Answer:
![v_s=0\ km/h](https://img.qammunity.org/2020/formulas/sat/middle-school/okc8598tnlz6ne5dvz9go9bvupc4gd50ta.png)
![v_s=3\ km/h](https://img.qammunity.org/2020/formulas/sat/middle-school/un7ygv491q8f39wjihoqpua2n8l59zm9j1.png)
Step-by-step explanation:
Relative Speed
A boat traveling in still waters has a speed v_b. If now an opposite stream appears, then the apparent speed of the boat will be less than before because the relative speed is the subtraction of both speeds. Thus we say
![v_(t1)=v_b-v_s](https://img.qammunity.org/2020/formulas/sat/middle-school/zg7s7m3il5k47qlnmo6m5jif3nr2aop18c.png)
Conversely, if the stream goes with the boat, both speeds are added.
![v_(t2)=v_b+v_s](https://img.qammunity.org/2020/formulas/sat/middle-school/96twohgvf7sgkzf3ozxp2shhfkckrgilfe.png)
where
it the total or real speed of the boat respect to the ground,
is the speed of the boat in still water, and
is the speed of the stream.
The boat goes 36 km with stream. The time it took for doing so is:
![\displaystyle t_1=(36)/(v_b+v_s)](https://img.qammunity.org/2020/formulas/sat/middle-school/tzxxaqmostagv6wtpy1a1fztuavxt6w5z7.png)
The boat goes 24 km against the stream in
![\displaystyle t_2=(24)/(v_b-v_s)](https://img.qammunity.org/2020/formulas/sat/middle-school/4fvmgmc29d6bz5bx3k14df2ba4rwf3mn4z.png)
The problem states that
![t_1+t_2=4\ hours](https://img.qammunity.org/2020/formulas/sat/middle-school/d3dhgd9kk41v6pga36vi7j5zjvsnqh9yd8.png)
![\displaystyle (36)/(v_b+v_s)+(24)/(v_b-v_s)=4](https://img.qammunity.org/2020/formulas/sat/middle-school/v1xzph63ox30ys4fotrqyta79dp4abm6ie.png)
Knowing
![v_b=15](https://img.qammunity.org/2020/formulas/sat/middle-school/dovnnc9zguq77v8f7d72tcjoe3ao33x7i5.png)
![\displaystyle (36)/(15+v_s)+(24)/(15-v_s)=4](https://img.qammunity.org/2020/formulas/sat/middle-school/ttihjm7y94ncuvur16oh4s5bmau6uaxqht.png)
Operating
![\displaystyle 36(15-v_s)+24(15+v_s)=4(15+v_s)(15-v_s)](https://img.qammunity.org/2020/formulas/sat/middle-school/ul1bhf9w2t522xjt60vxl112dj147ojolb.png)
Rearranging and simplifying
![225-3v_s=225-v_s^2](https://img.qammunity.org/2020/formulas/sat/middle-school/mg40vkz8l84nnw7re63lfi6gcbrvjh8sz0.png)
Factoring
![v_s(v_s-3)=0)](https://img.qammunity.org/2020/formulas/sat/middle-school/9slnghx20j2bense8rau14r262uk1n0yjl.png)
We get
![v_s=0,\ v_s=3](https://img.qammunity.org/2020/formulas/sat/middle-school/6gllulow0764qj4gavvpbtsmqbx56j4ipu.png)
Both solutions are possible, i.e.
1. The stream does not exist and the boat travels the total distance of 60 Km in 4 hours
2. The stream has a speed of 3 Km/h and the boat travels 36 km in 2 hours and 24 km in 2 hours