Answer:
Part a)
![v_(bw) = 2.1 mph](https://img.qammunity.org/2020/formulas/physics/high-school/ebz27dts1gtl4wv17q0vhx3hzedj32lquc.png)
Part b)
![t = 0.6 h](https://img.qammunity.org/2020/formulas/physics/high-school/t89dw4jas7fafb094uf1xet96zv3cvtgdw.png)
Part c)
![x = 0.4 mile](https://img.qammunity.org/2020/formulas/physics/high-school/qma7zo5bu04nhta2k80fa4dnepq6oc7otp.png)
Part d)
![\theta = 46.6 degree](https://img.qammunity.org/2020/formulas/physics/high-school/ihsl1u31zd86k2zkoakq66x8p2a6njxkpk.png)
Step-by-step explanation:
Part a)
Velocity of the boat with respect to water stream is given as
![v_(bw) = v_b + v_w](https://img.qammunity.org/2020/formulas/physics/high-school/d44khhcmbjlsygdzu21zjqcq1p4xojoa8b.png)
![v_(bw) = (1.60 - 2.20sin25) \hat i + 2.20 cos25\hat j](https://img.qammunity.org/2020/formulas/physics/high-school/a00gk78ruq2j9ncuwm20t7scrt47toqezn.png)
so we have
![v_(bw) = 0.67 \hat i + 2 \hat j](https://img.qammunity.org/2020/formulas/physics/high-school/dm6kr9qs3vpoa75kzkhy12qibg8wfnxeje.png)
magnitude of the speed is given as
![v_(bw) = √(0.67^2 + 2^2)](https://img.qammunity.org/2020/formulas/physics/high-school/szexj06tzmw5d7qatdlbzvvmlfsfa593ab.png)
![v_(bw) = 2.1 mph](https://img.qammunity.org/2020/formulas/physics/high-school/ebz27dts1gtl4wv17q0vhx3hzedj32lquc.png)
Part b)
Time to cross the river is given as
![t = (y)/(v_y)](https://img.qammunity.org/2020/formulas/physics/high-school/l77inm7drryl9x9ir48da2mesqxfs1wzwj.png)
![t = (1.20)/(2)](https://img.qammunity.org/2020/formulas/physics/high-school/nnitd90axjz8ofqv94815n7nru5og28073.png)
![t = 0.6 h](https://img.qammunity.org/2020/formulas/physics/high-school/t89dw4jas7fafb094uf1xet96zv3cvtgdw.png)
Part c)
Distance moved by the boat in downstream is given as
![x = v_x t](https://img.qammunity.org/2020/formulas/physics/college/9hi5gknaktz0yli62bxwf4q77wz9q70t6q.png)
![x = 0.67 * 0.6](https://img.qammunity.org/2020/formulas/physics/high-school/a5idbj4z7r8rmywvmswj8wdi4vwl4zbzbj.png)
![x = 0.4 mile](https://img.qammunity.org/2020/formulas/physics/high-school/qma7zo5bu04nhta2k80fa4dnepq6oc7otp.png)
Part d)
In order to go straight we must net speed along the stream must be zero
so we will have
![vsin\theta = v_w](https://img.qammunity.org/2020/formulas/physics/high-school/5fmfz9wk1xnyq39qejfe89bpuwqesr1qso.png)
![2.20 sin\theta = 1.60](https://img.qammunity.org/2020/formulas/physics/high-school/iqa6pyqnzbxsp45goqgol8kbl12hlxpdfn.png)
![\theta = 46.6 degree](https://img.qammunity.org/2020/formulas/physics/high-school/ihsl1u31zd86k2zkoakq66x8p2a6njxkpk.png)