Justin C. answered • 09/22/17

Tutor

New to Wyzant
Bachelors in Physics with 5 years tutoring experience

Hey there Gnarls,

This question is phrased as if the diver hasn't fully decelerated by the time he reaches that depth, but we can only do this problem if we assume that the diver has slowed to a stop when they reach 1.85 m.

If we do, we can use this equation

v

_{f}^{2}= v_{i}^{2}+ 2aΔxv

_{f}= 0m/sv

_{i}= 17.64m/sΔx = 1.85m

a = ?

Plug it all in we get

311.1696 m

^{2}/s^{2}= 0 + 2*a*(1.85 m)**Remove 0**311.1696 m

^{2}/s^{2}=2*a*(1.85 m)**Divide by 2**155.5848 m

^{2}/s^{2}=a*(1.85 m)**Divide by 1.85 m**84.09989 m/s

^{2}= a**Round appropriately**So there you have it

The diver's acceleration was 84.10 m/s

^{2}