Answer:
t_{out} =
t_{in}, t_{out} = 
Explanation:
This in a relative velocity exercise in one dimension,
let's start with the swimmer going downstream
its speed is

The subscripts are s for the swimmer, r for the river and g for the Earth
with the velocity constant we can use the relations of uniform motion
= D / 
D = v_{sg1} t_{out}
now let's analyze when the swimmer turns around and returns to the starting point

= D / 
D = v_{sg 2} t_{in}
with the distance is the same we can equalize

t_{out} = t_{in}
t_{out} =
t_{in}
This must be the answer since the return time is known. If you want to delete this time
t_{in}= D / 
we substitute
t_{out} = \frac{v_s - v_r}{v_s+v_r} ()
t_{out} = 
Answer:
C. The bug's change in momentum is equal to the car's change in momentum.
Explanation:
As we know by Newton's 2nd law

here we have also know that when car hits the bug then force applied by wind shield on the bug is same as the force applied by the bug on the car's wind shield as per Newton's III law

so we know that

so we have

so correct answer will be
C. The bug's change in momentum is equal to the car's change in momentum.