02Describing Motion: Kinematics in one Dimension

Look for a single counterexample rather than trying to argue the general case.

Velocity and acceleration are independent at any given instant: one is $dx/dt$, the other is $dv/dt$, and a function can be zero where its derivative is not.

Consider a ball thrown upward at the top of its flight: $v = 0$ there while $a = -9.80\ \mathrm{m/s^{2}}$.

False, and the ball at the top of a throw is the standard counterexample.

Ask what would have to be added to the statement before a verdict about the speed became possible.

The speed rises when $v$ and $a$ share a sign and falls when they differ, so a verdict needs both signs.

Only the sign of $a$ is given here; the sign of $v$ is unknown, and $v$ could even be zero.

All that follows is that the velocity is becoming more negative; whether the object speeds up cannot be decided.

Both quantities have the same denominator, so the whole question is about their numerators.

$\bar s = d/\Delta t$ and $\vert\bar v\vert = \vert\Delta x\vert/\Delta t$, and in general $d \ge \vert\Delta x\vert$.

For a 400 m lap run in 100 s, $d = 400$ m while $\Delta x = 0$, so the two numerators differ as much as they possibly can.

False in general; they agree exactly when the motion never reverses direction.

Take one narrow strip of the graph and ask what its area means, before worrying about the whole shape.

One strip has area $v\,\Delta t$, which is a displacement; summing signed strips gives the signed total.

Strips below the axis have negative $v$, so their contribution is a negative displacement.

The signed area is the displacement; counting the areas as positive would give the distance travelled.

Two of these four questions care only about the first and last rows of the log; the other two care about all three.

$\Delta x = x_{\text{end}} - x_{\text{start}}$, $d$ is the sum of the leg lengths, $\bar v = \Delta x/\Delta t$ and $\bar s = d/\Delta t$.

The three positions are $+8.0$, $-12.0$ and $+2.0$ m, and the total time is 9.0 s.

$\Delta x = -6.0$ m, $d = 34$ m, $\bar v = -0.67$ m/s and $\bar s = 3.8$ m/s.

Convert the clock to seconds first, and decide which of the three questions uses which time interval.

$\bar v = \Delta x/\Delta t$ and $\bar s = d/\Delta t$; part (c) changes the denominator, not the numerator.

Total time 17.0 min = 1020 s; riding time 13.0 min = 780 s; $\Delta x = 3000 - 1200 = 1800$ m; $d = 4200$ m.

$\bar v = 1.76$ m/s, $\bar s = 4.12$ m/s, and while riding $5.38$ m/s.

Two of the three parts need only the velocity function; the third needs the position function at two instants.

$v = dx/dt$ and $a = dv/dt$; average velocity is $\Delta x/\Delta t$ and uses the position function, not the velocity function.

With $x(t) = -4.00 + 12.0t - 3.00t^{2}$, differentiate term by term, and for part (c) evaluate $x$ at $t = 0$ and $t = 4.00$ s.

$v = 12.0 - 6.00t$, $a = -6.00\ \mathrm{m/s^{2}}$; rest at $t = 2.00$ s at $x = 8.00$ m; and the average velocity over $[0, 4.00]$ is zero.

Part (b) needs no new arithmetic at all once part (a) is done — only a comparison of two signs at each instant.

$\bar a = \Delta v/\Delta t$; the speed rises when $v$ and $a$ share a sign; the ball is at rest when $v = 0$ in $v = v_{0} + at$.

$\Delta v = (-1.30) - (+3.20) = -4.50$ m/s over 2.50 s, and $v_{0} = +3.20$ m/s.

$a = -1.80\ \mathrm{m/s^{2}}$; slowing at $t = 0$, speeding up at $t = 2.50$ s; at rest at $t = 1.78$ s.

List the five quantities. One of them is missing from the data and not being asked for in part (a).

With $v$ spare, use $x = x_{0} + v_{0}t + \tfrac{1}{2}at^{2}$; then $v = v_{0}+at$ for part (b).

$v_{0} = 0$, so $20.0 = \tfrac{1}{2}a(3.00)^{2} = 4.50\,a$.

$a = 4.44\ \mathrm{m/s^{2}}$ and $v = 13.3\ \mathrm{m/s}$.

Parts (a) and (b) each have a one step route; part (c) is part (b) with a different height and is worth doing without the time.

$y = y_{0} + v_{0}t + \tfrac{1}{2}at^{2}$ for the time, and $v^{2} = v_{0}^{2} + 2a\,\Delta y$ for the speeds.

$y_{0} = 14.0$ m and $v_{0} = 0$, so $0 = 14.0 - 4.90t^{2}$; and for part (c) the fall is $\Delta y = -7.00$ m.

$t = 1.69$ s, impact at 16.6 m/s, and at half height 11.7 m/s.

The flight starts and ends at the same height, which is the condition that makes one particular shortcut legitimate here.

For a flight returning to its launch height the time up equals the time down; then $v = v_{0} + at$ and $v^{2} = v_{0}^{2} + 2a\,\Delta y$.

Half of 4.40 s is 2.20 s to the top, where $v = 0$; and for part (c) $\Delta y = +5.00$ m.

$v_{0} = 21.6$ m/s, maximum height 23.7 m, and 19.2 m/s at 5.00 m.

