Can anyone help me to solve this?

 

A lifeguard who can swim at 1.2 m/s in still water wants to reach a dock positioned perpendicularly directly across a 550 m wide river.

a) If the current in the river is 0.80 m/s, how long will it take the lifeguard to reach the dock?

b) If instead she had decided to swim in such a way that will allow her to cross the river in a minimum amount of time, where would she land relative to the dock?

What have you done to try and answer this?

 

Drawn a diagram?

My attempt at solving it:

a)
1.2 sin(α)= 0.8
sin(α)= 0.8 / 1.2
sin(α)= 2/3
α= sin^-1(2/3)
α= 41.8 degrees

So his crossing velocity is:
1.2 cos41.80= 0.894 m/s

And the time taken is:
distance = speed * time
550= 0.894 * t
t= 550 / 0.894
t=615.21 seconds

b)
550 / 1.2 = 458.3 seconds
distance = speed x time
distance = 0.80 m/s x 458.3s = 366.6 m

Thank You AshBox, This will help me.

My attempt at solving it:

a)

1.2 sin(α)= 0.8

sin(α)= 0.8 / 1.2

sin(α)= 2/3

α= sin^-1(2/3)

α= 41.8 degrees

 

So his crossing velocity is:

1.2 cos41.80= 0.894 m/s

 

And the time taken is:

distance = speed * time

550= 0.894 * t

t= 550 / 0.894

t=615.21 seconds

 

b)

550 / 1.2 = 458.3 seconds

distance = speed x time

distance = 0.80 m/s x 458.3s = 366.6 m

We try not to give direct worked answers here but to lead people through doing it themselves correcting and pushing where necessary.

We try not to give direct worked answers here but to lead people through doing it themselves correcting and pushing where necessary.

I agree, but in this case we just have to put values in related equations which I have guided here.