Question

PLEASE ANSWER! IT'S SOO HARD FOR ME!!

a. 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.




b. 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.




c. 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

a) 18.68 minutes

b) 2.34%

c) 9.96 km/h

Explanation:

a)

The velocity of Surya in the 1st segment (swimming part) is

`v_1=4 km/h`

So, the distance covered by Surya in the swimming part is

`d_1=4t_1`

where
`t_1` is the time taken to cover this part.

The velocity of Surya in the 2nd segment (biking part) is

`v_2=40 km/h`

So the distance covered in the biking part is

`d_2=40t_2`

where
`t_2` is the time taken to cover this part.

We know that the total distance covered in these 2 segments is

`d=d_1+d_2=43.29`

So it can be rewritten as

`4t1+40t_2=43.29` (1)

The total time taken to cover these two parts is

`t=t_1+t_2=81.75 min =1.36 h`

So we have

`t_2=1.36-t_1`

And substituting in (1) we can find t1:

`4t_1 + 40(1.36-t_1)=43.29\\36t_1=11.21\\t_1=(11.21)/(36)=0.31 h = 18.68 min`

b)

The total distance covered in all 3 segments is

`d=d_1+d_2+d_3=52.95 km`

where

where

d1 is the length of the swimming part

d2 is the length of the biking part

d3 is the length of the running part

We know that the distance covered in part 1 (swimming part) is

`d_1=v_1 t_1 = (4)(0.31)=1.24 km`

Therefore we can find the percent that the swimming part correspond to the total race as:

`(d_1)/(d)=(1.24)/(52.95)=0.0234 = 2.34\%`

c)

Here the total time taken by Surya for the race must be

`t=140 min = 2.33 h`

And this is the sum of the times of the 3 segments:

`t=t_1+t_2+t_3`

where

t1 is the time taken to complete the swimming part

t2 is the time taken to complete the biking part

t3 is the time taken to complete the running part

We know already that

`t_1=0.31 h`

and

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

So the time for the 3rd section must be

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

The distance to cover in the last part is

`d_3=d-d_1-d_2=52.95-43.29=9.66 km`

So the average velocity of the last segment must be:

`v_3=(d_3)/(t_3)=(9.66)/(0.97)=9.96 km/h`