Kinematics - projectile with acceleration that depends on V

  • #1
603
6

Homework Statement


A ship is travelling 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?
 
Last edited:

Answers and Replies

  • #2
603
6
Wait. Did I take the integral wrong, does it become

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

? Is that the correct answer?
 
  • #3
Nathanael
Homework Helper
1,650
239
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
 
  • Like
Likes Feodalherren
  • #4
603
6
Yes indeed I did mean to say that.

Thank you Sir!
 

Related Threads on Kinematics - projectile with acceleration that depends on V

Replies
14
Views
6K
Replies
12
Views
12K
Replies
16
Views
5K
Replies
3
Views
826
  • Last Post
Replies
1
Views
3K
Replies
8
Views
5K
Replies
2
Views
1K
Replies
31
Views
12K
  • Last Post
Replies
8
Views
3K
  • Last Post
Replies
1
Views
4K
Top