Chapter 3: Motion in a Plane

Choose a coordinate system with the positive $$x$$-axis pointing east and the positive $$y$$-axis pointing vertically upward. The two velocities (with respect to the ground) are

$$\vec{v}_r = -35\,\hat{\mathbf{j}}\ \mathrm{m\,s^{-1}}$$ (rain, falling vertically downward),
$$\vec{v}_w = -12\,\hat{\mathbf{i}}\ \mathrm{m\,s^{-1}}$$ (wind, blowing from east to west).

The rain reaches the boy with the resultant velocity

$$\vec{v} = \vec{v}_r + \vec{v}_w = -12\,\hat{\mathbf{i}} - 35\,\hat{\mathbf{j}}\ \mathrm{m\,s^{-1}}.$$

Its magnitude is

$$|\vec{v}| = \sqrt{12^2 + 35^2} = \sqrt{144 + 1225} = \sqrt{1369} = 37\ \mathrm{m\,s^{-1}}.$$

If $$\theta$$ is the angle which $$\vec{v}$$ makes with the vertical (downward direction),

$$\tan\theta = \dfrac{12}{35} = 0.343 \quad\Rightarrow\quad \theta \approx 19^\circ.$$

The boy should hold the umbrella tilted about $$19^\circ$$ from the vertical, toward the west (the direction from which the wind blows the rain).

Draw $$\mathbf{A}$$ along $$OP$$ and $$\mathbf{B}$$ along $$OQ$$ so that the angle between them at $$O$$ is $$\theta$$. Complete the parallelogram $$OPSQ$$; its diagonal $$OS$$ represents the resultant $$\mathbf{R} = \mathbf{A} + \mathbf{B}$$. Drop a perpendicular from $$S$$ onto the line $$OP$$ extended, meeting it at $$N$$.

In the right triangle $$PNS$$, $$PS = B$$ and angle $$SPN = \theta$$, so

$$SN = B\sin\theta, \qquad PN = B\cos\theta.$$

From the right triangle $$ONS$$,

$$OS^2 = ON^2 + SN^2 = (OP + PN)^2 + SN^2 = (A + B\cos\theta)^2 + (B\sin\theta)^2.$$

Expanding and using $$\sin^2\theta + \cos^2\theta = 1$$:

$$R^2 = A^2 + 2AB\cos\theta + B^2\cos^2\theta + B^2\sin^2\theta = A^2 + B^2 + 2AB\cos\theta.$$

Hence the magnitude of the resultant is

$$R = \sqrt{A^2 + B^2 + 2AB\cos\theta}.$$

If $$\alpha$$ is the angle that $$\mathbf{R}$$ makes with $$\mathbf{A}$$, then in triangle $$ONS$$

$$\tan\alpha = \dfrac{SN}{ON} = \dfrac{B\sin\theta}{A + B\cos\theta}.$$

So

$$\alpha = \tan^{-1}\!\left(\dfrac{B\sin\theta}{A + B\cos\theta}\right).$$

These two formulae together give the magnitude and direction of the resultant.

Take east as the positive $$x$$-direction and north as the positive $$y$$-direction. The boat's velocity (relative to water) and the water-current velocity (relative to ground) are

$$\vec{v}_b = 25\,\hat{\mathbf{j}}\ \mathrm{km/h}.$$

The current makes $$60^\circ$$ with south toward east, so it points $$30^\circ$$ south of east:

$$\vec{v}_c = 10\sin 60^\circ\,\hat{\mathbf{i}} - 10\cos 60^\circ\,\hat{\mathbf{j}} = 5\sqrt{3}\,\hat{\mathbf{i}} - 5\,\hat{\mathbf{j}}\ \mathrm{km/h}.$$

The resultant velocity of the boat with respect to ground is

$$\vec{v} = \vec{v}_b + \vec{v}_c = 5\sqrt{3}\,\hat{\mathbf{i}} + 20\,\hat{\mathbf{j}}\ \mathrm{km/h}.$$

Its magnitude is

$$|\vec{v}| = \sqrt{(5\sqrt{3})^2 + 20^2} = \sqrt{75 + 400} = \sqrt{475} \approx 21.8\ \mathrm{km/h}.$$

If $$\phi$$ is the angle that $$\vec{v}$$ makes with north (the $$y$$-axis),

$$\tan\phi = \dfrac{5\sqrt{3}}{20} = \dfrac{\sqrt{3}}{4} \approx 0.4330 \quad\Rightarrow\quad \phi \approx 23.4^\circ.$$

Thus the resultant velocity of the boat is about $$21.8\ \mathrm{km/h}$$, directed about $$23.4^\circ$$ east of north.

