Corrected exercises: vector kinematics in two and three dimensions (203-SN1-RE)

“$20^{\circ}$ west of north” reads in order: you start from NORTH and turn $20^{\circ}$ toward the west. The angle is therefore measured from the $y$-axis, and it is the SINE that gives the $x$-component: $\Delta x_{2}=-3.00\sin 20.0^{\circ}=-1.03$ km (toward the west, so negative) and $\Delta y_{2}=3.00\cos 20.0^{\circ}=2.82$ km. The classic trap is to write $3\cos 20^{\circ}$ for $x$ out of reflex: you would get a leg pointing almost due east, which the figure contradicts at once. Safe rule: the cosine goes on the axis you START FROM when measuring the angle.

Leg 1, angle measured from east: $\Delta\vec{r}_{1}=(4.00\cos 30^{\circ}, 4.00\sin 30^{\circ})=(3.46, 2.00)$ km. Leg 3: $(0, -2.50)$ km. Add COMPONENT BY COMPONENT: $\Delta x=3.464-1.026+0=2.44$ km and $\Delta y=2.000+2.819-2.500=2.32$ km. Adding the magnitudes, $4+3+2.5=9.5$ km, would give the DISTANCE traveled, not the displacement.

Magnitude: $\sqrt{2.438^{2}+2.319^{2}}=3.36$ km. Direction: $\arctan\dfrac{2.319}{2.438}=43.6^{\circ}$, and since both components are positive, the vector is indeed in the northeast quadrant: $43.6^{\circ}$ north of east. Checking the quadrant is not a formality: $\arctan$ only returns angles between $-90^{\circ}$ and $90^{\circ}$, and a vector in the southwest quadrant would give the same result on a calculator.

Total time: $50+40+30=120$ min $=2.00$ h. Average velocity: $\dfrac{3.36}{2.00}=1.68$ km/h, directed like the displacement. Average speed: $\dfrac{9.50}{2.00}=4.75$ km/h. The first only keeps the starting point and the end point, the second counts every meter walked. Leg 3, which comes back toward the south, makes the distance grow while REDUCING the displacement: it is the one that opens the gap. The two are equal only for straight-line motion with no turning back.

The return is the opposite vector, $-\Delta\vec{r}=(-2.44, -2.32)$ km: same magnitude, $3.36$ km, direction $43.6^{\circ}$ south of west. You change neither the angle nor the reference, you only flip the two compass points.

The average velocity depends ONLY on the positions: $\vec{v}_{\mathrm{avg}}=\dfrac{\vec{r}_{2}-\vec{r}_{1}}{t_{2}-t_{1}}$. Component by component, $\Delta\vec{r}=(11-3, 7-(-1), 2-5)=(8.0, 8.0, -3.0)$ m, and $\Delta t=4.0$ s, so $\vec{v}_{\mathrm{avg}}=(2.0, 2.0, -0.75)$ m/s. The velocities $\vec{v}_{1}$ and $\vec{v}_{2}$ given in the statement are not used here: that is deliberate, a problem often gives more than the question asks for. Watch the $-(-1)$ in $y$: it is the most frequent sign error in this calculation.

In three dimensions, the magnitude takes the THREE squares: $\sqrt{2.0^{2}+2.0^{2}+0.75^{2}}=\sqrt{8.5625}=2.93$ m/s. Forgetting the $z$-component would give $2.83$ m/s, an error that goes unnoticed because the result stays plausible.

Same move with the velocities: $\vec{a}_{\mathrm{avg}}=\dfrac{\vec{v}_{2}-\vec{v}_{1}}{\Delta t}=\dfrac{(-4.0, 2.0, 2.0)}{4.0}=(-1.0, 0.50, 0.50)$ m/s². Magnitude: $\sqrt{1+0.25+0.25}=\sqrt{1.5}=1.22$ m/s². Notice that the speeds, $|\vec{v}_{1}|=4.58$ m/s and $|\vec{v}_{2}|=6.40$ m/s, do NOT recover this result: $\dfrac{6.40-4.58}{4}=0.46$ m/s² only measures the change in speed, whereas acceleration also measures the change in direction.

$\dfrac{\vec{v}_{1}+\vec{v}_{2}}{2}=(0, 5.0, 0)$ m/s, very different from $(2.0, 2.0, -0.75)$ m/s. Now, if the acceleration were constant, the two would be EQUAL: this is the fourth equation of constant-acceleration motion, $\Delta\vec{r}=\dfrac{\vec{v}_{i}+\vec{v}_{f}}{2}\,\Delta t$, which is true only under that assumption. Since the equality fails, we conclude with certainty that the acceleration VARIED between the two instants. The average acceleration of c) is therefore just an average, and nothing says the drone ever had that acceleration at a particular instant.

