Quiz Samples for Chapter 2 General Physics I Name...
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General Physics I Quiz Samples for Chapter 2 Motion Along a Straight Line March 16, 2020 Name: Department: Student ID #: Notice +2 (-1) points per correct (incorrect) answer. No penalty for an unanswered question. Fill the blank ( ) with (✘) if the statement is correct (incorrect). Textbook: Walker, Halliday, Resnick, Principles of Physics, Tenth Edition, John Wiley & Sons (2014). 2-1 Position, Displacement, and Average Velocity 1. () If we consider a motion of a particle along a straight line, then the object must not have a substructure or that can be neglected. A stiff pig slipping down a straight playground slide can be considered to be moving like a particle. 2. () In order to measure the position and displacement of a particle, we need to introduce a frame of reference . The frame of reference has a certain origin and certain axes. Each axis is scaled in units of a certain physical unit. As long as the distance is concerned, the unit is the base unit of length, meter m, or equivalent units like cm, km, and so on. 3. () (positive direction) (negative direction) The displacement is the change in position Δx = x 2 - x 1 , where x 1 and x 2 are the initial and final positions, respectively. The magnitude of the displacement is its absolute value, |Δx|. There are two displacements with the same magnitude in one dimension: one is in the positive direction and the other is in the negative direction. A vanishing displacement is zero (null) displacement that does not have the magnitude and direction, either. 4. () (position of ) The position x of a particle can be defined as the displacement of the particle from the origin. 5. () The displacement is the most fundamental vector quantity that has both magnitude and direction. 6. () When a particle has moved from position x 1 to x 2 during a time interval [t 1 ,t 2 = t 1 +Δt], then the average velocity of the particle during that time interval is defined by v average = Δx Δt = x 2 - x 1 t 2 - t 1 . 7. () When a particle has moved from position x 1 to x 2 during a time interval [t 1 ,t 2 = t 1 +Δt], then the average speed of the particle during that time interval is defined by s average = total distance Δt . 8. () The average velocity is independent of the actual distance but it depends only on the ratio of the displacement and the time interval. 2020 KPOPE All rights reserved. Korea University Page 1 of 6
Transcript of Quiz Samples for Chapter 2 General Physics I Name...
General Physics IQuiz Samples for Chapter 2Motion Along a Straight Line
March 16, 2020
Name: Department: Student ID #:
Notice
� +2 (−1) points per correct (incorrect) answer.
� No penalty for an unanswered question.
� Fill the blank ( ) with � (8) if the statement iscorrect (incorrect).
� Textbook: Walker, Halliday, Resnick, Principlesof Physics, Tenth Edition, John Wiley & Sons(2014).
2-1 Position, Displacement, and AverageVelocity
1. (�) If we consider a motion of a particle along astraight line, then the object must not have asubstructure or that can be neglected. A stiff pigslipping down a straight playground slide can beconsidered to be moving like a particle.
2. (�) In order to measure the position anddisplacement of a particle, we need to introduce aframe of reference. The frame of reference has acertain origin and certain axes. Each axis isscaled in units of a certain physical unit. As longas the distance is concerned, the unit is the baseunit of length, meter m, or equivalent units likecm, km, and so on.
3. (�)
(positive direction)
(negative direction)
The displacement is the change in position
∆x = x2 − x1,
where x1 and x2 are the initial and final positions,respectively. The magnitude of the displacementis its absolute value,
|∆x|.
There are two displacements with the samemagnitude in one dimension: one is in thepositive direction and the other is in thenegative direction. A vanishing displacement iszero (null) displacement that does not have themagnitude and direction, either.
4. (�)
(position of )
The position x of a particle can be defined as thedisplacement of the particle from the origin.
5. (�) The displacement is the most fundamentalvector quantity that has both magnitude anddirection.
6. (�) When a particle has moved from position x1 tox2 during a time interval [t1, t2 = t1 + ∆t], then theaverage velocity of the particle during that timeinterval is defined by
vaverage =∆x
∆t=x2 − x1
t2 − t1.
7. (�) When a particle has moved from position x1 tox2 during a time interval [t1, t2 = t1 + ∆t], then theaverage speed of the particle during that timeinterval is defined by
saverage =total distance
∆t.
8. (�) The average velocity is independent of theactual distance but it depends only on the ratio ofthe displacement and the time interval.
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General Physics IQuiz Samples for Chapter 2Motion Along a Straight Line
March 16, 2020
For path c1 and c2, the average velocity is same,but the average speed is not (c2 > c1).
9. Consider a particle moving along a straight linewhose position at time t is given by
x(t) = x0 + v0t,
where x0 and v0 are constants.
