Module 1: Kinematics

Average speed is total distance over total time, not the average of the two speeds. For a distance $d$ each half: $v_{av} = \dfrac{2d}{\frac{d}{20} + \frac{d}{30}} = \dfrac{2}{\frac{1}{20} + \frac{1}{30}} = \dfrac{2}{\frac{5}{60}} = 24\text{ m/s}$. The slower half takes more time, so it pulls the average below $25\text{ m/s}$ (the tempting trap).

Add the perpendicular parts separately. North-south: $3 - 3 = 0$. East-west: $4\text{ m}$ east. So the resultant is just $4\text{ m}$ east. ($10\text{ m}$ is the total distance walked, not the displacement.)

Use $v^2 = u^2 + 2as$: $10^2 = 30^2 + 2a(100)$, so $100 = 900 + 200a$ and $a = \dfrac{-800}{200} = -4\text{ m/s}^2$. The magnitude is $4\text{ m/s}^2$. (You cannot use $a = \frac{\Delta v}{\Delta t}$ here because the time is not given.)

Distance fallen is $s = \tfrac{1}{2}gt^2$. From $t = 2$ to $t = 3$: $s = \tfrac{1}{2}(9.8)(3^2 - 2^2) = 4.9 \times (9 - 4) = 4.9 \times 5 = 24.5\text{ m}$. The distance in each successive second grows (Galileo's odd-number rule 1 : 3 : 5), so it is not simply three times the first second.

They meet when their displacements are equal: $15t = \tfrac{1}{2}(3)t^2$, so $15t = 1.5t^2$ giving $t = 10\text{ s}$. Distance $= 15 \times 10 = 150\text{ m}$. (At that moment the police car is doing $30\text{ m/s}$, twice the other car's speed, which is a useful check.)

Using $s = ut - \tfrac{1}{2}gt^2$: $14.7 = 19.6t - 4.9t^2$, so $4.9t^2 - 19.6t + 14.7 = 0$, i.e. $t^2 - 4t + 3 = 0$ and $(t - 1)(t - 3) = 0$. The ball is at $14.7\text{ m}$ twice: at $t = 1\text{ s}$ on the way up and $t = 3\text{ s}$ on the way down.

Phase 1 reaches $v = 1.0 \times 20 = 20\text{ m/s}$, covering $\tfrac{1}{2}(1)(20)^2 = 200\text{ m}$. Phase 2 slows from $20\text{ m/s}$ to rest at $2.0\text{ m/s}^2$, taking $10\text{ s}$ and covering $\tfrac{20^2}{2 \times 2} = 100\text{ m}$. Total: $300\text{ m}$ in $30\text{ s}$, so $v_{av} = \dfrac{300}{30} = 10\text{ m/s}$.

To land directly opposite, the swimmer must angle upstream so the upstream part of their $5.0\text{ m/s}$ cancels the $3.0\text{ m/s}$ current. The part left over, directed straight across, is $\sqrt{5^2 - 3^2} = \sqrt{16} = 4.0\text{ m/s}$. (Here the $5.0\text{ m/s}$ is the hypotenuse, not a leg, so we subtract under the root.)

$\vec{v}_{AB} = \vec{v}_A - \vec{v}_B$. With $\vec{v}_A = (0, 4)$ and $\vec{v}_B = (3, 0)$, $\vec{v}_{AB} = (-3, 4)$, whose magnitude is $\sqrt{(-3)^2 + 4^2} = 5.0\text{ m/s}$. On the diagram this is the arrow from the head of $\vec{v}_B$ to the head of $\vec{v}_A$.

Acceleration is the gradient of a velocity-time graph. The curve gets steeper as time goes on, so the gradient is growing: the acceleration is increasing. (A straight line would mean constant acceleration; this curve means the motion is non-uniform.)

Reaction phase (constant speed): $20 \times 0.50 = 10\text{ m}$. Braking phase: $s = \dfrac{v^2}{2a} = \dfrac{20^2}{2 \times 5} = 40\text{ m}$. Total stopping distance $= 10 + 40 = 50\text{ m}$. (Forgetting the reaction distance gives the incorrect $40\text{ m}$.)

The distance travelled in the $n$th second is $u + a(n - \tfrac{1}{2})$. So $u + 1.5a = 24$ and $u + 3.5a = 32$. Subtracting: $2a = 8$, giving $a = 4\text{ m/s}^2$. (The gap of $8\text{ m}$ is spread over 2 seconds of extra acceleration.)

