must be high-ish. I would guess maybe 5 m/s2 whichis about half of g .

Given: t = 4.33 s, ∆x = 55.0 m, v = 0 m/sAsked for: aStrategy: use

∆x = vt −1

2at2


Sketch:






]



t = 6.0 s.t = 0¢v

-1.5 m/s2.2.0 m/s2, 0 m/s2,





v

a

(a)

x

v

a

(b)

x

v

a

(d)

x

v

a

x

(c)


▲


▲

WALKMC02_0131536311.QXD 12/9/05 3:26 Page 28



∆x = vt −1

2at2


Sketch:






]



t = 6.0 s.t = 0¢v

-1.5 m/s2.2.0 m/s2, 0 m/s2,





v

a

(a)

x

v

a

(b)

x

v

a

(d)

x

v

a

x

(c)


▲


▲

WALKMC02_0131536311.QXD 12/9/05 3:26 Page 28



∆x = vt −1

2at2


Sketch:






]



t = 6.0 s.t = 0¢v

-1.5 m/s2.2.0 m/s2, 0 m/s2,





v

a

(a)

x

v

a

(b)

x

v

a

(d)

x

v

a

x

(c)


▲


▲

WALKMC02_0131536311.QXD 12/9/05 3:26 Page 28


Given: t = 4.33 s, ∆x = 55.0 m, v = 0 m/sAsked for: a

Strategy: use

∆x = vt −1

2at2


Sketch:






]



t = 6.0 s.t = 0¢v

-1.5 m/s2.2.0 m/s2, 0 m/s2,





v

a

(a)

x

v

a

(b)

x

v

a

(d)

x

v

a

x

(c)


▲


▲

WALKMC02_0131536311.QXD 12/9/05 3:26 Page 28



∆x = vt −1

2at2

Example 5

A car driver sees an obstacle in the road and applies the brakes. Ittakes him 4.33 s to stop the car over a distance of 55.0 m.Assuming the car brakes with constant acceleration, what was thecar’s deceleration?

∆x = vt −1

2at2

1

2at2 = vt −∆x

a =2(vt −∆x)

t2

=2(0 − 55.0 mi)

(4.33 s)2

a = −5.87 m/s2 i

Or, the car’s acceleration is 5.87 m/s2, opposite the direction ofthe car’s travel.

Example 5

A car driver sees an obstacle in the road and applies the brakes. Ittakes him 4.33 s to stop the car over a distance of 55.0 m.Assuming the car brakes with constant acceleration, what was thecar’s deceleration?

∆x = vt −1

2at2

1

2at2 = vt −∆x

a =2(vt −∆x)

t2

=2(0 − 55.0 mi)

(4.33 s)2

a = −5.87 m/s2 i

Or, the car’s acceleration is 5.87 m/s2, opposite the direction ofthe car’s travel.

Example 5

Reasonable / Agrees with hypothesis: This is a large deceleration,but a car can manage it. It is reasonable considering he needs tostop before the obstacle and is breaking hard.

Example 6A car traveling at a constant speed of 45.0 m/s passes a trooperon a motorcycle hidden behind a billboard. One second after thespeeding car passes the billboard, the trooper sets out from thebillboard to catch the car, accelerating at a constant rate of 3.00m/s2. How long does it take the trooper to overtake the car?

Sketch:

Equation 2.13 is the only equation in the particle under constant acceleration model that does not involve position, so we use it to find the acceleration of the jet, modeled as a particle:

ax 5vxf 2 vxi

t<

0 2 63 m/s2.0 s

5 232 m/s2

(B) If the jet touches down at position xi 5 0, what is its final position?

S O L U T I O N

Use Equation 2.15 to solve for the final position: xf 5 xi 1 12 1vxi 1 vxf 2 t 5 0 1 1

2 163 m/s 1 0 2 12.0 s 2 5 63 m

Given the size of aircraft carriers, a length of 63 m seems reasonable for stopping the jet. The idea of using arresting cables to slow down landing aircraft and enable them to land safely on ships originated at about the time of World War I. The cables are still a vital part of the operation of modern aircraft carriers.

Suppose the jet lands on the deck of the aircraft carrier with a speed faster than 63 m/s but has the same acceleration due to the cable as that calculated in part (A). How will that change the answer to part (B)?

Answer If the jet is traveling faster at the beginning, it will stop farther away from its starting point, so the answer to part (B) should be larger. Mathematically, we see in Equation 2.15 that if vxi is larger, then xf will be larger.

WHAT IF ?

2.6 Analysis Model: Particle Under Constant Acceleration 39

Example 2.8 Watch Out for the Speed Limit!

A car traveling at a constant speed of 45.0 m/s passes a trooper on a motorcycle hidden behind a billboard. One sec-ond after the speeding car passes the billboard, the trooper sets out from the billboard to catch the car, accelerating at a constant rate of 3.00 m/s2. How long does it take the trooper to overtake the car?

