Practice · Unit 1 · Aula Física

Measurement tools

An expression proposes $x=vt+at$ . Without calculating numerical values, decide whether it can represent a position and justify your answer.

Request a hint

Compare $[v][t]$ with $[a][t]$ .

Review the solution

Dimensions
$[vt]=(L/T)T=L$ .
Second term
$[at]=(L/T^2)T=L/T$ .
Conclusion
They cannot be added as terms of a position because their dimensions differ.

Measurement tools

Convert 72.0 $\mathrm{km/h}$ to $\mathrm{m/s}$ and retain consistent precision.

Request a hint

Use 1 km = 1000 m and 1 h = 3600 s.

Review the solution

Conversion chain
72.0 $\mathrm{km/h}$ × (1000 m/1 km) × (1 h/3600 s).
Result
20.0 $\mathrm{m/s}$ ; the units km and h cancel.

Measurement tools

A person takes approximately 8× $10^3$ steps in one day, and each step is about 0.75 m long. Estimate the daily distance and, using the nearest-power-of-ten convention, its order of magnitude in metres.

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Multiply the number of steps by the average length of each step.

Review the solution

Estimate
(8× $10^3$ )(0.75 m) ≈ 6× $10^3$ m.
Scale
Under the nearest-power-of-ten convention, 6× $10^3$ is closer to $10^4$ than to $10^3$ .

Vectors

Two vectors have the same magnitude. Does this guarantee that they are equal?

Review the solution

Criterion
Vector equality requires the same magnitude and the same orientation.
Conclusion
Two arrows of equal length can point in different directions.

Vectors

A moving platform travels 7.50 m at 32.0° above $+x$ . Determine its Cartesian components.

Request a hint

The angle is measured from $+x$ ; identify the adjacent and opposite components.

Review the solution

Model
$\Delta x$ = Δr cos 32.0° and Δy = Δr sin 32.0°.
Result
$\Delta x$ ≈ 6.36 m and Δy ≈ 3.97 m; both are positive because of the quadrant.
Check
√(6.36²+3.97²) ≈ 7.50 m.

Vectors

A = 2i + λj − k and B = 3i − 2j + 4k. Find λ so that the vectors are perpendicular.

Request a hint

For perpendicular vectors, $\vec A\cdot\vec B$ = 0.

Review the solution

Product
$\vec A\cdot\vec B=2(3)+\lambda(-2)+(-1)(4)=2-2\lambda$ .
Condition
$2-2\lambda=0$ , therefore λ = 1.

Vectors

A = a i + 2j and B = 3i − j. Determine a so that $|A+B|=|A-B|$ .

Request a hint

Square both sides before expanding.

Review the solution

Condition
$|A+B|^2-|A-B|^2$ = 4 $\vec A\cdot\vec B$ , so $\vec A\cdot\vec B$ = 0.
Product
$\vec A\cdot\vec B=3a-2=0$ .
Result
a = $2/3$ .

1D motion

An object has $x=-20\,m$ . What can be inferred about its velocity and acceleration?

Review the solution

Interpretation
$x=-20\,m$ locates the object on the negative side of the origin.
Limit of the datum
It does not tell us how x changes or how v changes.

1D motion

A student goes from x = 0 to x = 10 m and returns to x = 4 m. Calculate distance and displacement.

Review the solution

Path
The student travels 10 m outward and 6 m on the return: distance = 16 m.
Endpoints
$\Delta x=4\,m-0\,m=+4\,m$ .

1D motion

At an instant, v < 0 and a < 0. Is the speed increasing or decreasing?

Review the solution

Directions
v and a both point toward $-x$ .
Speed
When they have the same direction, the magnitude $|v|$ increases.

1D motion

Read two points on the $x(t)$ line in the graph. Determine the constant velocity and explain what its sign indicates relative to the chosen axis.

The required information is in the axes and line; the prompt does not repeat their coordinates.

Request a hint

Calculate $\Delta x/\Delta t$ using two points on the line.