(a) Differentiate $$\mathbf{r}$$ with respect to $$t$$ to obtain the velocity, and again to obtain the acceleration.

$$\mathbf{v}(t) = \dfrac{d\mathbf{r}}{dt} = \dfrac{d}{dt}\big(3.0t\,\hat{\mathbf{i}} + 2.0t^2\,\hat{\mathbf{j}} + 5.0\,\hat{\mathbf{k}}\big) = 3.0\,\hat{\mathbf{i}} + 4.0t\,\hat{\mathbf{j}}\ \mathrm{m\,s^{-1}}.$$

$$\mathbf{a}(t) = \dfrac{d\mathbf{v}}{dt} = 4.0\,\hat{\mathbf{j}}\ \mathrm{m\,s^{-2}}.$$

The acceleration is constant in magnitude and direction (along the $$+y$$-axis).

(b) At $$t = 1.0\ \mathrm{s}$$,

$$\mathbf{v}(1) = 3.0\,\hat{\mathbf{i}} + 4.0\,\hat{\mathbf{j}}\ \mathrm{m\,s^{-1}}.$$

$$|\mathbf{v}(1)| = \sqrt{3.0^2 + 4.0^2} = \sqrt{9 + 16} = \sqrt{25} = 5.0\ \mathrm{m\,s^{-1}}.$$

If $$\theta$$ is the angle that $$\mathbf{v}(1)$$ makes with the $$x$$-axis,

$$\tan\theta = \dfrac{v_y}{v_x} = \dfrac{4.0}{3.0} \quad\Rightarrow\quad \theta = \tan^{-1}\!\left(\dfrac{4}{3}\right) \approx 53^\circ.$$

So at $$t = 1.0\ \mathrm{s}$$ the speed is $$5.0\ \mathrm{m\,s^{-1}}$$ directed about $$53^\circ$$ above the $$x$$-axis in the $$xy$$-plane.

Initial position $$\mathbf{r}_0 = 0$$, initial velocity $$\mathbf{u} = 5.0\,\hat{\mathbf{i}}\ \mathrm{m/s}$$, constant acceleration $$\mathbf{a} = 3.0\,\hat{\mathbf{i}} + 2.0\,\hat{\mathbf{j}}\ \mathrm{m/s^2}$$. Using $$\mathbf{r}(t) = \mathbf{u}t + \tfrac{1}{2}\mathbf{a}t^2$$ component-wise:

$$x(t) = 5.0\,t + \tfrac{1}{2}(3.0)t^2 = 5t + 1.5\,t^2,$$
$$y(t) = 0 + \tfrac{1}{2}(2.0)t^2 = t^2.$$

(a) Set $$x(t) = 84\ \mathrm{m}$$:

$$1.5\,t^2 + 5\,t - 84 = 0 \quad\Rightarrow\quad 3t^2 + 10t - 168 = 0.$$

By the quadratic formula,

$$t = \dfrac{-10 \pm \sqrt{100 + 12\cdot 168}}{6} = \dfrac{-10 \pm \sqrt{2116}}{6} = \dfrac{-10 \pm 46}{6}.$$

Taking the positive root, $$t = 36/6 = 6\ \mathrm{s}$$.

$$y(6) = 6^2 = 36\ \mathrm{m}.$$

(b) The velocity at time $$t$$ is

$$\mathbf{v}(t) = \mathbf{u} + \mathbf{a}\,t = (5.0 + 3.0\,t)\,\hat{\mathbf{i}} + 2.0\,t\,\hat{\mathbf{j}}.$$

At $$t = 6\ \mathrm{s}$$,

$$\mathbf{v} = (5 + 18)\,\hat{\mathbf{i}} + 12\,\hat{\mathbf{j}} = 23\,\hat{\mathbf{i}} + 12\,\hat{\mathbf{j}}\ \mathrm{m/s}.$$

$$|\mathbf{v}| = \sqrt{23^2 + 12^2} = \sqrt{529 + 144} = \sqrt{673} \approx 25.9\ \mathrm{m/s}.$$

