Kinematics - projectile with acceleration that depends on V

[Feodalherren]

Homework Statement

A ship is traveling across the sea when its engines are cut. The ship slows down by a = -kV². The ship gradually slows down from 5 knots/h to 12 knots/h over 20min. Determine the distance traveled in nautical miles. One knot is 1 nautical mile per hour.

Homework Equations

Kinematics

The Attempt at a Solution

Vi = .2 [nmiles / min]
Vf = .1167 [nmiles / min]

nmiles= nautical miles

a = dV/dt = -kV²

(1/V²)dV = -k dt

integrate over Vi to Vf and 0 to 20min.

therefore k = .1786 [1/nmiles]

units seem to add up if a = -kv^2 the units will become nmiles/min^2

confident that k is correct. Not sure though.

Now for V.

a= v dv/dx = -kv²

-kv² = v dv/dx

-kdx = (1/v) dv

integrating from Xi to Xf and Vi to Vf

-k(Xf - Xi) = ln(Vf - Vi)

Natural log of a negative number, impossible.
Where am I going wrong?

 

[Feodalherren]

Wait. Did I take the integral wrong, does it become

-k(Xf - Xi) = ln(Vf/Vi)

? Is that the correct answer?

 

[Nathanael]

The ship gradually slows down by 5 knots/h from 12 knots/h over 20min. Determine the distance traveled in nautical miles. One knot is 1 nautical mile per hour.
...
Vi = .2 [nmiles / min]
Vf = .1167 [nmiles / min]

Did you mean to say the ship slows from 12 knots/h to 7 knots/h?
If so, then I agree with your value of k

-k(Xf - Xi) = ln(Vf/Vi)

? Is that the correct answer?

Looks good to me :) I get the same

 

[Feodalherren]

Yes indeed I did mean to say that.

Thank you Sir!

 

[HalfLostAndSearching]

Your approach to solving this problem is correct, but there are a few small mistakes in your calculations.

First, when integrating from Vi to Vf, you should use the initial and final velocities in terms of nmiles/min, not nmiles/hour. So Vi = 5/60 = 0.0833 [nmiles/min] and Vf = 12/60 = 0.2 [nmiles/min].

Second, when integrating from 0 to 20min, you should use the time in terms of minutes, not hours. So your time interval should be 0 to 20, not 0 to 20min.

With these corrections, your final equation should be -k(Xf - Xi) = ln(0.2 - 0.0833) = ln(0.1167).

Solving for k, you should get k = 1.745 [1/nmiles].

Now, to find the distance traveled, you can use the equation d = (1/k)(ln(Vi) - ln(Vf)), where Vi = 0.0833 and Vf = 0.2.

Plugging in the values, you should get d = 0.052 [nmiles].

So the ship traveled approximately 0.052 nautical miles in 20 minutes.