A pictorial representation (Fig. 2.13) helps clarify the sequence of events. The car is modeled as a particle under con-stant velocity, and the trooper is modeled as a particle under constant acceleration. First, we write expressions for the position of each vehicle as a function of time. It is convenient to choose the posi-tion of the billboard as the origin and to set t! 5 0 as the time the trooper begins moving. At that instant, the car has already traveled a distance of 45.0 m from the billboard because it has traveled at a constant speed of vx 5 45.0 m/s for 1 s. Therefore, the initial position of the speeding car is x! 5 45.0 m.

AM

S O L U T I O N

Figure 2.13 (Example 2.8) A speeding car passes a hid-den trooper.

t" ! ?t! ! 0t# ! "1.00 s

# ! "

Using the particle under constant velocity model, apply Equation 2.7 to give the car’s position at any time t :

xcar 5 x! 1 vx cart

A quick check shows that at t 5 0, this expression gives the car’s correct initial position when the trooper begins to move: xcar 5 x! 5 45.0 m.

The trooper starts from rest at t! 5 0 and accelerates at ax 5 3.00 m/s2 away from the origin. Use Equation 2.16 to give her position at any time t :

xf 5 xi 1 vxit 1 12axt 2

x trooper 5 0 1 10 2 t 1 12axt 2 5 1

2axt 2

Set the positions of the car and trooper equal to repre-sent the trooper overtaking the car at position ":

x trooper 5 xcar

12axt 2 5 x ! 1 vx cart

▸ 2.7 c o n t i n u e d

continued

+x

1Serway & Jewett, “Physics for Scientists and Engineers”, pg 39.

Example 6

Hypothesis: The trooper accelerates at 3.00 m/s2. It will take 15 sfor him to reach a speed of 45.0 m/s, but then he still needs tocatch up to the car. I would guess it will take him about another15 s to catch up. Guess: about 30 s or so.

Given: vcar = 45.0 m/s, (acar = 0), atr = 3.00 m/s2, vtr,0 = 0 m/s,and the car has a 1 s head start

Asked for: t when they are at the same positition.

Strategy: Use the idea that when the trooper has caught up

∆xcar = ∆x tr

We can visualize this on a graph, or just do a bit of algebra.

Example 6





∆xcar = ∆x tr


Example 6





∆xcar = ∆x tr


Example 6

A car traveling at a constant speed of 45.0 m/s passes a trooperon a motorcycle hidden behind a billboard. One second after thespeeding car passes the billboard, the trooper sets out from thebillboard to catch the car, accelerating at a constant rate of 3.00m/s2. How long does it take the trooper to overtake the car?

∆xcar = ∆x tr

vcar(t + 1)�i =(��

��*0(vtr,0)t +

1

2atrt

2)�i

0 =1

2atrt

2 − vcart − vcar

This is a quadratic expression in t.

Example 6

A car traveling at a constant speed of 45.0 m/s passes a trooperon a motorcycle hidden behind a billboard. One second after thespeeding car passes the billboard, the trooper sets out from thebillboard to catch the car, accelerating at a constant rate of 3.00m/s2. How long does it take the trooper to overtake the car?

0 =1

2atrt


This is a quadratic expression in t.

0 =1

2atrt


t =vcar ±

√v2car + 2atrvcaratr

t = 31.0 s

Example 6

Answer: t = 31.0 s

Analysis: This answer is quite close to the hypothesis! The car isgoing 45.0 m/s, or 100 mi/h, which is very fast, but the trooperalso has a high acceleration of 3.00 m/s2. Motorcycles can haveaccelerations even higher than that, so the numbers in thisquestion are reasonable.

Free-Falling Objects

One common scenario of interest where acceleration is constant isobjects freely falling.

When we refer to free fall, we mean objects moving under theinfluence of gravity, and where we are ignoring resistive forces,eg. air resistance.

Galileo and the Leaning Tower of Pisa

Aristotle, an early Greek natural philosopher, said that heavierobjects fall faster than lighter ones.

Galileo tested this idea and found it was wrong. Any two massiveobjects accelerate at the same rate.

Galileo studied the motion of objects by experiment, as well as byabstract reasoning.

Galileo and Inertia

He considered balls rolling on inclined surfaces and developed thenotion of inertia.

Inertia is the tendency of objects to stay doing whatever they arealready doing, unless they are interfered with.

Galileo’s idea of inertia:

A body moving on a level surface will continue in the samedirection at a constant speed unless disturbed.

Galileo and Inertia

He considered balls rolling on inclined surfaces and developed thenotion of inertia.

Inertia is the tendency of objects to stay doing whatever they arealready doing, unless they are interfered with.

