A 52.0kg swimmer with an initial speed of 1.26m/s decides to coast until she comes to rest.
If she slows with constant acceleration and stops after coasting 2.05m , what was the magnitude of the force exerted on her by the water?

Question

A 52.0kg swimmer with an initial speed of 1.26m/s decides to coast until she comes to rest.
If she slows with constant acceleration and stops after coasting 2.05m , what was the magnitude of the force exerted on her by the water?

anonymous · Answer

to find a: $$v_f^2 = v_i^2 + 2a \Delta s$$
to find F:

F = m*a

anonymous · Answer

because of Newton's 3rd where every action has opposite reaction, her force on the water = the force of the water on her

anonymous · Answer

So I have to find acceleration first?
0=1.26m/s+2a(delta)2.05?

agent0smith · Answer

delta s just means change in displacement... it shouldn't be in there once you plug in the displacement which was 2.05m.

anonymous · Answer

so $$a=\frac{ -1.26 }{ 2(2.05) }$$

agent0smith · Answer

Yep

Then find F as Euler showed above.

anonymous · Answer

ok awesome thank you!

anonymous · Answer

So overall is the answer F=-15.98? The Mastering Physics says that answer is incorrect

agent0smith · Answer

You didn't square the velocity when finding a.

anonymous · Answer

Yes I just realized that!! I got F=20.3 and it is correct!

agent0smith · Answer

:)