(a) (�) The physical dimensions of x0 and v0 are
[x0] = [L], [v0] = [L][T ]−1.
(b) (�) During the time interval [t1, t2], thedisplacement of the particle is
∆x = v0(t2 − t1).
(c) (�) The average velocity of the particleduring the time interval [t1, t2] is
vaverage = v0.
The average velocity of this particle isindependent of the time interval.
(d) (�)
The slope of the curve x(t) is always v0 whichis constant.
10. There are two particles 1 and 2 moving along the xaxis. The positions of these particles at time t aregiven by
x1(t) = A sin2πt
C,
x2(t) = Bt
(1− t
C
),
where A, B, and C are constants and thesubscripts represent the particle identifier 1 or 2.
(a) (�) The physical dimensions of A, B, and Care
[A] = [L], [B] = [L][T ]−1, [C] = [T ].
(b) (�) At t = 0, both particles are at the origin.
(c) (�) During the time interval [0, C], thedisplacements of the particles are
∆x1 = x1(C)− x1(0) = 0,
∆x2 = x2(C)− x2(0) = 0.
(d) (�) The average velocity of each particleduring the time interval [0, C] is
vaverage = 0.
(e) (�)
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General Physics IQuiz Samples for Chapter 2Motion Along a Straight Line
March 16, 2020
The average speed of each particle during thetime interval [0, C] is
s1average(t) =4|A||C|
,
s2average(t) =|B|2.
2-2 Instantaneous Velocity and Speed
1. (�) The instantaneous velocity (velocity) of aparticle is defined by
lim∆t→0
∆x
∆t=dx
dt.
A common short-hand notation of the timederivative is the over-dot:
x ≡ dx
dt.
In a similar manner, the double time derivative iswritten in the form
x ≡ d2x
dt2.
2. (�) The instantaneous speed is the magnitudeof the instantaneous velocity.
3. Compute the instantaneous velocities of thefollowing particles with the position vectors attime t:
x1(t) = A cos2πt
C,
x2(t) = A sin2πt
C,
x3(t) = x0 + v0t+1
2gt2,
x4(t) = B(1− e−βt),
x5(t) = Be−βt sin2πt
C,
where A, B, C, x0, v0, g, and β are constants.
(a) (�) Leibniz rule for the derivatives of aproduct function states that
d
dt[A(t)B(t)] = AB +AB.
In a similar manner, we find that
d
dt[A(t)B(t)C(t)] = ABC +ABC +ABC.
(b) (�) The physical dimensions of the constantsare given by
[A] = [L],
[B] = [L],
[C] = [T ],
[x0] = [L],
[v0] = [L][T ]−1,
[g] = [L][T ]−2,
[β] = [T ]−1.
(c) (�) x1 can be computed as follows:
x1(t) =d
dt
[A cos
2πt
C
]= −2πA
Csin
2πt
C.
(d) (�) x2 can be computed as follows:
x2(t) =d
dt
[A sin
2πt
C
]=
2πA
Ccos
2πt
C.
(e) (�) x3 can be computed as follows:
x3(t) =d
dt
[x0 + v0t+
1
2gt2]
= v0 + gt.
(f) (�) x4 can be computed as follows:
x4(t) =d
dt
[B(1− e−βt)
]= βBe−βt.
(g) (�) x5 can be computed as follows:
x5(t) =d
dt
[Be−βt sin
2πt
C
]= B
d
dt
[e−βt
]sin
2πt
C
+Be−βtd
dt
[sin
2πt
C
]= Be−βt
(−β sin
2πt
C+
2π
Ccos
2πt
C
).
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General Physics IQuiz Samples for Chapter 2Motion Along a Straight Line
March 16, 2020
2-3 Acceleration
1. (�) When the velocity of a particle has changedfrom v1 to v2 during a time interval[t1, t2 = t1 + ∆t], then the average accelerationof the particle during that time interval is definedby
aaverage =∆v
∆t=v2 − v1
t2 − t1.
2. (�) The instantaneous acceleration(acceleration) of a particle is defined by
lim∆t→0
∆v
∆t=dv
dt=d2x
dt2.
A common short-hand notation of the double timederivative is the over-double-dot:
x ≡ d2x
dt2.
3. Compute the instantaneous accelerations of thefollowing particles with the position vectors attime t:
x1(t) = A cos2πt
C,
x2(t) = A sin2πt
C,
x3(t) = x0 + v0t+1
2gt2,
x4(t) = B(1− e−βt),
x5(t) = Be−βt sin2πt
C,
where A, B, C, x0, v0, g, and β are constants.