The horizontal part of $\vec{v}_{\mathrm{avg}}$ has magnitude $\sqrt{2.0^{2}+2.0^{2}}=2.83$ m/s; the vertical component is $-0.75$ m/s. The angle below the horizontal is $\arctan\dfrac{0.75}{2.83}=14.9^{\circ}$. Do NOT take the arctangent of $\dfrac{v_{z}}{v_{x}}$: the horizontal has two components, and you must combine them before comparing with the vertical.

Differentiate EACH component separately, like three independent straight-line motions: $\vec{v}(t)=(4.0-1.0t)\,\hat{\imath}+3.0t\,\hat{\jmath}-0.60t^{2}\,\hat{k}$ m/s, then $\vec{a}(t)=-1.0\,\hat{\imath}+3.0\,\hat{\jmath}-1.2t\,\hat{k}$ m/s². The $x$- and $y$-components of the acceleration are constant, but not the $z$-component, which depends on $t$: the motion is NOT uniformly accelerated, and the constant-acceleration equations are off limits here. ONE varying component is enough to invalidate everything, because the acceleration vector then changes.

$\vec{v}(2.0)=(4.0-2.0, 6.0, -0.60\times 4.0)=(2.0, 6.0, -2.4)$ m/s and $|\vec{v}|=\sqrt{4.0+36+5.76}=6.76$ m/s.

$v_{x}=4.0-t=0$ for $t=4.0$ s. At that instant, $\vec{a}=(-1.0, 3.0, -4.8)$ m/s² and $|\vec{a}|=\sqrt{1+9+23.04}=5.75$ m/s². A vanishing $v_{x}$ says NOTHING about $a_{x}$, which stays $-1.0$ m/s²: a zero velocity component is not a zero acceleration component, exactly as at the top of a trajectory where $v_{y}=0$ but $a_{y}=-g$.

At $t=2.0$ s, $\vec{a}=(-1.0, 3.0, -2.4)$ m/s², of magnitude $\sqrt{15.76}=3.97$ m/s². Dot product: $\vec{v}\cdot\vec{a}=(2.0)(-1.0)+(6.0)(3.0)+(-2.4)(-2.4)=-2.0+18+5.76=21.76$. Angle: $\cos\theta=\dfrac{21.76}{6.765\times 3.970}=0.810$, so $\theta=35.9^{\circ}$.

The angle is acute, the dot product positive: the acceleration has a component ALONG the velocity, so the speed increases. This component is the tangential acceleration, $a_{t}=\dfrac{\vec{v}\cdot\vec{a}}{|\vec{v}|}=\dfrac{21.76}{6.765}=3.22$ m/s²; the rest of the acceleration, perpendicular to $\vec{v}$, bends the path. This is the criterion that returns throughout the set: you do not judge whether an object speeds up or slows down by the sign of a component, but by the sign of $\vec{v}\cdot\vec{a}$. Here $a_{x}<0$, and yet the helicopter goes faster and faster.

$\vec{a}=2\vec{v}$: the two vectors are parallel and point the same way. $\vec{v}\cdot\vec{a}=18+32=50>0$, the speed increases; $3\times 8-4\times 6=0$, the direction does not change. This is straight-line motion that speeds up.

$\vec{v}\cdot\vec{a}=-24+24=0$: the acceleration is entirely PERPENDICULAR to the velocity. The speed does not change at this instant, and all of the acceleration goes into turning. This is the situation of uniform circular motion, where you accelerate constantly without ever going faster. The trap is to conclude “nonzero acceleration, so the object goes faster”.

$\vec{a}=-\vec{v}$: parallel, opposite directions. $\vec{v}\cdot\vec{a}=-25<0$, the speed decreases, and the direction does not change as long as the object has not stopped. A car braking in a straight line.

$\vec{v}\cdot\vec{a}=0+4\times(-10)=-40<0$: the speed decreases, the ball rises while losing speed. And $3\times(-10)-4\times 0=-30\neq 0$: the direction changes, the path curves downward. Tangential acceleration: $a_{t}=\dfrac{\vec{v}\cdot\vec{a}}{|\vec{v}|}=\dfrac{-40}{5}=-8$ m/s², the minus sign meaning “opposite to the motion”. Normal acceleration: $a_{n}=\sqrt{|\vec{a}|^{2}-a_{t}^{2}}=\sqrt{100-64}=6$ m/s². Check by the other route: $a_{n}=\dfrac{|v_{x}a_{y}-v_{y}a_{x}|}{|\vec{v}|}=\dfrac{30}{5}=6$ m/s². The same gravity slows the ball down AND bends its path, in a proportion that changes at every instant of the flight.

