Unit 1 review mini quiz · Basic Physics I

Question

Infer the units of a coefficient

Measurement tools

A position is proposed as $x(t)$ = $x_{0}$ +Bt+D/t, where $x_{0}$ is a position, B has units of $\mathrm{m/s}$ , and t is measured in seconds. Which SI unit must D have for the expression to be dimensionally homogeneous?

Choose an answer

\mathrm{m/s}

m·s

\mathrm{m/s^2}

m

Question

Dimensional homogeneity does not prove a model

Measurement tools

A physical model proposes τ=3√(ℓ/a), with [ℓ]=L and [a]=LT⁻². What can be concluded from dimensional analysis alone?

Choose an answer

The model is physically correct because the right-hand side has dimension time.It is incorrect because 3 has no units.It is dimensionally compatible with a time, but dimensional analysis cannot establish that either factor 3 or the physical model is correct.Dimensional analysis establishes that the numerical factor must be exactly 3.

Question

Magnitude versus component

Vectors

A displacement A has $A_{x}$ =−6 m and $A_{y}$ =8 m. Which statement is correct?

Choose an answer

Magnitude is −6 m because the x-component is negative.Magnitude is 10 m;

A_{x}

<0 means that component points toward

-x

.Magnitude is 8 m because the component with larger absolute value determines magnitude.Magnitude is −10 m because one component is negative.

Question

Components when the angle is not measured from +x

Vectors

A vector B of magnitude B lies in the fourth quadrant and makes an angle of 25° to the right of the −y axis. Which components match the geometry?

Choose an answer

Bₓ=B cos 25°, Bᵧ=

-\vec B

sin 25°Bₓ=B sin 25°, Bᵧ=

-\vec B

cos 25°Bₓ=

-\vec B

sin 25°, Bᵧ=

-\vec B

cos 25°Bₓ=B sin 25°, Bᵧ=B cos 25°

Question

Apply perpendicularity through the dot product

Vectors

Let A=(2,1). Which vector is perpendicular to A?

Choose an answer

(1,2)(1,−2)(−2,−1)(2,−1)

Question

Cross-product orientation

Vectors

In a right-handed Cartesian basis, A points along $+x$ and B along +y. Which statement is correct?

Choose an answer

Both

\vec A\times\vec B

and

\vec B\times\vec A

point toward +z.

\vec A\times\vec B

points toward +z, while

\vec B\times\vec A

points toward −z.Both products are zero because the vectors are perpendicular.

\vec A\times\vec B

points toward −z and

\vec B\times\vec A

toward +z.

Question

Position and displacement

1D motion

On an axis whose positive direction points east, a particle moves from xᵢ=−4 m to $x_{f}$ =3 m. Which statement is correct?

Choose an answer

Its final position is 7 m.Its displacement is +7 m.Its displacement is 3 m because that is the final position.Its displacement is −7 m because it started at a negative position.

Question

Compare distance and displacement along two paths

1D motion

Two particles start at x=−2 m and finish at x=6 m. P moves directly from −2 to 6 m. Q first reaches x=10 m and then returns to x=6 m. Which statement is correct?

Choose an answer

Both have the same displacement +8 m, but Q travels a greater distance.Both travel the same distance because they have the same initial and final positions.Q has greater displacement because it follows a longer path.P has displacement +8 m and Q has displacement +16 m.

Question

Average speed and average velocity

1D motion

In 20 s, a person travels 60 m east and then 20 m west. Take east as $+x$ . Which result is correct?

Choose an answer

Average speed 2

\mathrm{m/s}

; average velocity 4

\mathrm{m/s}

east.Average speed 4

\mathrm{m/s}

; average velocity 2

\mathrm{m/s}

east.Both 4

\mathrm{m/s}

.Both 2

\mathrm{m/s}

.

Question

Negative position does not determine direction

1D motion

At one instant, x=−5 m and v=+2 $\mathrm{m/s}$ . What follows?

Choose an answer

It is on the negative side of the origin and moving toward

+x

.It is on the negative side and must move toward

-x

.It is on the positive side because velocity is positive.Its position must immediately become positive.

