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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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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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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

).



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).



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.



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.
