Unit 1 review mini quiz

$[vt]=(L/T)T=L$ .

$[at]=(L/T^2)T=L/T$ .

They cannot be added as terms of a position because their dimensions differ.

72.0 $\mathrm{km/h}$ × (1000 m/1 km) × (1 h/3600 s).

20.0 $\mathrm{m/s}$ ; the units km and h cancel.

Vector equality requires the same magnitude and the same orientation.

Two arrows of equal length can point in different directions.

$\Delta x$ = Δr cos 32.0° and Δy = Δr sin 32.0°.

$\Delta x$ ≈ 6.36 m and Δy ≈ 3.97 m; both are positive because of the quadrant.

√(6.36²+3.97²) ≈ 7.50 m.

$\vec A\cdot\vec B=2(3)+\lambda(-2)+(-1)(4)=2-2\lambda$ .

$2-2\lambda=0$ , therefore λ = 1.

$|A+B|^2-|A-B|^2$ = 4 $\vec A\cdot\vec B$ , so $\vec A\cdot\vec B$ = 0.

$\vec A\cdot\vec B=3a-2=0$ .

a = $2/3$ .

$x=-20\,m$ locates the object on the negative side of the origin.

It does not tell us how x changes or how v changes.

The student travels 10 m outward and 6 m on the return: distance = 16 m.

$\Delta x=4\,m-0\,m=+4\,m$ .

v and a both point toward $-x$ .

When they have the same direction, the magnitude $|v|$ increases.

$v=\frac{-4-2}{3-0}=\frac{-6}{3}=-2\,m/s$ .

Position decreases by two metres per second relative to the chosen axis.

v = 4.0 + (−1.0)(3.0) = 1.0 $\mathrm{m/s}$ .

$\Delta x$ = 4.0(3.0)+ $1/2$ (−1.0)(3.0)² = 7.5 m.

$0=12-3t$ gives t = 4 s.

$v(6)=12-18=-6\,m/s$ : it is already moving toward $-x$ .

The vertical component changes from positive to negative and is zero at the turning instant.

Gravity continues to act: $a_{y}$ = −g throughout the ideal flight.

At the top, $v_{y}$ = 0 and $v_{x}$ remains 8 $\mathrm{m/s}$ .

$a_{x}$ = 0 and $a_{y}$ = −g throughout the ideal flight.

0 = 11.25− $1/2$ (10) $t^2$ gives $t^2$ = 2.25 and t = 1.5 s.

$\Delta x$ = $v_{x}$ t = 6.0(1.5) = 9.0 m.

Both components share the same time; $a_{x}$ = 0.

Its magnitude is fixed, but its tangent direction changes.

Changing the velocity vector requires acceleration toward the centre.

$a_{c}$ = $v^2/R$ .

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

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.

The ratios $\Delta x/\Delta t$ are +2 $\mathrm{m/s}$ , 0 $\mathrm{m/s}$ , and −3 $\mathrm{m/s}$ , respectively.

Position does not change between 2 s and 5 s; this is the rest interval.

Displacement is −2 m − 0 m = −2 m. Distance is 4 m + 0 m + 6 m = 10 m.

The successive slopes of the polyline are positive, zero, and negative. The curve crosses v = 0 at t = 7 s.

The slopes give +2 $\mathrm{m/s^2}$ , 0 $\mathrm{m/s^2}$ , −2 $\mathrm{m/s^2}$ , and −1 $\mathrm{m/s^2}$ .

The signed areas are 4 m, 12 m, 4 m, and −2 m; therefore $\Delta x$ = 18 m.

Distance adds the absolute values of the areas: 4 m + 12 m + 4 m + 2 m = 22 m.

After the start, direction changes when the curve crosses v = 0 at t = 7 s.

The head is three units left and four units up: $A_{x}$ = −3 and $A_{y}$ = +4, in quadrant II.

The 3–4–5 triangle gives |A| = 5.

atan2(4,−3) gives θ ≈ 126.9° from $+x$ , equivalent to 53.1° above $-x$ .

The angle must lie between 90° and 180° because both components place the vector in quadrant II.

The first arrow represents A = (4,1), and the second, translated to its head, represents B = (−1,3).

R = (4−1, 1+3) = (3,4).

|R| = √(3²+4²) = 5.

The final head of the construction lies three units right and four units above the origin.

The required resultant is vertical, so the boat's horizontal component must cancel the current.

The required westward component is 1.5 $\mathrm{m/s}$ . The northward component is √(2.5²−1.5²) = 2.0 $\mathrm{m/s}$ .

sin θ = 1.5/2.5 gives θ ≈ 36.9° west of north.