For a projectile launched from level ground with initial speed $$u$$ at angle $$\theta$$ to the horizontal, the horizontal range is

$$R(\theta) = \dfrac{u^2 \sin 2\theta}{g}.$$

Consider two angles that are symmetric about $$45^\circ$$:

$$\theta_1 = 45^\circ + \alpha \quad\text{and}\quad \theta_2 = 45^\circ - \alpha,$$

where $$0 < \alpha < 45^\circ$$. Then

$$R(\theta_1) = \dfrac{u^2 \sin\big(2(45^\circ + \alpha)\big)}{g} = \dfrac{u^2 \sin(90^\circ + 2\alpha)}{g} = \dfrac{u^2 \cos 2\alpha}{g},$$

$$R(\theta_2) = \dfrac{u^2 \sin\big(2(45^\circ - \alpha)\big)}{g} = \dfrac{u^2 \sin(90^\circ - 2\alpha)}{g} = \dfrac{u^2 \cos 2\alpha}{g}.$$

Hence $$R(45^\circ + \alpha) = R(45^\circ - \alpha)$$, proving Galileo's statement. (As a corollary, the maximum range occurs at $$\alpha = 0$$, i.e. $$\theta = 45^\circ$$.)

Choose the origin at the point of projection on the cliff's edge, with $$+x$$ horizontal in the direction of throw and $$+y$$ vertically upward. Then $$u_x = 15\ \mathrm{m\,s^{-1}}$$, $$u_y = 0$$, $$a_x = 0$$, $$a_y = -g = -9.8\ \mathrm{m\,s^{-2}}$$.

Time of flight. The vertical displacement when the stone strikes the ground is $$y = -490\ \mathrm{m}$$. Using $$y = u_y t + \tfrac{1}{2}a_y t^2$$:

$$-490 = 0 + \tfrac{1}{2}(-9.8)\,t^2 \quad\Rightarrow\quad t^2 = \dfrac{2(490)}{9.8} = 100.$$

$$t = 10\ \mathrm{s}.$$

Speed on landing. The horizontal velocity is unchanged: $$v_x = 15\ \mathrm{m\,s^{-1}}$$. The vertical velocity just before impact is

$$v_y = u_y + a_y t = 0 - 9.8(10) = -98\ \mathrm{m\,s^{-1}}.$$

Hence

$$v = \sqrt{v_x^2 + v_y^2} = \sqrt{15^2 + 98^2} = \sqrt{225 + 9604} = \sqrt{9829} \approx 99.1\ \mathrm{m\,s^{-1}}.$$

Here $$u = 28\ \mathrm{m\,s^{-1}}$$, $$\theta_0 = 30^\circ$$, $$g = 9.8\ \mathrm{m\,s^{-2}}$$. Use $$\sin 30^\circ = 1/2$$ and $$\sin 60^\circ = \sqrt{3}/2$$.

(a) Maximum height. Using $$h_{\max} = \dfrac{(u\sin\theta_0)^2}{2g}$$:

$$h_{\max} = \dfrac{(28\sin 30^\circ)^2}{2(9.8)} = \dfrac{(14)^2}{19.6} = \dfrac{196}{19.6} = 10.0\ \mathrm{m}.$$

(b) Time to return to same level (time of flight).

$$T = \dfrac{2\,u\sin\theta_0}{g} = \dfrac{2(28)(1/2)}{9.8} = \dfrac{28}{9.8} = 2.86\ \mathrm{s}.$$

(c) Horizontal range.

$$R = \dfrac{u^2 \sin 2\theta_0}{g} = \dfrac{(28)^2 \sin 60^\circ}{9.8} = \dfrac{784 \times 0.866}{9.8} \approx 69.3\ \mathrm{m}.$$

The radius is $$R = 12\ \mathrm{cm} = 0.12\ \mathrm{m}$$. Seven complete revolutions in $$100\ \mathrm{s}$$ give a time period

$$T = \dfrac{100}{7}\ \mathrm{s} \approx 14.3\ \mathrm{s}.$$

(a) Angular speed:

$$\omega = \dfrac{2\pi}{T} = \dfrac{2\pi \times 7}{100} = \dfrac{14\pi}{100} \approx 0.44\ \mathrm{rad\,s^{-1}}.$$

Linear speed:

$$v = R\,\omega = 0.12 \times 0.44 \approx 0.053\ \mathrm{m\,s^{-1}} = 5.3\ \mathrm{cm\,s^{-1}}.$$

