Solve Diving Drag Homework: Time to Reach 2% Speed

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SUMMARY

The discussion focuses on solving a physics homework problem involving drag force during a dive into water. The diver, with a mass of 65 kg and an initial speed of 7.0 m/s, experiences a drag force described by Fd=(-1.40x10^4)v. The correct approach requires setting up a differential equation due to the variable nature of acceleration, rather than assuming constant acceleration. The initial attempt incorrectly calculated time as 0.00046 seconds, which was identified as erroneous due to neglecting the proper application of forces and signs.

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  • Understanding of Newton's second law of motion
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  • Knowledge of differential equations
  • Basic principles of kinematics
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  • Learn about variable acceleration and its implications in motion
  • Explore the concept of drag forces in fluid dynamics
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Students studying physics, particularly those tackling problems involving forces, drag, and differential equations. This discussion is beneficial for anyone seeking to deepen their understanding of motion in fluid environments.

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Homework Statement


You dive straight down into a pool of water. You hit the water with a speed of 7.0m/s, and your mass is 65kg.
Assuming a drag force of the form Fd=(-1.40x10^4)v, how long does it take you to reach 2% of your original speed? (Ignore any effects of buoyancy.)

Homework Equations


Fd=(-1.40x10^4)v
Fg=mg
Fnet=ma
a=(vf-vi)/t

The Attempt at a Solution


Fnet=mg-(-1.40x10^4)v
a =(mg-(-1.40x10^4)v)/m
a = -1500= vf-vi/t=0.14-7/t
t= -6.86/-1500
t= o.oo46s

When I enter this number or 4.6x10^-3, says it is wrong?? could someone help me, thisi s really frustrating me. The physics all seems to make sense to me.
 
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Torrencio said:

The Attempt at a Solution


Fnet=mg-(-1.40x10^4)v
Careful with signs. The drag force acts up; gravity acts down.

Since the force varies with speed, the acceleration is not constant. You have to set up and solve a differential equation to get the time. Hint: a = dv/dt.
 

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