When the horizontal components cancel, the speed relative to the ground is 2.0 $\mathrm{m/s}$ due north.

t = 120 m/(2.0 $\mathrm{m/s}$ ) = 60 s.

The line crosses v = 0 at t = 2 s: maximum height occurs there.

The positive triangular area is $1/2$ (2 s)(20 $\mathrm{m/s}$ ) = 20 m.

Between 2 s and 4 s there is a triangle with area −20 m; the total signed area is 0 m.

Distance adds the magnitudes of both areas: 20 m + 20 m = 40 m.

Speed increases linearly, remains constant, and then decreases linearly to zero.

The accelerations are +1 $\mathrm{m/s^2}$ , 0 $\mathrm{m/s^2}$ , and −1 $\mathrm{m/s^2}$ .

The two triangles contribute 2 m each, and the central rectangle contributes 12 m.

Total displacement is 2 m + 12 m + 2 m = 16 m, positive in the chosen direction.

Between 0 s and 2 s, $\Delta v$ = (+2 $\mathrm{m/s^2}$ )(2 s) = +4 $\mathrm{m/s}$ ; thus v(2 s) = 3 $\mathrm{m/s}$ .

Since a = 0 between 2 s and 5 s, v(5 s) = 3 $\mathrm{m/s}$ .

Between 5 s and 9 s, $\Delta v$ = (−1 $\mathrm{m/s^2}$ )(4 s) = −4 $\mathrm{m/s}$ ; therefore v(9 s) = −1 $\mathrm{m/s}$ .

In the first segment, −1 + 2t = 0 gives t = 0.5 s. In the last, 3 − (t−5) = 0 gives t = 8 s.

The two crossings separate motion toward $-x$ , then $+x$ , and finally $-x$ .

The figure marks the highest point of the trajectory at t = 2 s; the tangent is horizontal there.

Substituting t = 2 s gives r(2 s) = (4 m)i + (4 m)j.

$v(t)$ = (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.

$a(t)$ = −(2.0 $\mathrm{m/s^2}$ )j throughout the interval.

At that instant velocity is horizontal toward $+x$ , consistent with the trajectory's tangent, while acceleration points toward −y.

The digits written in a measured result communicate its precision.

4.50 contains three significant figures; the final zero is deliberate.

Because both vectors are nonzero, |A||B| ≠ 0.

Therefore cos θ = 0 and θ = 90°.

The final displacement is zero.

v̄ = $\Delta x/\Delta t$ = 0.

If $\vec a=d\vec v/dt$ = 0, v does not change.

A constant velocity need not be zero.

Add the motion of O relative to P to the motion of P relative to S.

This gives the velocity of O relative to S.

The figure starts at x = 0 m and ends at x = −2 m.

$\Delta x$ = −2 − 0 = −2 m.

$\Delta x$ = +20 m − 2 m = +18 m.

$x_{f}$ = $x_{i}$ + $\Delta x$ = −3 m + 18 m = 15 m.

The areas are 2 m, 12 m, and 2 m.

Total displacement is 16 m.

The head lies left of and above the origin.

Therefore $A_{x}$ < 0, $A_{y}$ > 0: quadrant II.

Velocity is tangent.

Radial acceleration points toward the centre, perpendicular to the instantaneous velocity.

[T] = L^(p+q) T^(−2q).

−2q = 1 gives q = −1/2; p + q = 0 gives p = 1/2.

The dot product is linear in each argument.

A · (B + C) = A · B + A · C.

0 = 12² + 2(−3) $\Delta x$ .

$\Delta x$ = 24 m.

t = √(2h/g).

t₄ₕ/tₕ = √(4h/h) = 2.

d = |17 − 2| + |8 − 17| = 24 m.

$\Delta x$ = 8 − 2 = +6 m.

a = (24 − 0)/8.0.

a = 3.0 $\mathrm{m/s^2}$ .

t = √[2(1.25)/10] = 0.50 s.

x = $v_{x}$ t = 4.0(0.50) = 2.0 m.

v_person/ground = v_person/walkway + v_walkway/ground.

v = 1.5 + 0.8 = +2.3 $\mathrm{m/s}$ .

$\Delta x$ ₁ = 18 m and $\Delta x$ ₂ = −8 m.

$\Delta x$ = 10 m; $\Delta t$ = 10 s.

v̄ = 10/10 = 1 $\mathrm{m/s}$ .

T = 2π(2)/(4π) = 1 s.

$a_{c}$ = (4π)²/2 = 8π² ≈ 78.96 $\mathrm{m/s^2}$ .

Moving toward $-x$ implies v < 0.

As the magnitude of a negative velocity decreases, a points toward $+x$ : a > 0.