Review the solution

Slope
$v=\frac{-4-2}{3-0}=\frac{-6}{3}=-2\,m/s$ .
Interpretation
Position decreases by two metres per second relative to the chosen axis.

Equations

An object has $v_{0}$ = 4.0 $\mathrm{m/s}$ , a = −1.0 $\mathrm{m/s^2}$ , and evolves for 3.0 s. Calculate v and $\Delta x$ .

Review the solution

Velocity
v = 4.0 + (−1.0)(3.0) = 1.0 $\mathrm{m/s}$ .
Displacement
$\Delta x$ = 4.0(3.0)+ $1/2$ (−1.0)(3.0)² = 7.5 m.

Equations

An object starts with $v_{0}$ = 12 $\mathrm{m/s}$ and constant a = −3 $\mathrm{m/s^2}$ . When does it stop? If acceleration continues, what is v at 6 s?

Review the solution

Stopping
$0=12-3t$ gives t = 4 s.
Afterwards
$v(6)=12-18=-6\,m/s$ : it is already moving toward $-x$ .

Equations

A ball is at the highest point of an ideal vertical launch. Describe $v_{y}$ and $a_{y}$ if +y points upward.

Review the solution

Velocity
The vertical component changes from positive to negative and is zero at the turning instant.
Acceleration
Gravity continues to act: $a_{y}$ = −g throughout the ideal flight.

Equations

A particle has $a(t)$ = (2.0 $\mathrm{m/s^3}$ )t and $v(0)$ = 3.0 $\mathrm{m/s}$ . Obtain $v(t)$ for t ≥ 0.

Review the solution

Model
$v(t)$ = $v(0)$ +∫₀ᵗ(2.0 $\mathrm{m/s^3}$ )τ dτ.
Integration
The integral contributes (1.0 $\mathrm{m/s^3}$ ) $t^2$ , which has the dimension of velocity.
Result
$v(t)$ = 3.0 $\mathrm{m/s}$ + (1.0 $\mathrm{m/s^3}$ ) $t^2$ ; constant-acceleration equations were not used.

2D/3D motion

An ideal projectile reaches its highest point with $v_{x}$ = 8 $\mathrm{m/s}$ . Determine the velocity vector—or equivalently its magnitude and direction—at that instant, and describe the acceleration.

Review the solution

Velocity
At the top, $v_{y}$ = 0 and $v_{x}$ remains 8 $\mathrm{m/s}$ .
Acceleration
$a_{x}$ = 0 and $a_{y}$ = −g throughout the ideal flight.

2D/3D motion

An object is launched horizontally at 6.0 $\mathrm{m/s}$ from a height of 11.25 m. Use g = 10 $\mathrm{m/s^2}$ and the ideal model. Calculate the time to reach the ground and the horizontal displacement.

Review the solution

Vertical
0 = 11.25− $1/2$ (10) $t^2$ gives $t^2$ = 2.25 and t = 1.5 s.
Horizontal
$\Delta x$ = $v_{x}$ t = 6.0(1.5) = 9.0 m.
Model
Both components share the same time; $a_{x}$ = 0.

2D/3D motion

The position is $\vec r(t)$ = (2.0 $\mathrm{m/s}$ )t i + [(1.0 $\mathrm{m/s^2}$ ) $t^2$ − 1.0 m]j. Obtain $v(t)$ and $a(t)$ .

Review the solution

Velocity
Differentiating each component preserves its dimensions: $v_{x}$ = 2.0 $\mathrm{m/s}$ and $v_{y}$ = (2.0 $\mathrm{m/s^2}$ )t.
Acceleration
Differentiating again gives $a_{x}$ = 0 and $a_{y}$ = 2.0 $\mathrm{m/s^2}$ .

Circular and relative

A particle travels around a circle at constant speed. Is its velocity constant? Does it have acceleration?

Review the solution

Velocity
Its magnitude is fixed, but its tangent direction changes.
Acceleration
Changing the velocity vector requires acceleration toward the centre.

Circular and relative

A point moves at a speed of 3.0 $\mathrm{m/s}$ on a circle of radius 1.5 m. Calculate the magnitude of its radial acceleration.

