A diver of mass 60 kg: on entering water with an initial velocity of 14.87 ms-1,

Question

A diver of mass 60 kg: on entering water with an initial velocity of 14.87 ms-1, experiences a force of resistance, due to the water opposing motion, of magnitude 180v Newtons, where v is the velocity of the diver in m s-1. In addition to the force of gravity, the diver also experiences a constant upward buoyancy force of 720 N. Take the origin of motion to be at the surface of the water. The magnitude g of the acceleration due to gravity can be taken as 10 m s-1. Calculate the maximum depth, in metres, the diver will submerge. Give your answer to 3 decimal places.

Explanation / Answer

its a simple physics problem involving calculus.

net force is Y direction = (force of gravity) - (force of buoyoncy) -(resistance force by water)

= (60 x 10) - 720 - 180v

= 600 - 720 - 180v

so net acceleration = (-120 -180v)/60 ......... net acceleration= net force / mass

ie dv/dt = -(3v + 2)

or

dv/(3v + 2) = -dt

integration both sides and taking limits as: when t= 0 v= 14.8,

when t= t, v=v(t)

so we have final value as

[ln(3v + 2)]14.8v = [-3t]t0

=> ln[(3v + 2)/(3x 14.87 + 2)] = -3t

=> ln(3v+2/46.61) = -3t

or 3v + 2 = 46.61e-3t

or 3v = 46.61e-3t - 2

or v =  15.536e-3t - 2/3

now dx/dt = 15.536e-3t - 2/3

dx = (15.536e-3t - 2/3)dt

integrating both sides we get: and taking limits as: when t=0, x=0 and when t= t, x= x(t)

x(t) = -5.178e-3t - (2/3)t + 5.178   

now maximum depth will be when net velocity will be 0 in the water

ie v=0

which gives t = 1.05 s

so putting the value of t in depth vs time equation we get

x(1.05)=  -5.178e-3(1.05) - (2/3)(1.05) + 5.178 = 4.256m

so maximum depth the diver will go is 4.256m

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