Use \(\text{speed}=\dfrac{\text{distance}}{\text{time}}\).

So \(\text{speed}=\dfrac{120}{15}=8\text{ m s}^{-1}\).

Use \(\text{distance}=\text{speed}\times \text{time}\).

So distance \(=9\times 7=63\text{ m}\).

Use \(\text{time}=\dfrac{\text{distance}}{\text{speed}}\).

So \(t=\dfrac{150}{25}=6\text{ s}\).

To use \(t=\dfrac{s}{v}\) directly, we model the motion with a constant speed over a fixed distance.

A suitable assumption is that the prey does not move and the hunter runs straight at constant speed from the start.

First convert the distance to metres:

150 million km \(=150\times 10^6\times 1000=1.50\times 10^{11}\) m.

Then

\(t=\dfrac{1.50\times 10^{11}}{3.00\times 10^8}=500\text{ s}\).

500 s \(=8\) min 20 s.

Convert speed to m s^-1:

\(1223.657\times\dfrac{1000}{3600}\approx 339.905\text{ m s}^{-1}\).

Then

\(t=\dfrac{1000}{339.905}\approx 2.94\text{ s}\).

Total distance \(=5\times 7+7\times 13=35+91=126\text{ m}\).

Total time \(=20\text{ s}\).

Average speed \(=\dfrac{126}{20}=6.3\text{ m s}^{-1}\).

Forward displacement \(=6\times 10=60\text{ m}\).

Backward displacement \(=3\times 5=15\text{ m}\).

Net displacement \(=60-15=45\text{ m}\) forwards.

From the previous motion, net displacement \(=45\text{ m}\) forwards.

Total time \(=15\text{ s}\).

Average velocity \(=\dfrac{45}{15}=3\text{ m s}^{-1}\) forwards.

Total distance travelled \(=60+15=75\text{ m}\).

Total time \(=15\text{ s}\).

Average speed \(=\dfrac{75}{15}=5\text{ m s}^{-1}\).

Distance in first 5 s \(=11\times 5=55\text{ m}\).

Total distance needed for an overall average of 12 m s^-1 over 15 s is \(12\times 15=180\text{ m}\).

So distance needed in the next 10 s is \(180-55=125\text{ m}\).

Required speed \(=\dfrac{125}{10}=12.5\text{ m s}^{-1}\).

Time through wood \(=\dfrac{33}{3300}=0.01\text{ s}\).

Time through air \(=\dfrac{33}{330}=0.10\text{ s}\).

Difference \(=0.10-0.01=0.09\text{ s}\).

Total time \(=3\text{ min }10\text{ s}=190\text{ s}\).

Let sprinting time be \(t\) s. Then jogging time is \(190-t\) s.

Total distance:

\(7t+4(190-t)=1000\).

So \(7t+760-4t=1000\), hence \(3t=240\).

Therefore \(t=80\text{ s}\).

Let the race length be \(d\) m.

Times taken are \(\dfrac{d}{44}\) and \(\dfrac{d}{45}\).

The slower time minus the faster time is 8 s:

\(\dfrac{d}{44}-\dfrac{d}{45}=8\).

\(d\left(\dfrac{1}{44}-\dfrac{1}{45}\right)=8\).

\(d\left(\dfrac{1}{1980}\right)=8\), so \(d=8\times 1980=15840\text{ m}\).

In the first 0.2 s, only puck 1 moves:

distance \(=1.3\times 0.2=0.26\text{ m}\).

So remaining separation is \(2-0.26=1.74\text{ m}\).

After that, they move towards each other with closing speed \(1.3+1.7=3.0\text{ m s}^{-1}\).

Extra time to collide \(=\dfrac{1.74}{3.0}=0.58\text{ s}\).

Total time puck 1 has moved \(=0.2+0.58=0.78\text{ s}\).

Distance moved by puck 1 \(=1.3\times 0.78=1.014\text{ m}\approx1.01\text{ m}\).

If each part takes time \(t\), then the distances are \(v_1t\) and \(v_2t\).

Total distance \(=(v_1+v_2)t\).

Total time \(=2t\).

So average speed \(=\dfrac{(v_1+v_2)t}{2t}=\dfrac{v_1+v_2}{2}\).

The stage speeds are \(\dfrac{s}{t_1}\) and \(\dfrac{s}{t_2}\).

Overall average speed is \(\dfrac{2s}{t_1+t_2}\).

This equals the arithmetic mean only when \(t_1=t_2\).

So the condition is that the two times are equal.

If each one-way distance is \(s\), total distance is \(2s\).

Total time is \(\dfrac{s}{v_1}+\dfrac{s}{v_2}\).

So

\(v=\dfrac{2s}{\frac{s}{v_1}+\frac{s}{v_2}}=\dfrac{2v_1v_2}{v_1+v_2}\).

The overall average speed is \(\dfrac{2v_1v_2}{v_1+v_2}\).

If this were equal to \(2v_1\), then

\(\dfrac{2v_1v_2}{v_1+v_2}=2v_1\).

For a real journey with \(v_1>0\), dividing by \(2v_1\) gives

\(\dfrac{v_2}{v_1+v_2}=1\), which would imply \(v_1=0\).

That is impossible for a moving outward trip, so the overall average speed cannot be exactly twice \(v_1\).

The model assumes each stage has a constant speed and that the change between the two speeds happens immediately.