Question

Slope of a position graph

1D motion

On an x-versus-t graph, at t₀ the curve passes through x=10 m and the tangent has slope −3 $\mathrm{m/s}$ . Which statement is correct?

x(t)
Tangent at t₀

Choose an answer

Instantaneous velocity is +10

\mathrm{m/s}

.Instantaneous velocity is −3

\mathrm{m/s}

.Instantaneous acceleration is −3

\mathrm{m/s^2}

.Instantaneous velocity is zero because only x matters.

Question

Velocity and acceleration signs from a graph

1D motion

On a v-versus-t graph, at t₀ the curve lies below v=0 and the tangent has negative slope. What happens to speed at that instant?

v(t)
Tangent at t₀

Choose an answer

It increases because v<0 and a<0 have the same sign.It decreases because negative velocity always means slowing down.It decreases because negative acceleration always means slowing down.It remains constant because the graph lies below the axis.

Question

Integrated relationship among v(t), x(t), and a(t)

1D motion

For 0≤t≤4 s, $v(t)$ =6 $\mathrm{m/s}$ −(2 $\mathrm{m/s^2}$ )t. Which description is correct?

v(t)=6−2t

Choose an answer

a=+2

\mathrm{m/s^2}

and

x(t)

increases throughout the interval.a=−2

\mathrm{m/s^2}

;

x(t)

increases until t=3 s and then decreases.a=−2

\mathrm{m/s^2}

;

x(t)

decreases from t=0 because acceleration is negative.a=0 at t=3 s because v=0 there.

Question

When does constant acceleration apply?

Equations

An $a(t)$ graph shows a=2 $\mathrm{m/s^2}$ from 0 to 4 s; from 4 to 8 s, a increases linearly from 2 to 6 $\mathrm{m/s^2}$ . Over which interval can constant-acceleration equations be used directly with one value of a?

Choose an answer

Only from 0 to 4 s.Only from 4 to 8 s.Over the full 0–8 s interval using average acceleration.Over neither interval, because velocity changes.

Question

When averaging endpoint velocities is justified

Equations

A particle has velocities $v_{0}$ and $v_{f}$ at the beginning and end of an interval. Which additional condition justifies using v̄=( $v_{0}$ + $v_{f}$ )/2 over that interval?

Choose an answer

The time interval is positive.Acceleration is constant throughout the interval.

v_{0}

and

v_{f}

have the same sign.Displacement is positive.

Question

Turning point: v=0 does not imply a=0

Equations

For $v(t)$ =4 $\mathrm{m/s}$ −(2 $\mathrm{m/s^2}$ )t, what happens at t=2 s?

Choose an answer

v=0 and a=0; it remains permanently at rest.v=0 and a=−2

\mathrm{m/s^2}

; rest is instantaneous and direction reverses if the model continues.v=−2

\mathrm{m/s}

and a=0.v=0 and acceleration cannot be determined.

Question

The sign of g depends on the axis

Equations

For a vertical launch choose +y downward, and let g denote the positive magnitude 9.8 $\mathrm{m/s^2}$ . At launch the object moves upward. Which signs are correct?

Choose an answer

v_{0}

ᵧ>0,

a_{y}

=−g

v_{0}

ᵧ<0,

a_{y}

=+g

v_{0}

ᵧ<0,

a_{y}

=−g

v_{0}

ᵧ>0,

a_{y}

=+g

Question

Gravity at maximum height

Equations

A ball is thrown vertically upward with no air resistance and +y upward. At maximum height:

Choose an answer

v_{y}

=0 and

a_{y}

=0.

v_{y}

=0 and

a_{y}

=−g.

v_{y}

>0 and

a_{y}

=0.

v_{y}

<0 and

a_{y}

=−g.

Question

Projectile components share time

2D/3D motion

A projectile is launched from a platform. Air resistance is neglected and +y points upward. Which description is correct?

Choose an answer

x and y have independent times because different equations are used.

a_{x}

=0,

a_{y}

=−g, and

x(t)

, y(t) describe the same projectile with the same t.

a_{x}

=g,

a_{y}

=−g because motion is two-dimensional.

a_{x}

=0,

a_{y}

=0 while the projectile is rising.

