Kinematics Physics Question

[SophiaRivera007]

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?

[Klaynos]

What have you done to try and answer this?

 

Drawn a diagram?

[AshBox]

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

[SophiaRivera007]

Thank You AshBox, This will help me.

[Klaynos]

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.

[AshBox]

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.