(b) The motion is uniform circular motion, so the speed is constant but the direction of motion changes continuously. The centripetal acceleration always points from the insect to the centre of the groove, so its direction is changing all the time. Therefore the acceleration vector is not a constant vector (only its magnitude is constant).

$$|\mathbf{a}| = \omega^2 R = (0.44)^2 \times 0.12 \approx 0.0232\ \mathrm{m\,s^{-2}} \approx 2.3\ \mathrm{cm\,s^{-2}}.$$

A scalar has only magnitude (specified by a single real number with a unit); a vector additionally has direction and obeys the laws of vector addition. Classifying each quantity:

Volume, mass, density and number of moles are intrinsically directionless. Speed and angular frequency are magnitudes of the corresponding vector quantities (velocity and angular velocity), and so are scalars. Acceleration, velocity, displacement and angular velocity have a definite direction in space.

Work is the dot product $$W = \vec{F}\cdot\vec{d}$$, which is a scalar.

Electric current is defined as the rate of flow of charge $$I = dq/dt$$; although charge has a sign and current has a sense, it does not obey the vector law of addition (currents along wires meeting at a junction add algebraically along the wire, not by the parallelogram law), so current is treated as a scalar.

All other quantities listed (force, angular momentum, linear momentum, electric field, average velocity, magnetic moment, relative velocity) have direction and obey vector addition; they are vectors.

Impulse is defined as $$\vec{J} = \vec{F}\,\Delta t = \Delta \vec{p}$$ — the change in linear momentum. Since it inherits its direction from the force (or the change of momentum), it has both magnitude and direction and obeys vector addition; hence impulse is a vector.

All the other quantities in the list (temperature, pressure, time, power, total path length, energy, gravitational potential, coefficient of friction, charge) are scalars.

Displacement is a vector from the initial point to the final point and is the same for all three girls because they begin at $$P$$ and end at $$Q$$. Since $$P$$ and $$Q$$ are diametrically opposite,

$$|\vec{PQ}| = 2R = 2 \times 200\ \mathrm{m} = 400\ \mathrm{m}.$$

The magnitude of the displacement for each girl is $$400\ \mathrm{m}$$, directed from $$P$$ to $$Q$$ along the diameter.

Only the girl whose path is the straight line $$PQ$$ (the diameter) actually travels a distance equal to the magnitude of the displacement. For the other two girls the path is curved, so the path length is greater than $$400\ \mathrm{m}$$.

From the figure, the cyclist's path is: along the radius $$OP$$ to the edge, then along the quarter-arc $$PQ$$ on the circumference, then along the radius $$QO$$ back to the centre. The radius is $$R = 1\ \mathrm{km}$$ and the total time is $$\Delta t = 10\ \mathrm{min} = \tfrac{1}{6}\ \mathrm{h}$$.

(a) Net displacement. The cyclist starts at $$O$$ and returns to $$O$$, so the displacement vector is the null vector:

$$\vec{s}_{\text{net}} = \vec{0}.$$

(b) Average velocity.

$$\vec{v}_{\text{avg}} = \dfrac{\vec{s}_{\text{net}}}{\Delta t} = \vec{0}.$$

(c) Average speed. Total path length = $$OP$$ + arc $$PQ$$ + $$QO$$ = $$R + \tfrac{1}{4}(2\pi R) + R = 2R + \tfrac{\pi R}{2}.$$

$$\text{Path} = 2(1) + \tfrac{\pi(1)}{2} = 2 + \tfrac{\pi}{2} \approx 3.571\ \mathrm{km}.$$

$$\text{Average speed} = \dfrac{\text{Path}}{\Delta t} = \dfrac{3.571}{1/6}\ \mathrm{km\,h^{-1}} \approx 21.4\ \mathrm{km\,h^{-1}}.$$

An exterior angle of $$60^\circ$$ at every turn corresponds to an interior angle of $$120^\circ$$, so successive $$500\ \mathrm{m}$$ legs lie along the sides of a regular hexagon $$ABCDEF$$ with side $$s = 500\ \mathrm{m}$$. After six turns the motorist returns to the starting vertex, since $$6 \times 60^\circ = 360^\circ$$.

After 3 turns. The motorist has covered three consecutive sides $$AB$$, $$BC$$, $$CD$$ and is at vertex $$D$$, which is diametrically opposite to $$A$$ in the hexagon. So

$$|\vec{AD}| = 2s = 1000\ \mathrm{m}, \qquad \text{path length} = 3s = 1500\ \mathrm{m}.$$

Ratio of magnitudes: $$1000/1500 = 2/3$$.

