200m race

Algebra Level 2

In a 200 m race, the runner who finished first was ahead of the 2nd runner by 40 m and ahead of the 3rd runner by 80 m.

What was the distance (in meters) between the 2nd runner and the 3rd runner when the 2nd runner reached the finish line ?

It is assumed that the three runners maintain their (constant) speeds throughout the race.


The answer is 50.

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3 solutions

Ved Pradhan
Jun 18, 2020

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.

#2 #3
0 m 0 m
160 m 120 m
200 m x x

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.

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

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

Zakir Husain
Jun 18, 2020

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

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

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

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

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

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

Total time taken by 2 n d 2_{nd} runner to finish the race = D i s t a n c e s p e e d = 20 0 × t 16 0 = 5 4 t =\frac{Distance}{speed}=20\cancel{0}\times\frac{t}{16\cancel{0}}=\frac{5}{4}t

Distance covered by 3 r d 3_{rd} runner during this period of time = s p e e d × t i m e = 120 t × 5 4 t = 150 m =speed\times time=\frac{120}{\cancel{t}}\times\frac{5}{4}\cancel{t}={150m}

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

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

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

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

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

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

20 m s 1 × 10 s = 200 m 20ms^{-1} × 10s = \boxed{200m}

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

15 m s 1 × 10 s = 150 m 15ms^{-1} × 10s = \boxed{150m}

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

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