Respuesta :
a) 18.68 minutes
b) 2.34%
c) 9.96 km/h
Step-by-step explanation:
a)
The velocity of Surya in the 1st segment (swimming part) is
[tex]v_1=4 km/h[/tex]
So, the distance covered by Surya in the swimming part is
[tex]d_1=4t_1[/tex]
where [tex]t_1[/tex] is the time taken to cover this part.
The velocity of Surya in the 2nd segment (biking part) is
[tex]v_2=40 km/h[/tex]
So the distance covered in the biking part is
[tex]d_2=40t_2[/tex]
where [tex]t_2[/tex] is the time taken to cover this part.
We know that the total distance covered in these 2 segments is
[tex]d=d_1+d_2=43.29[/tex]
So it can be rewritten as
[tex]4t1+40t_2=43.29[/tex] (1)
The total time taken to cover these two parts is
[tex]t=t_1+t_2=81.75 min =1.36 h[/tex]
So we have
[tex]t_2=1.36-t_1[/tex]
And substituting in (1) we can find t1:
[tex]4t_1 + 40(1.36-t_1)=43.29\\36t_1=11.21\\t_1=\frac{11.21}{36}=0.31 h = 18.68 min[/tex]
b)
The total distance covered in all 3 segments is
[tex]d=d_1+d_2+d_3=52.95 km[/tex]
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
[tex]d_1=v_1 t_1 = (4)(0.31)=1.24 km[/tex]
Therefore we can find the percent that the swimming part correspond to the total race as:
[tex]\frac{d_1}{d}=\frac{1.24}{52.95}=0.0234 = 2.34\%[/tex]
c)
Here the total time taken by Surya for the race must be
[tex]t=140 min = 2.33 h[/tex]
And this is the sum of the times of the 3 segments:
[tex]t=t_1+t_2+t_3[/tex]
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
[tex]t_1=0.31 h[/tex]
and
[tex]t_2=1.36-t_1=1.36-0.31=1.05 h[/tex]
So the time for the 3rd section must be
[tex]t_3=t-t_1-t_2=2.33-0.31-1.05=0.97h[/tex]
The distance to cover in the last part is
[tex]d_3=d-d_1-d_2=52.95-43.29=9.66 km[/tex]
So the average velocity of the last segment must be:
[tex]v_3=\frac{d_3}{t_3}=\frac{9.66}{0.97}=9.96 km/h[/tex]