# Physicsship intercept

#### markosheehan

##### Member
ship A is travelling at 30 km hr^-1 in a direction 60 degrees E of N. Ship B is 20 km east of ship A travelling 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.

#### joypav

##### Active member
Ship B is 20 km east of ship A travelling at a constant speed.
Is ship A travelling at a constant speed north?

#### markosheehan

##### Member
Is ship A travelling at a constant speed north?
ship A is travelling at a constant speed of 30 km hr^-1 in a direction 60 degrees east of north

#### joypav

##### Active member
ship A is travelling 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 travelling 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 travelling north, this would be a simple related rates problem (Calculus 1).

#### markosheehan

##### Member
Yes, I'm sorry.
I meant is ship B travelling 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 travelling 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

#### skeeter

##### Well-known member
MHB Math Helper
$r_A = (15\sqrt{3} \cdot t)i + (15 \cdot t)j$

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

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

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

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

$\theta = \arctan\left(\dfrac{3}{3\sqrt{3}-4}\right) \approx 68.3^\circ \implies v \approx 16.15 \, km/hr$