$\vec{v}\cdot\vec{a}=8+0=8>0$, the speed increases; $4\times 3-0\times 2=12\neq 0$, the direction changes too. This is the general case: both effects at once. Only situations a), b) and c) are “pure”, and they are the three textbook cases that every textbook draws.

Since the acceleration is constant, each component follows the constant-acceleration equation: $x(t)=x_{0}+v_{0x}t+\frac{1}{2}a_{x}t^{2}=2.0+3.0t-0.50t^{2}$ and $y(t)=6.0-4.0t+1.0t^{2}$, then $v_{x}=3.0-1.0t$ and $v_{y}=-4.0+2.0t$. This is the vector equation $\vec{r}=\vec{r}_{0}+\vec{v}_{0}t+\frac{1}{2}\vec{a}t^{2}$ written line by line. At $t=1.0$ s: $\vec{v}=(2.0, -2.0)$ m/s. The writing trap is to forget the factor $\frac{1}{2}$ on just one of the two lines.

$y$ is a minimum when $v_{y}=0$: $-4.0+2.0t=0$ gives $t=2.0$ s. Then $y=6-8+4=2.0$ m and $x=2+6-2=6.0$ m. You touch ONLY the $y$ line: the extremum of a coordinate looks only at its own velocity component.

$\vec{v}(2.0)=(1.0, 0)$ m/s, of magnitude $1.0$ m/s. The speed is NOT a minimum at that instant, and that is the trap of the exercise. “$y$ minimum” and “minimum speed” are two different questions: the first makes $v_{y}$ vanish, the second requires the velocity to be perpendicular to the acceleration. At $t=2.0$ s, $\vec{v}\cdot\vec{a}=(1)(-1)+(0)(2)=-1<0$: the speed is still decreasing.

$|\vec{v}|^{2}=(3-t)^{2}+(2t-4)^{2}=5t^{2}-22t+25$, a parabola whose minimum is at the vertex $t=\dfrac{22}{10}=2.2$ s. Then $|\vec{v}|^{2}=24.2-48.4+25=0.80$ and $|\vec{v}|=0.89$ m/s. Check with the criterion of exercise 4: $\vec{v}(2.2)=(0.8, 0.4)$ and $\vec{v}\cdot\vec{a}=-0.8+0.8=0$. The velocity never vanishes: a constant-acceleration motion whose $\vec{a}$ is not parallel to $\vec{v}_{0}$ can never stop.

Crossing the $y$-axis means $x=0$: $2.0+3.0t-0.50t^{2}=0$, that is $t^{2}-6t-4=0$ and $t=3\pm\sqrt{13}$. The negative root, $-0.61$ s, is before the start: reject it. So $t=3+\sqrt{13}=6.61$ s. For $y$, the equation gives $t^{2}=6t+4$, hence $y=6-4t+t^{2}=2t+10=23.2$ m. The puck first heads to the right, stops in $x$ at $t=3.0$ s (where $x=6.5$ m) then turns back to the left while climbing: its path is a parabola whose axis is parallel to $\vec{a}$, not vertical.

The velocity points at $+45.0^{\circ}$ from east, the acceleration at $-15.0^{\circ}$: the angle between them is $45+15=60.0^{\circ}$. It is acute, so $\vec{v}\cdot\vec{a}>0$ and the speed starts by increasing. But the effect is not pure: the gust ALSO turns the drone toward the east.

The constant-acceleration relation holds for vectors: $\vec{v}=\vec{v}_{0}+\vec{a}\,\Delta t$. $\vec{v}_{0}=(12.0\cos 45^{\circ}, 12.0\sin 45^{\circ})=(8.49, 8.49)$ m/s. $\vec{a}\,\Delta t$ has magnitude $3.20\times 2.50=8.00$ m/s, and components $(8.00\cos 15^{\circ}, -8.00\sin 15^{\circ})=(7.73, -2.07)$ m/s, with the minus sign of “south”. Sum: $\vec{v}=(16.21, 6.41)$ m/s, of magnitude $\sqrt{16.21^{2}+6.41^{2}}=17.4$ m/s, directed $\arctan\dfrac{6.41}{16.21}=21.6^{\circ}$ north of east. The figure in the solution shows the triangle: $\vec{a}\,\Delta t$ is placed at the tip of $\vec{v}_{0}$.

$\Delta\vec{r}=\vec{v}_{0}\,t+\frac{1}{2}\vec{a}\,t^{2}$. First term: $(8.485\times 2.50, 8.485\times 2.50)=(21.21, 21.21)$ m. Second term, of magnitude $\frac{1}{2}\times 3.20\times 2.50^{2}=10.0$ m at $-15^{\circ}$: $(9.66, -2.59)$ m. Total $(30.87, 18.63)$ m, that is $36.1$ m at $31.1^{\circ}$ north of east. The direction of the displacement, $31.1^{\circ}$, lies between the starting direction, $45^{\circ}$, and the final one, $21.6^{\circ}$: this is consistent with a path that turns gradually.