(a) (�) Leibniz rule for the derivatives of aproduct function states that
d2
dt2[A(t)B(t)] = AB + 2AB +AB.
(b) (�) x1 can be computed as follows:
x1(t) =d2
dt2
[A cos
2πt
C
]= −
(2π
C
)2
A cos2πt
C.
(c) (�) x2 can be computed as follows:
x2(t) =d2
dt2
[A sin
2πt
C
]= −
(2π
C
)2
A sin2πt
C.
(d) (�) x3 can be computed as follows:
x3(t) =d2
dt2
[x0 + v0t+
1
2gt2]
= g.
(e) (�) x4 can be computed as follows:
x4(t) =d2
dt2
[B(1− e−βt)
]= −β2Be−βt.
(f) (�) x5 can be computed as follows:
x5(t) =d2
dt2
[Be−βt sin
2πt
C
]= B
d2
dt2
[e−βt
]sin
2πt
C
+2Bd
dt
[e−βt
] ddt
[sin
2πt
C
]+Be−βt
d2
dt2
[sin
2πt
C
]= Be−βt
[(β2 − 4π2
C2
)sin
2πt
C
−4πβ
Ccos
2πt
C
].
2-4 Constant Acceleration
1. (�) The position of a particle under a constantacceleration moving along the x axis is given by
x(t) = x0 + v0t+1
2at2,
where x0, v0, and a are constants.
(a) (�) x0 is of dimension [L] and it is the initialposition:
x0 = x(0).
(b) (�) v0 is of dimension [L][T ]−1 and it is theinitial velocity:
v0 = v(0).
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General Physics IQuiz Samples for Chapter 2Motion Along a Straight Line
March 16, 2020
(c) (�) a is of dimension [L][T ]−2 and it is theacceleration at any t:
a = a(t) = constant.
2. (�) Consider the motion of a particle underuniform acceleration a whose position at t alongthe x axis is given by
x(t) = x0 + v0t+1
2at2,
where x0 and v0 are the initial position andvelocity, respectively.
(a) (�) The displacement from t = t1 to t2 is
∆x = x(t2)− x(t1)
= v0(t2 − t1) +1
2a(t22 − t21).
(b) (�)
The average velocity during the time interval[t1, t2] is
vaverage =x(t2)− x(t1)
t2 − t1= v0 +
1
2a(t2 + t1)
=1
2(v2 + v1).
Here, v1 = v(t1) and v2 = v(t2). Thus theaverage velocity is the arithmetic average ofthe initial and the final velocities. The reasonis that the slope of v versus t graph has auniform slope.
(c) (�) The displacement during the time interval[t1, t2] is
∆x = x(t2)− x(t1)
= (t2 − t1)
[v0 +
1
2a(t2 + t1)
]=
(v0 + at2)− (v0 + at1)
a
×(v0 + at2) + (v0 + at1)
2
=v2 − v1
a× v2 + v1
2
=v2
2 − v21
2a.
(d) (�) v0 is of dimension [L][T ]−1 and it is theinitial velocity:
v0 = v(0).
(e) (�) a is of dimension [L][T ]−2 and it is theacceleration at any t:
a = a(t) = constant.
2-5 Free-Fall Acceleration
1. (�) Near the Earth’s surface the gravitationalacceleration g is approximately given by
g ≈ 9.8 m/s2.
The direction of the acceleration is approximatelytowards the center of mass of the Earth. Theacceleration due to the gravitational accelerationneglecting other external forces or fictitious forcesis called the free-fall acceleration.
2. (�) The gravitational acceleration depends on thelatitude because the Earth rotates about its ownaxis. The surface of the Earth is an acceleratingframe of reference in which a fictitious force, thatis called the centrifugal force, applies. Theperpendicular distance from the rotation axisincreases as the latitude decreases to the equator.The altitude of a position also affects the value ofthe gravitational acceleration because thegravitational force is attenuated as the distancefrom the center of mass of the Earth increases.
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General Physics IQuiz Samples for Chapter 2Motion Along a Straight Line
March 16, 2020
2-6 Graphical Integration in Motion Analysis
1. (�)
The displacement can be read off from the v versust graph as
∆x = x(t2)− x(t1)
=
∫ t2
t1
vdt
=area between velocity curve
and the time axis, from t1 to t2.
Note that the area is negative if the curve is belowthe time axis.
2. (�)
The change in the velocity can be read off from thea versus t graph as
∆v = v(t2)− v(t1)
=
∫ t2
t1
adt
=area between the acceleration curve
and the time axis, from t1 to t2.
Note that the area is negative if the curve is belowthe time axis.
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