Galileo’s idea of inertia:

A body moving on a level surface will continue in the samedirection at a constant speed unless disturbed.

Acceleration and Free-Fall

Galileo reasoned about the acceleration due to gravity by thinkingmore about inclined surfaces.

θ

h

Ov

The steeper the incline the larger the acceleration.

The final velocity of the ball is also larger. Starting from rest:

v = at

Free-Fall

When the ball drops straight downward, it gains approximately 10m/s of speed in each second.

Time of fall (s) Velocity acquired (m/s)

0 01 102 203 30...

...

This is a constant acceleration! We call this acceleration g .

g = 9.8 m s−2 ≈ 10 m s−2

After 6.5 s, (roughly) what is the ball’s speed? 65 m/s

Free-Fall



0 01 102 203 30...

...


g = 9.8 m s−2 ≈ 10 m s−2

After 6.5 s, (roughly) what is the ball’s speed?

65 m/s

Free-Fall



0 01 102 203 30...

...


g = 9.8 m s−2 ≈ 10 m s−2

After 6.5 s, (roughly) what is the ball’s speed? 65 m/s


The important point is that at the surface of the Earth, all objectsexperience this same acceleration due to gravity: g = 9.8 m s−2.

In the absence of air resistance, the acceleration does not dependon an object’s mass!38 CHAPTER 2 ONE-DIMENSIONAL KINEMATICS

move with constant acceleration. His conclusions were based on experimentsdone by rolling balls down inclines of various steepness. By using an incline,Galileo was able to reduce the acceleration of the balls, thus producing motionslow enough to be timed with the rather crude instruments available.

Galileo also pointed out that objects of different weight fall with the same con-stant acceleration—provided air resistance is small enough to be ignored.Whether he dropped objects from the Leaning Tower of Pisa to demonstrate thisfact, as legend has it, will probably never be known for certain, but we do knowthat he performed extensive experiments to support his claim.

Today it is easy to verify Galileo’s assertion by dropping objects in a vacuumchamber, where the effects of air resistance are essentially removed. In a standardclassroom demonstration, a feather and a coin are dropped in a vacuum, and bothfall at the same rate. In 1971, a novel version of this experiment was carried out onthe Moon by astronaut David Scott. In the near perfect vacuum on the Moon’ssurface he dropped a feather and a hammer and showed a worldwide televisionaudience that they fell to the ground in the same time.

To illustrate the effect of air resistance in everyday terms, consider dropping asheet of paper and a rubber ball (Figure 2–17). The paper drifts slowly to theground, taking much longer to fall than the ball. Now, wad the sheet of paper intoa tight ball and repeat the experiment. This time the ball of paper and the rubberball reach the ground in nearly the same time. What was different in the two ex-periments? Clearly, when the sheet of paper was wadded into a ball, the effect ofair resistance on it was greatly reduced, so that both objects fell almost as theywould in a vacuum.

Before considering a few examples, let’s first discuss exactly what is meant by“free fall.” To begin, the word free in free fall means free from any effects otherthan gravity. For example, in free fall we assume that an object’s motion is notinfluenced by any form of friction or air resistance.

• Free fall is the motion of an object subject only to the influence of gravity.

Though free fall is an idealization—which does not apply to many real-worldsituations—it is still a useful approximation in many other cases. In the followingexamples we assume that the motion may be considered as free fall.

Next, it should be realized that the word fall in free fall does not mean theobject is necessarily moving downward. By free fall, we mean any motion underthe influence of gravity alone. If you drop a ball, it is in free fall. If you throw a ballupward or downward, it is in free fall as soon as it leaves your hand.

• An object is in free fall as soon as it is released, whether it is dropped fromrest, thrown downward, or thrown upward.

Finally, the acceleration produced by gravity on the Earth’s surface (sometimescalled the gravitational strength) is denoted with the symbol g. As a shorthandname, we will frequently refer to g simply as “the acceleration due to gravity.” Infact, as we shall see in Chapter 12, the value of g varies according to one’s locationon the surface of the earth, as well as one’s altitude above it. Table 2–5 shows howg varies with latitude.

▲ Whether she is on the wayup, at the peak of her flight, oron the way down, this girl is infree fall, accelerating downwardwith the acceleration of gravity.Only when she is in contact withthe blanket does her accelerationchange.

(a) (b)

FIGURE 2–17 Free fall and airresistance(a) Dropping a sheet of paper and a rub-ber ball compared with (b) dropping awadded-up sheet of paper and a ball.

▲

TABLE 2–5 Values of g at DifferentLocations on Earth

Location Latitude g

North Pole 90° N 9.832Oslo, Norway 60° N 9.819Hong Kong 30° N 9.793Quito, Ecuador 0° 9.780

1m>s22

▲ In the absence of air resistance, allbodies fall with the same acceleration,regardless of their mass.

