200m race

When #1 finished 200 m, #2 had finished 160 m, and #3 had finished 120 m.

Here is a table of values that shows where the last two runners from the start line are at different key points in time.

Because it is given that the two runners run at a constant rate, and that they both start at the start line (nobody has a head start), the distances of the two runners must be proportionate. Using this, let's solve for x.

$\dfrac{x}{200 \text{ m}}=\dfrac{120 \text{ m}}{160 \text{ m}}$ $x=\dfrac{200 \text{ m} \times 120}{160}$ $x=150 \text{ m}$

Thus, when #2 crosses the finish line, #3 is 150 m away from the start line, and $\boxed{\text{50 m}}$ away from the finish line.

Let the time taken by the first runner to end the race be $t$ seconds

Speed of first runner $=\frac{200}{t}$

Distance covered by $2_{nd}$ runner in $t$ seconds $=200-40=160m$

Speed of $2_{nd}$ runner $=\frac{160}{t}$

Distance covered by $3_{rd}$ runner in $t$ seconds $=200-80=120m$

Speed of $3_{rd}$ runner $=\frac{120}{t}$

Total time taken by $2_{nd}$ runner to finish the race $=\frac{Distance}{speed}=20\cancel{0}\times\frac{t}{16\cancel{0}}=\frac{5}{4}t$

Distance covered by $3_{rd}$ runner during this period of time $=speed\times time=\frac{120}{\cancel{t}}\times\frac{5}{4}\cancel{t}={150m}$

Distance between $2_{nd}$ and $3_{rd}$ runner $=200-150=\boxed{50m}$

Assume the racer who was placed first took $8s$ to finish the $200m$ race

Calculating the velocities of the second and third racer, we can see that:

$\text{Racer two's velocity is =} \frac{160m}{8s} = 20ms^{-1}$

$\text{Racer three's velocity is =} \frac{120m}{8s} = 15ms^{-1}$

For $\text{Racer Two}$ to finish the race, he should take $10s$ , as:

$20ms^{-1} × 10s = \boxed{200m}$

$\text{Racer Three}$ would have then run a distance of:

$15ms^{-1} × 10s = \boxed{150m}$

Therefore, the difference between the running distances of $\text{racer three}$ and $\text{racer two}$ when racer two finishes the race is $\boxed{50m}$