Question

Vector state at the apex

2D/3D motion

An ideal projectile is launched obliquely with $v_{0}$ ₓ≠0. At its highest point, with +y upward:

Choose an answer

v=0 and a=0.

v_{x}

=0,

v_{y}

=0,

a_{y}

=−g.

v_{x}

=

v_{0}

ₓ≠0,

v_{y}

=0, a=(0,−g).

v_{x}

=

v_{0}

ₓ≠0,

v_{y}

=0, a=0.

Question

Vector kinematics from r(t)

2D/3D motion

A particle has $\vec r(t)$ =(1 $\mathrm{m/s}$ )t i+(1 $\mathrm{m/s^2}$ ) $t^2$ j, with t in seconds. At t=1 s, which pair is correct?

Choose an answer

v=(1,1)

\mathrm{m/s}

, a=(0,1)

\mathrm{m/s^2}

.v=(1,2)

\mathrm{m/s}

, a=(0,2)

\mathrm{m/s^2}

.v=(0,2)

\mathrm{m/s}

, a=(1,2)

\mathrm{m/s^2}

.v=(1,2)

\mathrm{m/s}

, a=(1,2)

\mathrm{m/s^2}

.

Question

Parallel and perpendicular acceleration components

2D/3D motion

At one instant velocity points east. Acceleration has one component east and one north. Which interpretation is correct?

Choose an answer

The eastward component changes speed; the northward component changes velocity direction.The northward component changes only speed and the eastward one only direction.Both components change only speed.A perpendicular acceleration component implies that speed must decrease.

Question

Constant speed does not imply constant velocity

Circular and relative

A particle moves around a circle at constant speed. Which statement is correct?

Choose an answer

Velocity is constant because speed is constant.Acceleration is zero because speed does not change.Velocity changes direction and radial acceleration points toward the centre.Acceleration is tangent to the circle in the direction of motion.

Question

Circular motion with increasing speed

Circular and relative

A particle is at the rightmost point of a circle. At that instant it moves upward and its speed is increasing. Where does total acceleration point?

Choose an answer

Left only.Up only.Up and left.Down and right.

Question

Meaning of subscript order

Circular and relative

What does $v_{A/B}$ mean?

Choose an answer

Velocity of B as observed from A.Velocity of A as observed from B.Sum of the absolute velocities of A and B.Acceleration of A relative to B.

Question

Vector composition with explicit frames

Circular and relative

A boat moves relative to water at 4 $\mathrm{m/s}$ north. Water moves relative to the ground at 3 $\mathrm{m/s}$ east. What is the boat's velocity relative to the ground?

Choose an answer

1

\mathrm{m/s}

north because 4−3=1.7

\mathrm{m/s}

northeast because magnitudes add.(3i+4j)

\mathrm{m/s}

, magnitude 5

\mathrm{m/s}

, northeast.(−3i+4j)

\mathrm{m/s}

, northwest.

Question

A possible dimension does not prove a model

Measurement tools

A speed v is to be constructed only from acceleration a and length d. Which statement is correct?

Choose an answer

v=ad is possible because it contains acceleration and length.v=√(ad) has the correct dimension for speed, but dimensional analysis alone does not prove that it is the correct physical model.v=a/d has the correct dimension for speed.If √(ad) has the dimension of speed, its numerical coefficient must be exactly 1.

Question

Components, magnitude, and quadrant

Vectors

A vector has components $A_{x}$ =−4 and $A_{y}$ =−3 in the same units. Which statement is correct?

Choose an answer

Magnitude −5, third quadrant.Magnitude 5, third quadrant.Magnitude 7, second quadrant.Magnitude 1, fourth quadrant.

Question

x(t), v(t), a(t), and speed

1D motion

The position is $x(t)$ =12 m−(4 $\mathrm{m/s}$ )t+(1 $\mathrm{m/s^2}$ ) $t^2$ . Which description is correct at t=1 s?