Review the solution

Relation
$a_{c}$ = $v^2/R$ .
Result
$a_{c}$ = (3.0)²/1.5 = 6.0 $\mathrm{m/s^2}$ , directed toward the centre.

Circular and relative

A person walks at 1.2 $\mathrm{m/s}$ toward $+x$ relative to a walkway that moves at 0.8 $\mathrm{m/s}$ toward $+x$ relative to the ground. Determine the person's velocity relative to the ground. What changes if the person walks at 1.2 $\mathrm{m/s}$ toward $-x$ relative to the walkway?

Review the solution

Chain
v_person/ground = v_person/walkway + v_walkway/ground.
Same direction
1.2+0.8 = +2.0 $\mathrm{m/s}$ .
Opposite directions
−1.2+0.8 = −0.4 $\mathrm{m/s}$ .

Polar coordinates

In the polar expression for acceleration, take r = R constant and $\dot\theta$ = ω constant. Simplify a and explain its direction.

Review the solution

Conditions
$\dot r$ = $\ddot r$ = 0 and $\ddot\theta$ = 0.
Substitution
The transverse component vanishes and the radial component becomes $-R\omega^2\hat r$ .
Interpretation
− $\hat r$ points toward the origin: this is centripetal acceleration.

1D motion

Read the position-time graph. Determine the velocity in each segment, the rest interval, the total displacement, and the total distance travelled.

The slope of each $x(t)$ segment represents its velocity.

Read coordinates and slopes directly from the polyline before distinguishing net change from total path length.

Request a hint

On $x(t)$ , each slope is the segment's velocity; a horizontal section represents rest.

Review the solution

Observation
The graph rises from 0 m to 4 m between 0 s and 2 s, remains horizontal until 5 s, and falls to −2 m at 7 s.
Slopes
The ratios $\Delta x/\Delta t$ are +2 $\mathrm{m/s}$ , 0 $\mathrm{m/s}$ , and −3 $\mathrm{m/s}$ , respectively.
Rest
Position does not change between 2 s and 5 s; this is the rest interval.
Overall results
Displacement is −2 m − 0 m = −2 m. Distance is 4 m + 0 m + 6 m = 10 m.

1D motion

From the velocity-time graph, determine the acceleration in each segment, total displacement, distance travelled, and the instant after the start at which the direction of motion changes.

Slope: acceleration. Signed area: displacement.

The segment below the axis contributes negative displacement; add its magnitude when calculating distance.

Request a hint

The slope of $v(t)$ is a; for distance, areas below the axis also count with a positive sign.

Review the solution

Observation
The successive slopes of the polyline are positive, zero, and negative. The curve crosses v = 0 at t = 7 s.
Accelerations
The slopes give +2 $\mathrm{m/s^2}$ , 0 $\mathrm{m/s^2}$ , −2 $\mathrm{m/s^2}$ , and −1 $\mathrm{m/s^2}$ .
Signed area
The signed areas are 4 m, 12 m, 4 m, and −2 m; therefore $\Delta x$ = 18 m.
Distance
Distance adds the absolute values of the areas: 4 m + 12 m + 4 m + 2 m = 22 m.
Interpretation
After the start, direction changes when the curve crosses v = 0 at t = 7 s.

Vectors

Read the vector from the grid and determine $A_{x}$ , $A_{y}$ , its magnitude, quadrant, and direction measured from $+x$ . You may also express the equivalent angle relative to $-x$ if you state that reference.

The grid is the source of the components; no numerical values are shown beside the arrow.

Request a hint

First read the horizontal and vertical displacement from the origin to the head.

Review the solution

Reading
The head is three units left and four units up: $A_{x}$ = −3 and $A_{y}$ = +4, in quadrant II.
Magnitude
The 3–4–5 triangle gives |A| = 5.
Direction
atan2(4,−3) gives θ ≈ 126.9° from $+x$ , equivalent to 53.1° above $-x$ .
Check
The angle must lie between 90° and 180° because both components place the vector in quadrant II.

