Question

Surya is competing in a triathlon that includes swimming, bicycling and running. For the first two segments of the race, Surya swam at an average rate of 4 kilometers per hour and bicycled at an average rate of 40 kilometers per hour. It took 81.75 minutes for Surya to swim and cycle a combined total of 43.29 kilometers, completing the first two segments of the race. How long, in minutes, did it take him to complete the swimming portion of the race? Express your answer to the nearest hundredth.
If an athlete who completes this triathlon swims, bicycles and runs a combined total of 52.95 kilometers, the distance the athlete swam is what percent of the total race distance? Express your answer as a percent to the nearest hundredth.

At what average rate, in kilometers per hour, must Surya run the third, and final, portion of the race if he wishes to complete the entire race in no more than 140 minutes? Express your answer to the nearest hundredth.

Answer 1

1) 18.68 minutes

2) 2.34 %

3) 10.01 km/h

Step-by-step explanation:

1)

In the first part (swimming part), Surya's velocity is

$v_1 = \frac{d_1}{t_1}= 4 km/h$

where

$d_1$ is the length of the swimming part

$t_1$ is the time taken to complete the swimming part

We can rewrite this equation as

$d_1 = v_1 t_1\\d_1 = 4t_1$ (1)

In the second part (biking part), Surya's velocity is

$v_2 = \frac{d_2}{t_2}=40 km/h$

where

$d_2$ is the length of the biking part

$t_2$ is the time taken to complete the biking part

We can rewrithe this equation as

$d_2=v_2 t_2\\d_2 = 40 t_2$ (2)

The total length of the two segments of the race is:

$d=d_1+d_2=43.29 km$ (3)

And the total time taken to complete them is

$t=t_1 + t_2 = 81.75 min \cdot \frac{1}{60}=1.36 h$  (4)

Substituting (1) and (2) into (3),

$4t_1+40t_2 = 43.29$ (5)

From (4), we find:

$t_2=1.36-t_1$

And substituting into (5), we can find t1, the time taken to complete the swimming part of the race:

$4t_1 +40(1.36-t_1) = 43.29\\4t_1 +54.5 -40t_1 = 43.29\\36t_1 = 11.21\\t_1=\frac{11.21}{36}=0.31 h \cdot 60 = 18.68 min$

2)

In this part, we are told that the total distance covered by an athlete completing the swim + bycicle + run is

d = 52.95 km

This distance can be written as

$d=d_1+d_2+d_3$

where

$d_1$ is the length of the swimming part

$d_2$ is the length of the biking part

$d_3$ is the length of the running part

From the previous part, we can write the length of the swimming part as

$d_1 = v_1 t_1$

where:

$v_1=4 km/h$ is the velocity in the swimming part

$t_1=0.31 h$ is the time taken for the swimming part

So we have

$d_1 = (4)(0.31)=1.24 km$

So, the percent of the swimming part over the total is:

$\frac{d_1}{d}\cdot 100 = \frac{1.24}{52.95}\cdot 100 =2.34\%$

3)

The total time for the entire race must be

$t=140 min \cdot \frac{1}{60}=2.33 h$

And this total time can be written as

$t=t_1+t_2+t_3$

where

$t_1$ is the time taken to complete the swimming part

$t_2$ is the time taken to complete the biking part

$t_3$ is the time taken to complete the running part

From part 1) we know that

$t_1=0.31 h$

and

$t_2=1.36-t_1=1.36-0.31=1.05 h$

So the time to complete the running part must be

$t_3=t-t_1-t_2=2.33-0.31-1.05=0.97 h$

The distance of the last segment is

$d_3=d-d_1 -d_2$

where $d_2$ is given by

$d_2=v_2 t_2 = (40)(1.05)=42 km$

So,

$d_3=52.95-1.24-42=9.71 km$

So the average velocity in the 3rd segment is

$v_3=\frac{d_3}{t_3}=\frac{9.71}{0.97}=10.01 km/h$