Choose an answer

v=−2

\mathrm{m/s}

, a=+2

\mathrm{m/s^2}

; motion is toward

-x

and speed is decreasing.v=+2

\mathrm{m/s}

, a=−2

\mathrm{m/s^2}

; motion is toward

+x

and slowing.v=−2

\mathrm{m/s}

, a=+2

\mathrm{m/s^2}

; motion is toward

-x

and speed is increasing.v=9

\mathrm{m/s}

because x(1)=9 m.

Question

Reversal, displacement, and distance

Equations

For 0≤t≤4 s, $v(t)$ =4 $\mathrm{m/s}$ −(2 $\mathrm{m/s^2}$ )t. What are displacement and distance traveled?

v(t)=4−2t

Choose an answer

Displacement 0 m; distance 8 m.Displacement 8 m; distance 0 m.Displacement 0 m; distance 0 m.Displacement −8 m; distance 8 m.

Question

Projectile: shared time and apex

2D/3D motion

An ideal projectile is launched with $v_{0}$ ₓ=6 $\mathrm{m/s}$ and $v_{0}$ ᵧ=8 $\mathrm{m/s}$ , with +y upward and g=10 $\mathrm{m/s^2}$ . Which state corresponds to the apex?

Choose an answer

t=0.8 s,

v_{x}

=6

\mathrm{m/s}

,

v_{y}

=0,

a_{y}

=−10

\mathrm{m/s^2}

.t=0.6 s,

v_{x}

=0,

v_{y}

=8

\mathrm{m/s}

,

a_{y}

=0.t=0.8 s,

v_{x}

=0,

v_{y}

=0,

a_{y}

=−10

\mathrm{m/s^2}

.t=0.8 s,

v_{x}

=6

\mathrm{m/s}

,

v_{y}

=0,

a_{y}

=0.

Question

Circular motion: tangent velocity and radial acceleration

Circular and relative

A particle moves around a circle at constant speed. At the leftmost point it moves downward. What are the directions of v and a?

Choose an answer

v downward and a rightward.v downward and a leftward.Both downward.v rightward and a downward.

Question

Relative velocity as an inverse problem

Circular and relative

Wind moves relative to the ground as $\vec v_{a/g}$ =3i $\mathrm{m/s}$ east. An aircraft wants $\vec v_{p/g}$ =4j $\mathrm{m/s}$ north. What velocity must the aircraft have relative to the air?

Choose an answer

(3i+4j)

\mathrm{m/s}

(−3i+4j)

\mathrm{m/s}

(4i−3j)

\mathrm{m/s}

7

\mathrm{m/s}

northeast

Question

r(t) → v(t) → a(t) and change in speed

2D/3D motion

A particle has $\vec r(t)$ =(2 $\mathrm{m/s}$ )t i+(1 $\mathrm{m/s^2}$ ) $t^2$ j. At t=1 s, which statement is correct?

Choose an answer

v=(2,2)

\mathrm{m/s}

, a=(0,2)

\mathrm{m/s^2}

, and speed is increasing.v=(2,1)

\mathrm{m/s}

, a=(0,0)

\mathrm{m/s^2}

, and speed is constant.v=(0,2)

\mathrm{m/s}

, a=(2,2)

\mathrm{m/s^2}

.v=(2,2)

\mathrm{m/s}

, a=(0,2)

\mathrm{m/s^2}

, but speed cannot change because

a_{x}

=0.

Question

Variable acceleration: integrate rather than using final a as constant

Equations

A particle has $a(t)$ =(2 $\mathrm{m/s^3}$ )t with $v(0)$ =1 $\mathrm{m/s}$ . What is v(3 s)?

Choose an answer

10

\mathrm{m/s}

because v(3 s)=1

\mathrm{m/s}

+∫₀^(3 s) (2

\mathrm{m/s^3}

)t dt.19

\mathrm{m/s}

using v=

v_{0}

+a(3)t=1+6(3).7

\mathrm{m/s}

because a(3)=6

\mathrm{m/s^2}

.1

\mathrm{m/s}

because acceleration is not constant.