Vectors

The figure shows A and B in a head-to-tail construction. Obtain the components of R = A + B and its magnitude. The resultant is not drawn.

Both arrows retain their components in the head-to-tail translation; the resultant is left for the student.

Request a hint

Read the horizontal and vertical change of each arrow separately.

Review the solution

Reading
The first arrow represents A = (4,1), and the second, translated to its head, represents B = (−1,3).
Sum
R = (4−1, 1+3) = (3,4).
Magnitude
|R| = √(3²+4²) = 5.
Check
The final head of the construction lies three units right and four units above the origin.

2D/3D motion

The figure marks positions of an ideal projectile at equal time intervals. Explain what the horizontal spacing indicates, what happens to $v_{y}$ at the top, whether the velocity vector is zero there, and the direction of acceleration during the flight.

The markers correspond to equal time intervals; the figure does not draw vectors that would reveal the answers.

Request a hint

Separate the horizontal and vertical behaviour.

Review the solution

Observation
The points retain uniform horizontal spacing and form a curved trajectory.
Horizontal component
Equal time intervals and equal horizontal separations indicate constant $v_{x}$ in the ideal model.
Top
At the top $v_{y}$ = 0, but the velocity vector remains horizontal if $v_{x}$ ≠ 0.
Acceleration
Gravitational acceleration remains constant and directed downward throughout the flight.

Circular and relative

The particle moves counterclockwise around the circle shown. Describe the instantaneous direction of velocity and centripetal acceleration at the indicated point.

The circle and point establish the geometry; the velocity and acceleration arrows that the student must infer are not drawn.

Request a hint

Mentally draw the tangent and the radius connecting the point to the centre; compare their directions.

Review the solution

Observation
The point is in the first quadrant and the direction of travel is counterclockwise.
Velocity
Velocity is tangent to the circle and points counterclockwise—upward and to the left at that point.
Acceleration
Centripetal acceleration is radially inward: it points from the particle to the centre.
Check
The two directions are perpendicular in circular motion.

Circular and relative

A boat moves at 2.5 $\mathrm{m/s}$ relative to the water, and the current flows east at 1.5 $\mathrm{m/s}$ . Its velocity relative to the ground must point due north. Use the figure to determine the boat's direction relative to the water, its speed relative to the ground, and the crossing time for a 120 m-wide river.

The construction separates velocity relative to the water, current, and velocity relative to the ground through object and frame.

Request a hint

The westward component relative to the water must exactly cancel the eastward current.

Review the solution

Observation
The required resultant is vertical, so the boat's horizontal component must cancel the current.
Components
The required westward component is 1.5 $\mathrm{m/s}$ . The northward component is √(2.5²−1.5²) = 2.0 $\mathrm{m/s}$ .
Direction
sin θ = 1.5/2.5 gives θ ≈ 36.9° west of north.
Result relative to the ground
When the horizontal components cancel, the speed relative to the ground is 2.0 $\mathrm{m/s}$ due north.
Crossing
t = 120 m/(2.0 $\mathrm{m/s}$ ) = 60 s.

Equations

Read the $v(t)$ graph. Knowing that the signed area under $v(t)$ represents displacement, determine the instant of maximum height, the height gained to the top, displacement at t = 4 s, and total distance travelled.

The signed area under $v(t)$ represents displacement.

The line and axes contain the velocity law; the prompt does not provide it as the main expression.

Request a hint

The top occurs when vertical velocity crosses zero; separate the positive and negative areas.

Review the solution

Observation
The line crosses v = 0 at t = 2 s: maximum height occurs there.
Ascent
The positive triangular area is $1/2$ (2 s)(20 $\mathrm{m/s}$ ) = 20 m.
Displacement
Between 2 s and 4 s there is a triangle with area −20 m; the total signed area is 0 m.
Distance
Distance adds the magnitudes of both areas: 20 m + 20 m = 40 m.

Equations

The graph shows an elevator's vertical velocity. Determine the acceleration in each phase and the total displacement during the 10 s shown.