Displacement is the signed area. The velocity is zero at $t = 4\text{ s}$ (where the particle turns around). Forward area (0 to $4\text{ s}$): $\tfrac{1}{2}(4)(8) = +16\text{ m}$. Backward area (4 to $6\text{ s}$): $\tfrac{1}{2}(2)(4) = -4\text{ m}$. Net $= 16 - 4 = +12\text{ m}$. (The total distance travelled is $16 + 4 = 20\text{ m}$, a common trap.)

Magnitude: $\sqrt{40^2 + 30^2} = \sqrt{2500} = 50\text{ m/s}$. Direction measured up from east: $\tan\theta = \dfrac{30}{40}$, so $\theta = \tan^{-1}(0.75) \approx 37^\circ$ north of east. (Measuring $37^\circ$ from north instead, i.e. "east of north", points the wrong way.)

In the cyclist's frame the rain gains a horizontal velocity of $6.0\text{ m/s}$ backward (equal and opposite to the cyclist's motion), while still falling at $8.0\text{ m/s}$. The angle from the vertical is $\tan\theta = \dfrac{6.0}{8.0} = 0.75$, so $\theta \approx 37^\circ$. This is why you tilt an umbrella forward when moving.

Both balls accelerate downward at the same $g$, so the gap between them closes at the constant relative speed of $20\text{ m/s}$ (the thrown ball's initial speed). Time to close the $20\text{ m}$ gap: $t = \dfrac{20}{20} = 1.0\text{ s}$. (The $g$ terms cancel when you subtract the two position equations, so the meeting time does not depend on $g$.)

Average velocity between markers $= \dfrac{50}{2.0} = 25\text{ m/s}$, which for uniform acceleration equals $\dfrac{u + v}{2}$. So $\dfrac{u + 30}{2} = 25$, giving $u = 20\text{ m/s}$. Then $a = \dfrac{v - u}{t} = \dfrac{30 - 20}{2.0} = 5\text{ m/s}^2$.

The two legs are perpendicular, so the straight-line distance is the hypotenuse: $\sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13\text{ km}$. ($17\text{ km}$ is the total distance walked; the straight-line displacement is shorter.)

At the top $v = 0$, so $v^2 = u^2 - 2gs$ gives $0 = u^2 - 2(9.8)(5.0)$, hence $u = \sqrt{98} \approx 9.9\text{ m/s}$. (49 comes from forgetting the square root; the units alone rule it out.)

Speed is the magnitude of the gradient. From 0 to $4\text{ s}$ the gradient is $\frac{20}{4} = 5\text{ m/s}$. From 6 to $8\text{ s}$ the graph drops from $20\text{ m}$ to $0\text{ m}$, a gradient of $\frac{-20}{2} = -10\text{ m/s}$, so the speed is $10\text{ m/s}$. The steepest section, and therefore the fastest motion, is $6\text{ s}$ to $8\text{ s}$, even though the object is moving backward there.

By symmetry the time up is half the total: $2.0\text{ s}$. The height is the distance fallen from rest at the top in that time: $s = \tfrac{1}{2}(9.8)(2.0)^2 = 19.6\text{ m}$. (Using the full $4.0\text{ s}$ gives four times too much.)

Total distance = $120\text{ km}$. Time out $= \frac{60}{60} = 1$ h; time back $= \frac{60}{40} = 1.5$ h; total 2.5 h. Average speed $= \dfrac{120}{2.5} = 48\text{ km/h}$. (Averaging the two speeds to get $50\text{ km/h}$ ignores that more time is spent going slowly.)

The ground velocity is the vector sum of the aircraft's velocity (200 north) and the wind (150 east), which are perpendicular: $\sqrt{200^2 + 150^2} = \sqrt{62500} = 250\text{ km/h}$. (Simply adding to $350\text{ km/h}$ would only be right if the wind blew from directly behind.)

Free fall means the only force is gravity, giving a constant acceleration of $9.8\text{ m/s}^2$ downward, independent of mass. Constant acceleration means the velocity increases steadily (by $9.8\text{ m/s}$ each second). Acceleration and velocity are different quantities: a constant acceleration still produces a changing velocity.

Find each area of the velocity-time graph. Speed-up: $\tfrac{1}{2}(20)(8) = 80\text{ m}$. Cruise: $20 \times 12 = 240\text{ m}$. Brake: $\tfrac{1}{2}(20)(5) = 50\text{ m}$. Total $= 370\text{ m}$ over $8 + 12 + 5 = 25\text{ s}$, so $v_{av} = \dfrac{370}{25} = 14.8\text{ m/s}$.