Question

The polar basis rotates even though its unit vectors have unit magnitude

Polar coordinates

In the xy plane, θ is measured counterclockwise from $+x$ . At one instant a particle is at θ=90° and $\dot\theta$ >0. Which statement is correct at that instant?

Choose an answer

\hat r

=+ĵ,

\hat\theta

=−î, and d

\hat r

/dt points toward

-x

.

\hat r

=+ĵ,

\hat\theta

=+î, and d

\hat r

/dt points toward

+x

.

\hat r

=+ĵ,

\hat\theta

=−î, but d

\hat r

/dt=0 because

\hat r

has unit magnitude.

\hat r

=−ĵ,

\hat\theta

=+î, and d

\hat r

/dt points toward

-x

.

Question

From Cartesian to polar coordinates without losing the quadrant

Polar coordinates

Use the convention r≥0 and 0°≤θ<360°, with θ measured counterclockwise from $+x$ . A particle is at x=−3 m, y=3√3 m. What are its polar coordinates?

Choose an answer

r=6 m, θ=60°r=6 m, θ=120°r=6 m, θ=240°r=3√3 m, θ=120°

Question

Velocity: moving toward the origin while the angle increases

Polar coordinates

At one instant, r=4 m, $\dot r$ =−1 $\mathrm{m/s}$ , and $\dot\theta$ =0.50 rad/s. Which velocity expression and interpretation are correct?

Choose an answer

v=(−1

\hat r

+2

\hat\theta

)

\mathrm{m/s}

: the particle moves inward while advancing in the direction of increasing θ.v=−1

\hat r

\mathrm{m/s}

: only the change in r matters.v=(+1

\hat r

+2

\hat\theta

)

\mathrm{m/s}

: the particle moves outward while θ increases.v=(−1

\hat r

−2

\hat\theta

)

\mathrm{m/s}

: the particle moves inward while θ decreases.

Question

Translate a polar velocity into the Cartesian basis

Polar coordinates

At one instant, θ=90° and the velocity written in the local polar basis is v=(2 $\hat r$ +3 $\hat\theta$ ) $\mathrm{m/s}$ . What is the same velocity in the Cartesian basis (î,ĵ)?

Choose an answer

(−3 î+2 ĵ)

\mathrm{m/s}

(+3 î+2 ĵ)

\mathrm{m/s}

(+2 î+3 ĵ)

\mathrm{m/s}

(−2 î+3 ĵ)

\mathrm{m/s}

Question

The transverse term 2ṙθ̇

Polar coordinates

Polar acceleration can be written as a=( $\ddot r$ −r $\dot\theta$ ²) $\hat r$ +(r $\ddot\theta$ +2 $\dot r$ $\dot\theta$ ) $\hat\theta$ . At one instant, $\dot r$ >0, $\dot\theta$ >0, and $\ddot\theta$ =0. Which statement about the transverse component $a_{θ}$ is justified?

Choose an answer

a_{θ}

=0 necessarily because

\ddot\theta

=0.

a_{θ}

>0 because of the contribution 2

\dot r

\dot\theta

, even though

\ddot\theta

=0.

a_{θ}

<0 because increasing r always produces acceleration opposite to

\hat\theta

.If

\ddot\theta

=0, the entire acceleration must be radial.

Question

Integration: velocity and acceleration of a spiral

Polar coordinates

For t>0, a particle follows the polar law $\vec r(t)$ =(1.0 $\mathrm{m/s}$ )t, θ(t)=(1.0 rad/s)t. At t=2.0 s, which pair (v,a) is correct in the instantaneous polar basis?

Choose an answer

v=(1

\hat r

+2

\hat\theta

)

\mathrm{m/s}

, a=(−2

\hat r

+2

\hat\theta

)

\mathrm{m/s^2}

.v=1

\hat r

\mathrm{m/s}

, a=0.v=(1

\hat r

+2

\hat\theta

)

\mathrm{m/s}

, a=−2

\hat r

\mathrm{m/s^2}

.v=(2

\hat r

+1

\hat\theta

)

\mathrm{m/s}

, a=(2

\hat r

−2

\hat\theta

)

\mathrm{m/s^2}

.

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