Part (a) should produce two answers and part (b) only one. Decide before solving why the counts differ.

$y = v_{0}t + \tfrac{1}{2}at^{2}$ in both parts, with different values of $y$; then $v^{2} = v_{0}^{2} + 2a\,\Delta y$ for the speed.

For (a) set $y = +8.00$ m with $v_{0} = 14.0$ m/s; for (b) set $y = -30.0$ m; the acceleration is $-9.80\ \mathrm{m/s^{2}}$ throughout.

(a) $t = 0.789$ s and $t = 2.07$ s; (b) $t = 4.29$ s; (c) 28.0 m/s.

The motion has two phases with different accelerations, so decide where to cut it before writing any equation.

Phase 1 has $a = 0$, so distance is speed times time. Phase 2 has no time given or wanted, so use $v^{2} = v_{0}^{2} + 2a\,(x-x_{0})$.

Phase 1: 25.0 m/s for 0.80 s. Phase 2: from 25.0 m/s to 0 at $6.50\ \mathrm{m/s^{2}}$. For part (d), only $55.0$ m minus the reaction distance is available for braking.

(a) 20.0 m, (b) 48.1 m, (c) 68.1 m, (d) 13.0 m/s.

Three phases means three separate applications of the same skeleton, with the ending velocity of each carried into the next.

Within each phase use $v = v_{0}+at$ and $x = x_{0}+v_{0}t+\tfrac{1}{2}at^{2}$; the average velocity at the end is total displacement over total time.

Phase 1 lasts 2.50 s from rest at $1.20\ \mathrm{m/s^{2}}$; phase 2 lasts 6.00 s at the resulting speed; phase 3 decelerates at $1.50\ \mathrm{m/s^{2}}$ from that speed to zero.

Cruise 3.00 m/s; heights 3.75 m, 18.0 m and 3.00 m; total 24.75 m in 10.5 s; average velocity 2.36 m/s.

Sketch the graph from the description, then mark the instant where it crosses the axis; everything else follows from areas.

Signed area under a velocity graph is displacement; the object is furthest from the start at the instant the velocity changes sign.

The graph crosses zero at $t = 4.00$ s. The three patches are a rectangle, a triangle above the axis and a triangle plus rectangle below it.

The areas above and below cancel exactly, so the object is back at the origin at $t = 8.00$ s.

The trap is in the initial velocity. Ask what the sandbag was doing in the instant before it was released.

Free fall from release onwards: $v = v_{0}+at$, $y = y_{0}+v_{0}t+\tfrac{1}{2}at^{2}$ and $v^{2} = v_{0}^{2} + 2a\,\Delta y$, with $v_{0} = +3.00$ m/s.

$y_{0} = 90.0$ m, $v_{0} = +3.00$ m/s, $a = -9.80\ \mathrm{m/s^{2}}$, and the balloon keeps going up at 3.00 m/s.

(a) 0.459 m at $t = 0.306$ s; (b) 4.60 s; (c) 42.1 m/s; (d) the balloon is at 104 m.

Check each step against the conditions the equation it uses requires, rather than against the arithmetic.

Distance equals speed times time only when the velocity is constant; under constant acceleration use $x = \tfrac{1}{2}(v_{0}+v)t$ or $v = v_{0}+at$.

The car speeds up from 8.00 to 20.0 m/s over the 84.0 m, so it is never travelling at a constant 8.00 m/s.

Step 3 is wrong; the time is $t = (20.0-8.00)/2.00 = 6.00$ s.

Decide which quantities have to be converted and which do not before writing anything down.

$a = (v-v_{0})/t$ and $x-x_{0} = \tfrac{1}{2}(v_{0}+v)t$; converting a speed uses a factor of 3.6, converting an acceleration uses it once as well.

36.0 km/h is 10.0 m/s and 108 km/h is 30.0 m/s, over 5.00 s.

$a = 4.00\ \mathrm{m/s^{2}}$, distance 100 m, and 14.4 km/h gained per second.

You can rule out candidates without knowing any kinematics at all. Ask what units each expression produces.

Both sides of a physical equation must have the same units; a distance must come out in metres.

$[v] = \mathrm{m/s}$ and $[a] = \mathrm{m/s^{2}}$, so $[v/a] = \mathrm{s}$ and $[v^{2}/a] = \mathrm{m}$.

Only candidate (ii) has units of metres, and it is also the correct formula.

This is an estimate, so decide first how many significant figures the data can honestly support.

$v^{2} = v_{0}^{2} + 2a\,(x-x_{0})$ with $v_{0} = 0$, then $t = v/a$.

270 km/h is 75 m/s and 2.0 km is 2000 m, from rest.

$a \approx 1.4\ \mathrm{m/s^{2}}$, about $0.14g$, and the run lasts about 53 s.

Look at the input data and ask which one is the least precise, because that one sets the ceiling for everything computed from it.

A result carries no more significant figures than the crudest datum it was computed from; and $h$ depends on $t^{2}$, so a fractional error in $t$ roughly doubles when it reaches $h$.

The time 1.2 s has two significant figures and an uncertainty of about $\pm 0.1$ s, which is roughly 8 percent.

Two significant figures, so $h = 7.1$ m, with a plausible range from 5.9 m to 8.3 m.