WALKMC02_0131536311.QXD 12/9/05 3:27 Page 38

The fact that acceleration due to gravity is independent of masscan be seen in airless environments...

1Figure from Walker, “Physics”, page 39.


The important point is that at the surface of the Earth, all objectsexperience this same acceleration due to gravity: g = 9.8 m s−2.

In the absence of air resistance, the acceleration does not dependon an object’s mass!38 CHAPTER 2 ONE-DIMENSIONAL KINEMATICS

move with constant acceleration. His conclusions were based on experimentsdone by rolling balls down inclines of various steepness. By using an incline,Galileo was able to reduce the acceleration of the balls, thus producing motionslow enough to be timed with the rather crude instruments available.

Galileo also pointed out that objects of different weight fall with the same con-stant acceleration—provided air resistance is small enough to be ignored.Whether he dropped objects from the Leaning Tower of Pisa to demonstrate thisfact, as legend has it, will probably never be known for certain, but we do knowthat he performed extensive experiments to support his claim.

Today it is easy to verify Galileo’s assertion by dropping objects in a vacuumchamber, where the effects of air resistance are essentially removed. In a standardclassroom demonstration, a feather and a coin are dropped in a vacuum, and bothfall at the same rate. In 1971, a novel version of this experiment was carried out onthe Moon by astronaut David Scott. In the near perfect vacuum on the Moon’ssurface he dropped a feather and a hammer and showed a worldwide televisionaudience that they fell to the ground in the same time.

To illustrate the effect of air resistance in everyday terms, consider dropping asheet of paper and a rubber ball (Figure 2–17). The paper drifts slowly to theground, taking much longer to fall than the ball. Now, wad the sheet of paper intoa tight ball and repeat the experiment. This time the ball of paper and the rubberball reach the ground in nearly the same time. What was different in the two ex-periments? Clearly, when the sheet of paper was wadded into a ball, the effect ofair resistance on it was greatly reduced, so that both objects fell almost as theywould in a vacuum.

Before considering a few examples, let’s first discuss exactly what is meant by“free fall.” To begin, the word free in free fall means free from any effects otherthan gravity. For example, in free fall we assume that an object’s motion is notinfluenced by any form of friction or air resistance.

• Free fall is the motion of an object subject only to the influence of gravity.

Though free fall is an idealization—which does not apply to many real-worldsituations—it is still a useful approximation in many other cases. In the followingexamples we assume that the motion may be considered as free fall.

Next, it should be realized that the word fall in free fall does not mean theobject is necessarily moving downward. By free fall, we mean any motion underthe influence of gravity alone. If you drop a ball, it is in free fall. If you throw a ballupward or downward, it is in free fall as soon as it leaves your hand.

• An object is in free fall as soon as it is released, whether it is dropped fromrest, thrown downward, or thrown upward.

Finally, the acceleration produced by gravity on the Earth’s surface (sometimescalled the gravitational strength) is denoted with the symbol g. As a shorthandname, we will frequently refer to g simply as “the acceleration due to gravity.” Infact, as we shall see in Chapter 12, the value of g varies according to one’s locationon the surface of the earth, as well as one’s altitude above it. Table 2–5 shows howg varies with latitude.

▲ Whether she is on the wayup, at the peak of her flight, oron the way down, this girl is infree fall, accelerating downwardwith the acceleration of gravity.Only when she is in contact withthe blanket does her accelerationchange.

(a) (b)

FIGURE 2–17 Free fall and airresistance(a) Dropping a sheet of paper and a rub-ber ball compared with (b) dropping awadded-up sheet of paper and a ball.

▲

TABLE 2–5 Values of g at DifferentLocations on Earth

Location Latitude g

North Pole 90° N 9.832Oslo, Norway 60° N 9.819Hong Kong 30° N 9.793Quito, Ecuador 0° 9.780

1m>s22

▲ In the absence of air resistance, allbodies fall with the same acceleration,regardless of their mass.

WALKMC02_0131536311.QXD 12/9/05 3:27 Page 38

The fact that acceleration due to gravity is independent of masscan be seen in airless environments...

1Figure from Walker, “Physics”, page 39.


Objects near the Earth’s surface have a constant acceleration ofg = 9.8 ms−2. (Or, about 10 ms−2)

The kinematics equations for constant acceleration all apply.

Summary

• practice using the kinematics equations

• free fall (?)

Quiz Monday, start of class.

First Test next Thursday, Feb 1.

Homework

• Please bring a ruler to class tomorrow!

Walker Physics:

• Ch 2, onward from page 14. Problems: 61, 63

Documents

Introduction to Mechanics Practice using the Kinematics ...nebula2.deanza.edu/~lanasheridan/P50/Phys50-Lecture10.pdfIntroduction to Mechanics Practice using the Kinematics Equations