MHB Ship A & B: Find Min. Speed to Intercept

  • Thread starter Thread starter markosheehan
  • Start date Start date
  • Tags Tags
    Ship Speed
AI Thread Summary
Ship A is traveling at a constant speed of 30 km/hr at an angle of 60 degrees east of north, while Ship B is positioned 20 km east of Ship A and must intercept it. The direction of Ship B's travel is not specified, but it is assumed to be north for interception to occur. Calculations indicate that Ship B must travel at a speed greater than 15 km/hr to successfully intercept Ship A, with a minimum speed of approximately 16.15 km/hr determined through trigonometric relationships. The angle of Ship B's trajectory required for interception is approximately 68.3 degrees.
markosheehan
Messages
133
Reaction score
0
ship A is traveling at 30 km hr^-1 in a direction 60 degrees E of N. Ship B is 20 km east of ship A traveling at a constant speed. find the minimum speed of ship B to intercept to intercept ship A

i know there distance in the j direction must be the same but when i let 30cos60=xsiny it does not help me.
 
Mathematics news on Phys.org
markosheehan said:
Ship B is 20 km east of ship A traveling at a constant speed.

Is ship A traveling at a constant speed north?
 
joypav said:
Is ship A traveling at a constant speed north?

ship A is traveling at a constant speed of 30 km hr^-1 in a direction 60 degrees east of north
 
markosheehan said:
ship A is traveling at a constant speed of 30 km hr^-1 in a direction 60 degrees east of north

Yes, I'm sorry.
I meant is ship B traveling north. You did not indicate in what direction ship B is moving. I assume it is moving north, otherwise it wouldn't be able to intercept ship A.
If it is traveling north, this would be a simple related rates problem (Calculus 1).
 
joypav said:
Yes, I'm sorry.
I meant is ship B traveling north. You did not indicate in what direction ship B is moving. I assume it is moving north, otherwise it wouldn't be able to intercept ship A.
If it is traveling north, this would be a simple related rates problem (Calculus 1).

in the question i am not told what direction ship B is travelling. I can send you a pic of the question if you want
 
$r_A = (15\sqrt{3} \cdot t)i + (15 \cdot t)j$

$r_B = (20+v\cos{\theta} \cdot t)i + (v\sin{\theta} \cdot t) j$

$v\sin{\theta} = 15 \implies \sin{\theta} = \dfrac{15}{v} \implies v \ge 15$, however, if $v=15$ ship B will miss the intercept $\implies v > 15$

$20 + v\cos{\theta} = 15\sqrt{3} \implies \cos{\theta} = \dfrac{15\sqrt{3}-20}{v}$

$\tan{\theta} = \dfrac{\sin{\theta}}{\cos{\theta}} = \dfrac{3}{3\sqrt{3}-4}$

$\theta = \arctan\left(\dfrac{3}{3\sqrt{3}-4}\right) \approx 68.3^\circ \implies v \approx 16.15 \, km/hr$
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
Is it possible to arrange six pencils such that each one touches the other five? If so, how? This is an adaption of a Martin Gardner puzzle only I changed it from cigarettes to pencils and left out the clues because PF folks don’t need clues. From the book “My Best Mathematical and Logic Puzzles”. Dover, 1994.

Similar threads

Back
Top