After 6 turns. The motorist has returned to $$A$$.

$$|\text{displacement}| = 0, \qquad \text{path length} = 6s = 3000\ \mathrm{m}.$$

After 8 turns. Eight turns = six turns (one full hexagon) + two more. After two further sides (covering $$AB$$ and $$BC$$) the motorist is at $$C$$. In a regular hexagon, $$AC$$ is found using the law of cosines on triangle $$ABC$$ with $$AB = BC = s$$ and angle $$ABC = 120^\circ$$:

$$AC^2 = s^2 + s^2 - 2s^2\cos 120^\circ = 2s^2 + s^2 = 3s^2 \quad\Rightarrow\quad AC = s\sqrt{3} = 500\sqrt{3}\ \mathrm{m} \approx 866\ \mathrm{m}.$$

Path length $$= 8s = 4000\ \mathrm{m}$$.

Ratio: $$500\sqrt{3}/4000 = \sqrt{3}/8 \approx 0.217$$.

Directions: at the 3rd turn the displacement is along $$AD$$ (the diameter opposite the start); at the 6th turn it is zero (back at the start); at the 8th turn it is along $$AC$$, the line joining the starting vertex to the vertex two sides away.

Total path length $$= 23\ \mathrm{km}$$. Displacement magnitude $$= 10\ \mathrm{km}$$ (straight line from station to hotel). Time $$\Delta t = 28\ \mathrm{min} = \tfrac{28}{60}\ \mathrm{h}$$.

(a) Average speed.

$$v_{\text{avg}} = \dfrac{\text{Path}}{\Delta t} = \dfrac{23}{28/60}\ \mathrm{km\,h^{-1}} = \dfrac{23 \times 60}{28} \approx 49.3\ \mathrm{km\,h^{-1}}.$$

(b) Magnitude of average velocity.

$$|\vec{v}_{\text{avg}}| = \dfrac{|\text{displacement}|}{\Delta t} = \dfrac{10}{28/60}\ \mathrm{km\,h^{-1}} = \dfrac{10 \times 60}{28} \approx 21.4\ \mathrm{km\,h^{-1}}.$$

The two are not equal because the path length ($$23\ \mathrm{km}$$) is greater than the magnitude of displacement ($$10\ \mathrm{km}$$). They would be equal only if the cab travelled along the straight road from the station to the hotel.

For a fixed speed $$u$$, the projectile of maximum range that just grazes the ceiling has its maximum height exactly equal to the ceiling height, $$H = 25\ \mathrm{m}$$. (Throwing at a larger angle would still go higher and hit the ceiling; smaller angle would yield a smaller range while leaving the ceiling untouched.)

From the projectile formula $$H = \dfrac{u^2 \sin^2\theta}{2g}$$, solve for $$\sin^2\theta$$ with $$u = 40\ \mathrm{m\,s^{-1}}$$ and $$g = 9.8\ \mathrm{m\,s^{-2}}$$:

$$\sin^2\theta = \dfrac{2gH}{u^2} = \dfrac{2(9.8)(25)}{(40)^2} = \dfrac{490}{1600} = 0.30625.$$

$$\sin\theta = 0.5534, \qquad \cos\theta = \sqrt{1 - 0.30625} = \sqrt{0.69375} = 0.8329.$$

$$\theta = \sin^{-1}(0.5534) \approx 33.6^\circ.$$

Horizontal range:

$$R = \dfrac{u^2 \sin 2\theta}{g} = \dfrac{u^2 \cdot 2\sin\theta\cos\theta}{g} = \dfrac{1600 \times 2(0.5534)(0.8329)}{9.8}.$$

$$R = \dfrac{1600 \times 0.9220}{9.8} \approx \dfrac{1475}{9.8} \approx 150.5\ \mathrm{m}.$$

The maximum horizontal distance is approximately $$150.5\ \mathrm{m}$$.

The maximum horizontal range for a given launch speed $$u$$ occurs at angle $$45^\circ$$ and equals $$R_{\max} = \dfrac{u^2}{g}$$. Given $$R_{\max} = 100\ \mathrm{m}$$,

$$u^2 = R_{\max}\,g = 100 \times 9.8 = 980\ \mathrm{m^2\,s^{-2}}.$$

When the ball is thrown straight up (vertically), all the initial speed is along the upward direction and the maximum height attained is

$$H = \dfrac{u^2}{2g} = \dfrac{980}{2 \times 9.8} = \dfrac{980}{19.6} = 50\ \mathrm{m}.$$

So the cricketer can throw the ball to a height of $$50\ \mathrm{m}$$ above the ground. (Equivalently, $$H_{\max} = R_{\max}/2$$.)

