Motion on a Plane

A vector needs a direction as well as a size. Velocity is speed in a stated direction, so it is a vector. Speed, distance, time and temperature have magnitude only, so they are scalars.

The two legs are perpendicular, so the straight-line displacement is the hypotenuse: $\sqrt{3^2 + 4^2} = \sqrt{25} = 5\text{ m}$. ($7\text{ m}$ is the distance walked; you cannot just add vectors that point in different directions.)

A vector has no component in a direction perpendicular to itself. A purely eastward velocity has $0\text{ m/s}$ in the north direction. ($7.1\text{ m/s}$ would be the trap of assuming a $45^\circ$ vector.)

Measured from the east direction toward north, $\tan\theta = \dfrac{\text{north}}{\text{east}} = \dfrac{4}{3}$, so $\theta = \tan^{-1}(1.33) \approx 53^\circ$. (Using $\tan^{-1}(3/4) = 37^\circ$ measures the angle from north instead, a common mix-up.)

The horizontal component uses cosine: $F_x = F\cos\theta = 50 \times 0.87 \approx 43\text{ N}$. ($25\text{ N}$ is the vertical component, using sine.)

The vertical component uses sine: $F_y = F\sin\theta = 50 \times 0.5 = 25\text{ N}$. Horizontal uses cosine, vertical uses sine, provided the angle is measured from the horizontal.

At $60^\circ$: horizontal $= F\cos 60^\circ = 0.5F$, vertical $= F\sin 60^\circ = 0.87F$. The steeper the angle, the larger the vertical share, so the vertical component is bigger. (This ratio is the same whatever the size of $F$.)

The two velocities are perpendicular, so add them with Pythagoras: $\sqrt{6^2 + 8^2} = \sqrt{100} = 10\text{ m/s}$. (Perpendicular vectors are never simply added to $14$.)

Take east as positive: $v_{AB} = v_A - v_B = (+20) - (-15) = 35\text{ m/s east}$. Two cars approaching close on each other at the sum of their speeds, which is why head-on approaches feel so fast.

$v_{AB} = v_A - v_B = 30 - 20 = 10\text{ m/s east}$. Relative to B, car A creeps forward at only $10\text{ m/s}$, which is why overtaking on a highway takes so long.

Add components separately. East-west: $+4 - 4 = 0$. North-south: $+3$. The east and west legs cancel, leaving $3\text{ km}$ due north. ($11\text{ km}$ is the total distance walked.)

Magnitude: $\sqrt{4^2 + 3^2} = 5\text{ m/s}$. Direction, measured from north toward east: $\tan\theta = \dfrac{3}{4}$, so $\theta = \tan^{-1}(0.75) \approx 37^\circ$ east of north.

Only the across-river velocity ($4\text{ m/s}$) moves the boat toward the far bank, and the current does not change it: $t = \dfrac{60}{4} = 15\text{ s}$. The current only sweeps the boat sideways, not across. (Using $5\text{ m/s}$ gives the $12\text{ s}$ trap.)

The downstream drift is the current speed times the crossing time: $3 \times 15 = 45\text{ m}$.

Resolve each leg. East-west: $20\cos 30^\circ - 20\cos 30^\circ = 0$ (they cancel). North: $20\sin 30^\circ + 20\sin 30^\circ = 10 + 10 = 20\text{ m}$. The resultant is $20\text{ m}$ due north.

Velocity is a vector, so $\Delta \vec{v} = \vec{v} - \vec{u}$ is a vector subtraction, not a subtraction of speeds. With the two velocities perpendicular, $|\Delta v| = \sqrt{20^2 + 20^2} = \sqrt{800} \approx 28\text{ m/s}$. (The speed is unchanged, but the velocity certainly changed, so the answer is not zero.)

The airspeed and the wind are perpendicular: $\sqrt{200^2 + 150^2} = \sqrt{62500} = 250\text{ km/h}$. (Adding to $350$ would only be right if the wind blew from directly behind.)

The upstream part of the boat's velocity must exactly cancel the current: $5\sin\theta = 3$, so $\sin\theta = 0.6$ and $\theta = \sin^{-1}(0.6) \approx 37^\circ$ upstream of straight across.

The across-river component is what is left after the upstream part cancels the current: $\sqrt{5^2 - 3^2} = \sqrt{16} = 4\text{ m/s}$ (equivalently $5\cos 37^\circ = 4\text{ m/s}$). Aiming upstream costs some speed, so it is slower than $5\text{ m/s}$.

To cancel the eastward wind, the aircraft must aim partly west: $\sin\theta = \dfrac{70}{250} = 0.28$, so $\theta = \sin^{-1}(0.28) \approx 16^\circ$ west of north. You aim into the wind so the wind pushes you back onto the north line.

$R^2 = 6^2 + 8^2 + 2(6)(8)\cos 60^\circ = 36 + 64 + 96(0.5) = 148$, so $R = \sqrt{148} \approx 12.2\text{ N}$. ($14$ assumes they are parallel; $10$ assumes they are perpendicular. Only when the angle is $90^\circ$ does plain Pythagoras apply.)

$R^2 = 10^2 + 10^2 + 2(10)(10)\cos\theta = 200 + 200\cos\theta$. Setting $R = 10$: $100 = 200 + 200\cos\theta$, so $\cos\theta = -0.5$ and $\theta = 120^\circ$. The three vectors form an equilateral-triangle arrangement.

The rain's velocity relative to you is $\vec{v}_{\text{rain}} - \vec{v}_{\text{you}}$: down $4$ minus east $3$, i.e. down $4$ and west $3$. Its magnitude is $\sqrt{3^2 + 4^2} = 5\text{ m/s}$, and it slants toward you from in front, which is why you tilt an umbrella forward when walking.

Resolve the north-east leg: $100 \times 0.71 = 71\text{ km}$ east and $71\text{ km}$ north. Totals: east $= 100 + 71 = 171\text{ km}$, north $= 71\text{ km}$. Then $R = \sqrt{171^2 + 71^2} \approx \sqrt{34140} \approx 185\text{ km}$. (Simply adding gives $200$, which ignores the change of direction.)

$\vec{v}_{AB} = \vec{v}_A - \vec{v}_B$ = east $30$ minus north $40$ = east $30$ and south $40$. The magnitude is $\sqrt{30^2 + 40^2} = \sqrt{2500} = 50\text{ m/s}$. Subtracting the speeds to $10$ ignores that the velocities are perpendicular.