Slope indicates acceleration; area indicates displacement.

The profile distinguishes acceleration, constant-speed travel, and braking without giving the area decomposition.

Request a hint

Acceleration is the slope; displacement is the area under the complete graph.

Review the solution

Observation
Speed increases linearly, remains constant, and then decreases linearly to zero.
Slopes
The accelerations are +1 $\mathrm{m/s^2}$ , 0 $\mathrm{m/s^2}$ , and −1 $\mathrm{m/s^2}$ .
Areas
The two triangles contribute 2 m each, and the central rectangle contributes 12 m.
Result
Total displacement is 2 m + 12 m + 2 m = 16 m, positive in the chosen direction.

1D motion

The figure shows positions recorded every 1 s. State the direction of motion, whether speed increases, decreases, or remains constant, the sign of acceleration, and the visual evidence supporting your answers. An exact value of a is not required.

Each mark corresponds to one additional second; use the sequence's geometry as the source of information.

Request a hint

Compare the separations travelled during equal time intervals.

Review the solution

Observation
The marks advance toward increasing x values, and their separations grow.
Velocity
Greater distances are covered in equal intervals, so speed increases toward $+x$ .
Acceleration
An increasing positive velocity implies positive acceleration over the observed interval.
Limit
The figure supports a qualitative conclusion; without an additional model, it does not require an exact value of a.

Equations

The graph shows piecewise $a(t)$ , and $v(0)$ = −1 $\mathrm{m/s}$ is known. Determine v(2 s), v(5 s), v(9 s), and the instants when velocity changes sign.

The signed area under $a(t)$ is the change in velocity.

The jumps separate three constant-acceleration intervals; the initial velocity must be combined with the accumulated areas.

Request a hint

The signed area under $a(t)$ between two instants is the change in velocity.

Review the solution

First segment
Between 0 s and 2 s, $\Delta v$ = (+2 $\mathrm{m/s^2}$ )(2 s) = +4 $\mathrm{m/s}$ ; thus v(2 s) = 3 $\mathrm{m/s}$ .
Second segment
Since a = 0 between 2 s and 5 s, v(5 s) = 3 $\mathrm{m/s}$ .
Third segment
Between 5 s and 9 s, $\Delta v$ = (−1 $\mathrm{m/s^2}$ )(4 s) = −4 $\mathrm{m/s}$ ; therefore v(9 s) = −1 $\mathrm{m/s}$ .
Sign changes
In the first segment, −1 + 2t = 0 gives t = 0.5 s. In the last, 3 − (t−5) = 0 gives t = 8 s.
Interpretation
The two crossings separate motion toward $-x$ , then $+x$ , and finally $-x$ .

2D/3D motion

A particle obeys $\vec r(t)$ = [(2.0 $\mathrm{m/s}$ )t]i + [(4.0 $\mathrm{m/s}$ )t − (1.0 $\mathrm{m/s^2}$ ) $t^2$ ]j. The figure shows its trajectory for 0 ≤ t ≤ 4 s. At t = 2 s, determine position, velocity vector, acceleration vector, and interpret the instantaneous orientation.

The curve preserves physical scale in x and y; its local orientation is interpreted through its tangent, not by distorting the viewBox.

Request a hint

Differentiate both components with respect to time; velocity is tangent to the trajectory.

Review the solution

Observation
The figure marks the highest point of the trajectory at t = 2 s; the tangent is horizontal there.
Position
Substituting t = 2 s gives r(2 s) = (4 m)i + (4 m)j.
Velocity
$v(t)$ = (2.0 $\mathrm{m/s}$ )i + [(4.0 $\mathrm{m/s}$ ) − (2.0 $\mathrm{m/s^2}$ )t]j; therefore v(2 s) = (2.0 $\mathrm{m/s}$ )i.
Acceleration
$a(t)$ = −(2.0 $\mathrm{m/s^2}$ )j throughout the interval.
Interpretation
At that instant velocity is horizontal toward $+x$ , consistent with the trajectory's tangent, while acceleration points toward −y.