Length of string (radius of the circle): $$r = 80\ \mathrm{cm} = 0.80\ \mathrm{m}$$. Frequency:

$$\nu = \dfrac{14}{25}\ \mathrm{rev\,s^{-1}}.$$

Angular speed:

$$\omega = 2\pi\nu = 2\pi \times \dfrac{14}{25} = \dfrac{28\pi}{25}\ \mathrm{rad\,s^{-1}} \approx 3.52\ \mathrm{rad\,s^{-1}}.$$

For uniform circular motion the acceleration is centripetal, of magnitude

$$a = \omega^2 r = \left(\dfrac{28\pi}{25}\right)^2 \times 0.80.$$

Numerically, $$\omega^2 = (3.52)^2 \approx 12.38\ \mathrm{rad^2\,s^{-2}}$$, so

$$a \approx 12.38 \times 0.80 \approx 9.91\ \mathrm{m\,s^{-2}}.$$

The acceleration is directed along the string, from the stone toward the centre of the circle (centripetal direction). Since the speed is constant there is no tangential acceleration.

Convert the data to SI units.

$$r = 1.00\ \mathrm{km} = 1000\ \mathrm{m}, \qquad v = 900\ \mathrm{km/h} = \dfrac{900 \times 1000}{3600}\ \mathrm{m/s} = 250\ \mathrm{m/s}.$$

Centripetal acceleration:

$$a_c = \dfrac{v^2}{r} = \dfrac{(250)^2}{1000} = \dfrac{62500}{1000} = 62.5\ \mathrm{m\,s^{-2}}.$$

Comparing with $$g = 9.8\ \mathrm{m\,s^{-2}}$$:

$$\dfrac{a_c}{g} = \dfrac{62.5}{9.8} \approx 6.38.$$

The centripetal acceleration is about $$6.4$$ times the acceleration due to gravity.

(a) Differentiate $$\vec{r}$$ with respect to $$t$$:

$$\vec{v}(t) = \dfrac{d\vec{r}}{dt} = 3.0\,\hat{\mathbf{i}} - 4.0\,t\,\hat{\mathbf{j}}\ \mathrm{m\,s^{-1}}.$$

Differentiate again:

$$\vec{a}(t) = \dfrac{d\vec{v}}{dt} = -4.0\,\hat{\mathbf{j}}\ \mathrm{m\,s^{-2}}.$$

The acceleration is constant, of magnitude $$4.0\ \mathrm{m\,s^{-2}}$$, directed along the $$-y$$ axis.

(b) At $$t = 2.0\ \mathrm{s}$$:

$$\vec{v}(2) = 3.0\,\hat{\mathbf{i}} - 8.0\,\hat{\mathbf{j}}\ \mathrm{m\,s^{-1}}.$$

$$|\vec{v}(2)| = \sqrt{3.0^2 + (-8.0)^2} = \sqrt{9 + 64} = \sqrt{73} \approx 8.54\ \mathrm{m\,s^{-1}}.$$

The angle $$\theta$$ that $$\vec{v}(2)$$ makes with the $$+x$$ axis:

$$\tan\theta = \dfrac{v_y}{v_x} = \dfrac{-8.0}{3.0} \approx -2.667 \quad\Rightarrow\quad \theta \approx -69.4^\circ.$$

The velocity points about $$69.4^\circ$$ below the positive $$x$$-axis (in the fourth quadrant of the $$xy$$-plane).

With $$\vec{r}_0 = \vec{0}$$, $$\vec{u} = 10.0\,\hat{\mathbf{j}}\ \mathrm{m/s}$$, and $$\vec{a} = 8.0\,\hat{\mathbf{i}} + 2.0\,\hat{\mathbf{j}}\ \mathrm{m\,s^{-2}}$$, the equations of motion (constant acceleration) give component-wise:

$$x(t) = 0 + 0\cdot t + \tfrac{1}{2}(8.0)\,t^2 = 4.0\,t^2,$$
$$y(t) = 0 + 10.0\,t + \tfrac{1}{2}(2.0)\,t^2 = 10.0\,t + t^2.$$

