how fast is it moving and in what direction?

Question

Two radar stations at A and B, with B 6km east of A, are tracking a ship. At certain instant, the ship is 5km from A, and this distance is increasing at the rate of 28km/h. At the same instant, the ship is 5km from B, but this distance is increasing at only 4km/h where is the ship, how fast is it moving and in what direction is it moving?

Answer 1

The initial conditions are represented by an isosceles triangle ASB where S is the initial position of the ship.

If we drop a perpendicular from S to N on AB, then SN is the height of the triangle and AN=BN=3km, AS=BS=5km so:

SN²=AS²-AN²=25-9=16, making SN=4km. Position of the ship is 3km East of A (3km West of B) and 4km North of A and B.

Let's assume that the ship is moving with constant speed v and direction from point S. If the angle of direction is θ as measured on the easterly line AB then after a time t hours the ship reaches a point P. The distance of the ship from A is AP and from B is BP. SP=vt. Point P will be vtsinθ further north than S and vtcosθ further east. This puts the ship vtsinθ+4 km north of both A and B.

The ship at P is vtcosθ+3 km east of A and vtcosθ-3 km east of B.

AP²=(vtcosθ+3)²+(vtsinθ+4)²=v²t²cos²θ+6vtcosθ+9+v²t²sin²θ+8vtsinθ+16,

AP²=v²t²+6vtcosθ+8vtsinθ+25.

BP²=(vtcosθ-3)²+(vtsinθ+4)²=v²t²-6vtcosθ+8vtsinθ+25.

The rate of change of AP is given as 28km/h and that of BP as 4km/h.

Let a=AP and b=BP, then 2ada/dt=2v²t+6vcosθ+8vsinθ, 2bdb/dt=2v²t-6vcosθ+8vsinθ.

ada/dt=v²t+3vcosθ+4vsinθ, bdb/dt=v²t-3vcosθ+4vsinθ.

da/dt=28, db/dt=4. ada/dt-bdb/dt=6vcosθ, 28a-4b=6vcosθ, 14a-2b=3vcosθ.

[We can also use the Cosine Rule on the two triangles ASP and BSP.

AS=BS=5km. AŜP=180-(φ-θ), BŜP=φ+θ, where tanφ=4/3 (sinφ=⅘, cosφ=⅗). SP=vt.

a²=AP²=5²+v²t²+10vtcos(φ-θ), b²=BP²=5²+v²t²-10vtcos(φ+θ).

2ada/dt=2v²t+10vcos(φ-θ), ada/dt=v²t+5vcos(φ-θ); bdb/dt=v²t-5vcos(φ+θ).

28a=v²t+5vcos(φ-θ); 4b=v²t-5vcos(φ+θ). When the compound angles are expanded we get the same equations as before.]

But this rate of change is constant and starts at t=0 when the ship leaves point S.

Plugging in t=0, a=b=5. So 28a=140=5vcos(φ-θ) and 4b=20=-5vcos(φ+θ).

vcos(φ-θ)=28, vcos(φ+θ)=-4; vcosφcosθ+vsinφsinθ=28, vcosφcosθ-vsinφsinθ=-4.

2vcosφcosθ=24, 2v(⅗)cosθ=24, vcosθ=20; 2vsinφsinθ=32, 2v(⅘)sinθ=32, vsinθ=20.

vsinθ/vcosθ=tanθ=1, θ=π/4 (45°). v²sin²θ+v²cos²θ=v²=800, v=20√2 km/h.

The position of the ship initially: 3km East of A (3km West of B) and 4km North of A and B.

Speed: 20√2=28.28km/h approx.

Direction: northeast: 45° north of east.