Kinematics: Average Deceleration Underwater

In summary, the conversation is about a person jumping off a diving board and trying to estimate the average deceleration while under water. The person attempted to solve the problem using two different equations but was unsure of the correct answer. With some help, the person was able to find their error in algebra and solve for the average deceleration.
  • #1
RedBarchetta
50
1
[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:
 
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  • #2
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.
 
  • #3
Thank you! That worked. I knew I had a fault somewhere in my Algebra. :cool:
 

Related to Kinematics: Average Deceleration Underwater

What is kinematics?

Kinematics is the branch of physics that deals with the motion of objects without considering the forces that cause the motion.

What is average deceleration?

Average deceleration is the measure of how much an object's velocity decreases over a certain amount of time. It is calculated by dividing the change in velocity by the change in time.

How does deceleration occur underwater?

Deceleration underwater occurs due to the resistance of water against an object's motion. This resistance, also known as drag, causes the object to slow down as it moves through the water.

What factors affect average deceleration underwater?

The factors that affect average deceleration underwater include the speed of the object, its surface area, the density of the water, and the shape and size of the object.

How is average deceleration calculated in underwater kinematics?

In underwater kinematics, average deceleration is calculated using the equation a = (vf - vi) / t, where a is average deceleration, vf is final velocity, vi is initial velocity, and t is time. This equation can be applied to any object moving through water, including marine animals or underwater vehicles.

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