Measurement tools

A length is reported as 4.50 m. What does the final zero communicate?

Request a hint

Compare 4.5 m with 4.50 m as measured results.

Review the solution

Set-up
The digits written in a measured result communicate its precision.
Development
4.50 contains three significant figures; the final zero is deliberate.

Vectors

Two nonzero vectors satisfy A · B = 0. Which geometric conclusion is justified?

Request a hint

Use A · B = |A||B| cos θ.

Review the solution

Set-up
Because both vectors are nonzero, |A||B| ≠ 0.
Development
Therefore cos θ = 0 and θ = 90°.

1D motion

A person travels along a path and finishes exactly where they started. The total time interval is nonzero. What is their average velocity?

Request a hint

Average velocity uses displacement, not distance.

Review the solution

Set-up
The final displacement is zero.
Development
v̄ = $\Delta x/\Delta t$ = 0.

Equations

A particle has zero acceleration over an interval. What can be stated about its velocity during that interval?

Request a hint

Acceleration is change in velocity per unit time.

Review the solution

Set-up
If $\vec a=d\vec v/dt$ = 0, v does not change.
Development
A constant velocity need not be zero.

Circular and relative

Which expression preserves the correct order of frames when relating object O, platform P, and ground S?

Request a hint

The inner frames must form a chain: O relative to P and P relative to S.

Review the solution

Set-up
Add the motion of O relative to P to the motion of P relative to S.
Development
This gives the velocity of O relative to S.

1D motion

On the piecewise graph, use only the initial and final positions to determine the net displacement over the complete interval.

The slope of each $x(t)$ segment represents its velocity.

Read coordinates and slopes directly from the polyline before distinguishing net change from total path length.

Request a hint

Calculate $x_{f}$ − $x_{i}$ ; do not add the distance in each segment.

Review the solution

Set-up
The figure starts at x = 0 m and ends at x = −2 m.
Development
$\Delta x$ = −2 − 0 = −2 m.

1D motion

The $v(t)$ graph has a total positive area of +20 m and a negative area of −2 m. If x(0) = −3 m, determine x at the end of the interval.

Slope: acceleration. Signed area: displacement.

The segment below the axis contributes negative displacement; add its magnitude when calculating distance.

Request a hint

The net area is displacement, not final position.

Review the solution

Set-up
$\Delta x$ = +20 m − 2 m = +18 m.
Development
$x_{f}$ = $x_{i}$ + $\Delta x$ = −3 m + 18 m = 15 m.

Equations

For the elevator profile shown, calculate only the total displacement by adding the geometric areas under $v(t)$ .

Slope indicates acceleration; area indicates displacement.

The profile distinguishes acceleration, constant-speed travel, and braking without giving the area decomposition.

Request a hint

Add two triangles and the central rectangle.

Review the solution

Set-up
The areas are 2 m, 12 m, and 2 m.
Development
Total displacement is 16 m.

Vectors

Observe vector A drawn on the grid. In which quadrant does it lie, and what signs do its components have?

The grid is the source of the components; no numerical values are shown beside the arrow.

Request a hint

Read the horizontal and vertical direction from the origin to the head.

Review the solution

Set-up
The head lies left of and above the origin.
Development
Therefore $A_{x}$ < 0, $A_{y}$ > 0: quadrant II.

Circular and relative

At the position shown on the circular path, which direction must radial acceleration have?

The circle and point establish the geometry; the velocity and acceleration arrows that the student must infer are not drawn.

Request a hint

Centripetal means directed toward the centre.

Review the solution

Set-up
Velocity is tangent.
Development
Radial acceleration points toward the centre, perpendicular to the instantaneous velocity.

Measurement tools

A time scale is proposed as T = k L^p g^q, where k is dimensionless, [L] = L, and [g] = $L/T^2$ . Determine p and q.

Request a hint

Equate the exponents of L and T separately.

Review the solution

Set-up
[T] = L^(p+q) T^(−2q).
Development
−2q = 1 gives q = −1/2; p + q = 0 gives p = 1/2.

