Kinematics: Average Deceleration Underwater

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SUMMARY

The discussion focuses on calculating the average deceleration of a person underwater after jumping from a diving board. The participant initially used the equations of motion, specifically V^2 = Vo^2 - 2gy and V^2 = Vo^2 + 2a(x - xo), to determine the velocity upon entering the water and subsequently the deceleration underwater. The correct approach involves recognizing the total distance traveled (6.6 m) and correctly applying algebra to solve for deceleration. The final calculation confirms that the average deceleration is derived from rearranging the equation to isolate 'a'.

PREREQUISITES
  • Understanding of kinematic equations, specifically V^2 = Vo^2 - 2gy and V^2 = Vo^2 + 2a(x - xo).
  • Basic algebra skills for rearranging equations and solving for variables.
  • Knowledge of gravitational acceleration, specifically 9.8 m/s².
  • Familiarity with concepts of initial and final velocity in motion problems.
NEXT STEPS
  • Review kinematic equations in physics, focusing on their applications in vertical motion.
  • Practice solving problems involving deceleration and acceleration using real-world scenarios.
  • Explore the concept of buoyancy and its effects on motion in fluids.
  • Learn about the principles of energy conservation in motion, particularly in free fall and underwater scenarios.
USEFUL FOR

This discussion is beneficial for physics students, educators, and anyone interested in understanding motion dynamics, particularly in fluid environments. It is especially relevant for those studying kinematics and problem-solving in physics.

RedBarchetta
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[SOLVED] Kinematics: Average Deceleration Underwater

Hello! First poster here. :cool: Here is a tough question from my online HW:

Homework Statement


A person jumps off a diving board 5.0 m above the water's surface into a deep pool. The person’s downward motion stops 1.6 m below the surface of the water. Estimate the average deceleration (magnitude) of the person while under the water.

Answer: _blank_ m/s^2.

Homework Equations


V^2=Vo^2-2gy
V^2=Vo^2+2a(x-xo)
Vf=0

The Attempt at a Solution


I asked my physics professor for assistance and tried the problem on my own but came to no avail. I started off by using the first V squared formula and plugging in the known variables to find the velocity of the person as they approach the surface of the water.

Then, I plugged the inital velocity into the 2nd V squared equation, as this is the velocity as the person struck the water, and still couldn't find the correct answer. I must be having an algebraic issue.

Okay, there are two distances: 5 m above the water and 1.6 m below the surface of the water. So there is a combined total of 6.6 meters traveled. For above the water:

Vo or inital velocity is zero.

V^2=Vo^2-2gy
V^2=-2gy
V^2=-2(9.8m/s^2)(5m)
V^2=-98m/s

Then take the absolute value of the the square root?

|V|=Sqroot(98m/s)
V=9.8895 m/s

Now, use the second equation to find out the deceleration under the water:

Xo equals zero.
Vf or final velocity is zero.

V^2=Vo^2+2a(X-Xo)
(Plug in inital velocity when the person struck the water and underwater distance for X.)
V^2=(9.8995m/s)^2+2a(1.6m)
V^2=98 m^2/s^2 + (3.2m)a
(Plug in zero for final velocity and move a to the other side)
a=98 m^2/s^2 + 3.2m

Then...? I am not sure. :confused: Any help would be great. Thank you! :smile:
 
Last edited:
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RedBarchetta said:
V^2=(9.8995m/s)^2+2a(1.6m)
V^2=98 m^2/s^2 + (3.2m)a
(Plug in zero for final velocity and move a to the other side)
a=98 m^2/s^2 + 3.2m

Then...? I am not sure. :confused: Any help would be great. Thank you! :smile:
Welcome aboard! Check your algebra. You have 0 = 98 + 3.2a. You've got to subtract 98 from both sides before you solve for a.
 
Thank you! That worked. I knew I had a fault somewhere in my Algebra. :cool:
 

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