Answer:
t ’= [tex]\frac{1450}{0.6499 + 2 v_r}[/tex], v_r = 1 m/s t ’= 547.19 s
Explanation:
This is a relative velocity exercise in a dimesion, since the river and the boat are going in the same direction.
By the time the boat goes up the river
v_b - v_r = d / t
By the time the boat goes down the river
v_b + v_r = d '/ t'
let's subtract the equations
2 v_r = d ’/ t’ - d / t
d ’/ t’ = 2v_r + d / t
[tex]t' = \frac{d'}{ \frac{d}{t}+ 2 v_r }[/tex]
In the exercise they tell us
d = 1.22 +1.45 = 2.67 km= 2.67 10³ m
d ’= 1.45 km= 1.45 1.³ m
at time t = 69.1 min (60 s / 1min) = 4146 s
the speed of river is v_r
t ’= [tex]\frac{1.45 \ 10^3}{ \frac{ 2670}{4146} \ + 2 \ v_r}[/tex]
t ’= [tex]\frac{1450}{0.6499 + 2 v_r}[/tex]
In order to complete the calculation, we must assume a river speed
v_r = 1 m / s
let's calculate
t ’= [tex]\frac{ 1450}{ 0.6499 + 2 \ 1}[/tex]
t ’= 547.19 s