Vectors

Which expansion of A · (B + C) is correct?

Request a hint

Distribute A over each addend while retaining the dot product.

Review the solution

Set-up
The dot product is linear in each argument.
Development
A · (B + C) = A · B + A · C.

Equations

An object with initial speed $v_{0}$ = 12 $\mathrm{m/s}$ brakes with constant acceleration a = −3 $\mathrm{m/s^2}$ . Use a time-independent relation to find the displacement until it stops.

Request a hint

At the stop, v = 0.

Review the solution

Set-up
0 = 12² + 2(−3) $\Delta x$ .
Development
$\Delta x$ = 24 m.

2D/3D motion

Two objects are launched horizontally from heights h and 4h under the same g. What is the ratio of their fall times t₄ₕ/tₕ?

Request a hint

Vertically, h = $1/2$ g $t^2$ .

Review the solution

Set-up
t = √(2h/g).
Development
t₄ₕ/tₕ = √(4h/h) = 2.

1D motion

Starting at x = 2 m, a person walks to x = 17 m and then returns to x = 8 m. Determine total distance and displacement.

Request a hint

Add 15 m outward and 9 m back.

Review the solution

Set-up
d = |17 − 2| + |8 − 17| = 24 m.
Development
$\Delta x$ = 8 − 2 = +6 m.

Equations

A test vehicle starts from rest and reaches 24 $\mathrm{m/s}$ in 8.0 s with constant acceleration. Calculate the acceleration.

Request a hint

Use a = (v − $v_{0}$ )/t.

Review the solution

Set-up
a = (24 − 0)/8.0.
Development
a = 3.0 $\mathrm{m/s^2}$ .

2D/3D motion

A package leaves a 1.25 m-high table horizontally at 4.0 $\mathrm{m/s}$ . Use g = 10 $\mathrm{m/s^2}$ and neglect air resistance. Calculate the horizontal range.

Request a hint

First calculate t = √(2h/g).

Review the solution

Set-up
t = √[2(1.25)/10] = 0.50 s.
Development
x = $v_{x}$ t = 4.0(0.50) = 2.0 m.

Circular and relative

A person walks at +1.5 $\mathrm{m/s}$ relative to a walkway that moves at +0.8 $\mathrm{m/s}$ relative to the ground. Calculate the person's velocity relative to the ground.

Request a hint

Both velocities point toward $+x$ .

Review the solution

Set-up
v_person/ground = v_person/walkway + v_walkway/ground.
Development
v = 1.5 + 0.8 = +2.3 $\mathrm{m/s}$ .

1D motion

A particle moves for 6 s at +3 $\mathrm{m/s}$ and then for 4 s at −2 $\mathrm{m/s}$ . Determine total displacement and average velocity over the complete interval.

Request a hint

Add the displacements of both segments algebraically.

Review the solution

Set-up
$\Delta x$ ₁ = 18 m and $\Delta x$ ₂ = −8 m.
Development
$\Delta x$ = 10 m; $\Delta t$ = 10 s.
Result
v̄ = 10/10 = 1 $\mathrm{m/s}$ .

Circular and relative

A particle travels around a circle of radius 2.0 m at a constant speed of 4π $\mathrm{m/s}$ . Calculate the period and the magnitude of radial acceleration. Use π ≈ 3.1416 for the numerical value.

Request a hint

T = 2πR/v and $a_{c}$ = $v^2/R$ .

Review the solution

Set-up
T = 2π(2)/(4π) = 1 s.
Development
$a_{c}$ = (4π)²/2 = 8π² ≈ 78.96 $\mathrm{m/s^2}$ .

1D motion

A particle moves toward $-x$ and its speed decreases. Which combination of signs is compatible with that instant?

Request a hint

To slow down, velocity and acceleration must have opposite signs.

Review the solution

Set-up
Moving toward $-x$ implies v < 0.
Development
As the magnitude of a negative velocity decreases, a points toward $+x$ : a > 0.

Vectors

1D motion

Equations

2D/3D motion

Circular and relative