Ship A is located 4.2 km north and 2.6 km east of ship B. Ship A has a velocity

Question

Ship A is located 4.2 km north and 2.6 km east of ship B. Ship A has a velocity of 22 km/h toward the south and ship B has a velocity of 40 km/h in a direction 37 degree north of east, What is the velocity of A relative to B (Express your answer in terms of the unit vectors ? and j, where i is toward the east.) Write an expression (in terms of and ) for the position of A relative to B as a function of t, where t = 0 when the ships are in the positions described as above. (Use the following as necessary: t.) At what time is the separation between the ships least? What is that least separation?

Explanation / Answer

(a) Apply Relative velocity addition law

V(A,B)= V(A,E)+V(E,B); V(A,B)- velocity of A relative to B. V(A,E)- velocity of A relative to Earth. V(E,B)- velocity of Earth relative to B

Apply in direction i(east) , V(A,B)= 0 + (-40 cos(37)); Vx= -32 i

Apply in direction j(north) , V(A,B)= -220 + (-40 sin(37)); Vx= -46 j

V(A,B)=-32 i -46j

(b)Apply s=ut relative to B

s= V(A,B)*t=-32 i -46j; But at t=0, displacement of A= 2.6i + 4.2j

So, position vector(r) = (2.6-32 )i + (4.2-46) j ----(A)

(c) seperation= magnitude of r. Take D= r2 ;

D= (2.6-32t)2 + (4.2-46t)2 ; To find the time(T) when sepration is minimunm(min D), derivative of D w.r.t. t should be equal to 0

dD/dt = 2* (2.6-32T)+2*(4.2-46T)=0

T=0.087 hours= 5 minutes 14 seconds

(d) Apply t= 0.087 to expression (A)

r= [2.6-32*(0.087)]i +[ 4.2-46*(0.087)]j

r= -0.184i+ 0.198 j ;

minimum seperation= magnitude of r= sqrt(0.1842+ 0.1982)= 0.27km

Related Questions

Navigate

• Browse (All)
• Browse S
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.