He adds the MAGNITUDES as if the two vectors were parallel. This is true only if the acceleration has the direction of the velocity; here they make $60^{\circ}$ and part of $\vec{a}\,\Delta t$ goes into turning instead of speeding up. The true answer, $17.4$ m/s, is smaller than $20.0$ m/s, and it always will be: one side of a triangle is shorter than the sum of the other two.

For the direction to stay that of $\vec{v}_{0}$, $\vec{a}$ must be PARALLEL to $\vec{v}_{0}$ and in the same direction, so at $45.0^{\circ}$ north of east. Only then do the magnitudes add: $12.0+3.20\times 2.50=20.0$ m/s. The classmate's calculation was correct, but for a different gust.

The horizontal launch is the simplest case of a projectile, because $v_{0y}=0$: the marble starts its vertical fall EXACTLY as if it had been dropped. With the $y$-axis pointing down for this question, $0.920=\frac{1}{2}(9.80)t^{2}$ gives $t=\sqrt{\dfrac{2\times 0.920}{9.80}}=0.433$ s. The horizontal velocity plays no role at all: this is the reflex to acquire, the height alone sets the fall time of a horizontal launch.

Horizontally, the motion is uniform: $v_{0}=\dfrac{\Delta x}{t}=\dfrac{1.15}{0.4333}=2.65$ m/s. This is an INVERSE problem, the range is given and you work back to the speed; it is handled in the same order, the time first from the vertical motion, the speed next from the horizontal.

At impact, $v_{x}=2.65$ m/s, unchanged, and $v_{y}=gt=9.80\times 0.4333=4.25$ m/s downward. Magnitude: $\sqrt{2.654^{2}+4.246^{2}}=5.01$ m/s. Angle below the horizontal: $\arctan\dfrac{4.246}{2.654}=58.0^{\circ}$. Another route to the magnitude, through the energy methods seen later: $v^{2}=v_{0}^{2}+2gh=7.04+18.03$, same result.

The opening of the cup is $0.120$ m above the floor: the marble therefore only has $0.920-0.120=0.800$ m to fall. Time: $\sqrt{\dfrac{2\times 0.800}{9.80}}=0.404$ s, and distance $2.654\times 0.4041=1.07$ m. The trap is to compute $1.15$ m minus something proportional to $0.12$: the path is a PARABOLA, not a straight line, and the last stretch of the fall is the fastest. Redo the time calculation with the correct height.

They hit the floor AT THE SAME TIME. Vertically, both marbles start with $v_{0y}=0$ and undergo the same acceleration $g$: their vertical motions are identical, and the horizontal motion of the first changes nothing about it. This is the independence of the components, the thread of the whole set, seen in its barest form.

The top is recognized by $v_{y}=0$, not by a “nearly horizontal” velocity. Here $v_{y}=+1.0$ m/s: the ball is still RISING. It reaches the top when $1.0-9.80\,t=0$, that is in $t=0.10$ s, after rising a further $\dfrac{1.0^{2}}{2\times 9.80}=5.1$ cm. An angle of $18^{\circ}$ above the horizontal is still a climbing angle.

At the top, only the VERTICAL component of the velocity vanishes; the horizontal component $v_{0x}$ has never changed, so $\vec{v}=v_{0x}\,\hat{\imath}$. And the acceleration equals $-g\,\hat{\jmath}$ throughout the flight, the top included: if it vanished at the top, the ball would hang there. Only a perfectly vertical throw has zero velocity at the top, and even that one keeps $\vec{a}=-g\,\hat{\jmath}$.

Acceleration measures the change of the velocity VECTOR, direction included. An acceleration perpendicular to the velocity makes it turn without changing its magnitude: this is uniform circular motion, and it is situation b) of exercise 4. A constant speed means $\vec{v}\cdot\vec{a}=0$, not $\vec{a}=\vec{0}$.

The relation $\vec{v}_{\mathrm{avg}}=\dfrac{\vec{v}_{i}+\vec{v}_{f}}{2}$ is a consequence of constant-acceleration motion and holds only if the acceleration is CONSTANT. Exercise 2 gives a numerical counterexample: $(0, 5, 0)$ m/s on one side, $(2, 2, -0.75)$ m/s on the other. The definition that always holds is $\dfrac{\Delta\vec{r}}{\Delta t}$.

At the moment it is dropped, the package has the horizontal velocity of the plane, and nothing changes it horizontally afterward. Package and plane therefore advance by the same distance at every instant: the package stays directly below the plane and falls under it. The intuition “it falls behind” comes from air resistance, which slows the package down and which the statement excludes; it is also the error of Aristotle, who thought a motion stops as soon as you stop pushing.