(a) Set $$x(t) = 16$$ m:

$$4.0\,t^2 = 16 \quad\Rightarrow\quad t^2 = 4 \quad\Rightarrow\quad t = 2.0\ \mathrm{s}.$$

$$y(2) = 10.0(2) + (2)^2 = 20 + 4 = 24\ \mathrm{m}.$$

(b) Velocity at time $$t$$: $$\vec{v}(t) = \vec{u} + \vec{a}\,t = 8.0\,t\,\hat{\mathbf{i}} + (10.0 + 2.0\,t)\,\hat{\mathbf{j}}.$$

At $$t = 2.0\ \mathrm{s}$$:

$$\vec{v}(2) = 16.0\,\hat{\mathbf{i}} + 14.0\,\hat{\mathbf{j}}\ \mathrm{m/s}.$$

Speed:

$$|\vec{v}(2)| = \sqrt{16^2 + 14^2} = \sqrt{256 + 196} = \sqrt{452} \approx 21.3\ \mathrm{m/s}.$$

Magnitudes.

$$|\hat{\mathbf{i}} + \hat{\mathbf{j}}| = \sqrt{1^2 + 1^2} = \sqrt{2},$$
$$|\hat{\mathbf{i}} - \hat{\mathbf{j}}| = \sqrt{1^2 + (-1)^2} = \sqrt{2}.$$

Directions. Let $$\alpha$$ be the angle of $$\hat{\mathbf{i}} + \hat{\mathbf{j}}$$ with the $$x$$-axis:

$$\tan\alpha = \dfrac{1}{1} = 1 \quad\Rightarrow\quad \alpha = 45^\circ.$$

For $$\hat{\mathbf{i}} - \hat{\mathbf{j}}$$: $$\tan\beta = -1$$, so $$\beta = -45^\circ$$ (i.e. $$45^\circ$$ below the $$+x$$-axis).

Unit vectors along these directions.

$$\hat{\mathbf{n}}_1 = \dfrac{\hat{\mathbf{i}} + \hat{\mathbf{j}}}{\sqrt{2}}, \qquad \hat{\mathbf{n}}_2 = \dfrac{\hat{\mathbf{i}} - \hat{\mathbf{j}}}{\sqrt{2}}.$$

Components of $$\vec{A} = 2\hat{\mathbf{i}} + 3\hat{\mathbf{j}}$$ along these directions are obtained from the dot product:

$$A_1 = \vec{A}\cdot\hat{\mathbf{n}}_1 = \dfrac{(2)(1) + (3)(1)}{\sqrt{2}} = \dfrac{5}{\sqrt{2}} = \dfrac{5\sqrt{2}}{2} \approx 3.54.$$

$$A_2 = \vec{A}\cdot\hat{\mathbf{n}}_2 = \dfrac{(2)(1) + (3)(-1)}{\sqrt{2}} = \dfrac{-1}{\sqrt{2}} = -\dfrac{\sqrt{2}}{2} \approx -0.707.$$

The negative sign for $$A_2$$ indicates that the component is directed opposite to $$\hat{\mathbf{n}}_2$$ (i.e. along $$-\hat{\mathbf{n}}_2$$).

Let $$O$$ be the ground observation point and $$A$$, $$B$$ the two positions of the aircraft separated by $$\Delta t = 10.0\ \mathrm{s}$$. The aircraft flies horizontally at height $$h = 3400\ \mathrm{m}$$. Because the aircraft is on the same horizontal line, by symmetry the perpendicular from $$O$$ to $$AB$$ lands at the midpoint $$M$$ of $$AB$$, and $$OM = h$$. The given angle is $$\angle AOB = 30^\circ$$, so each half-angle is

$$\angle AOM = \angle MOB = 15^\circ.$$

In the right triangle $$OMA$$,

$$AM = OM\,\tan 15^\circ = 3400 \times \tan 15^\circ.$$

Using $$\tan 15^\circ \approx 0.2679$$:

$$AM \approx 3400 \times 0.2679 \approx 911\ \mathrm{m}.$$

The total horizontal distance covered between the two positions is

$$AB = 2\,AM \approx 2 \times 911 \approx 1822\ \mathrm{m}.$$

Hence the speed is

$$v = \dfrac{AB}{\Delta t} = \dfrac{1822}{10.0} \approx 182\ \mathrm{m\,s^{-1}}.$$

(Equivalently, $$v \approx 656\ \mathrm{km/h}$$.)