Solving the Ships P and Q Problem

[mathsmathsmaths]

Two ships P and Q are traveling at night with constant velocities. At midnight,P is at point with position vector (20i + 10j) km relative to fixed origin 0. At the same time, q is at the point with position vectro (14i - 6J)Km. Three hours later, P is at point with position vector (29i + 34j) Km . the ship Q travels with velocity 12jkm h^-1 . At time t hours after midnight, the position vector of P and Q are p km and qKm respectively. find

a) the velocity of P, in terms of i and j.

i got

t=3= (20i + 10j)Km + 3( i + J)

t = 3 = (20i + 10j)Km + 3( i + J) = (29 i + 34j)

= 20 + 3i = 29

i = 3

then 10 + 3b = 34j

so b = 8j

Then velocity will be = 3i + 8j

(ok now i need someone to see if this part is right please.)

B) Then i got to find an expression for P and Q in terms of t, i and j

I got

P= (20i + 10j)+ ( 3i + 8J)t

and For Q = (14i - 6j) + (12j)t

then the questions goes :

At time t hours after midnight, the disatnce between p and Q is d Km.

c) by finding an expression For PQ show that

d^2 = 25t^2 - 92t + 292.

Now for this one i know that i have to take away Q from P but does it mean i take away the following :

Q = (14i - 6j) + (12j)t - P= (20i + 10j)+ ( 3i + 8J)t

Because i get the worng answer, so what i want from someone is to kindly, see if i have done part A and B right and to show me how to do C becuase i ahve been goign in circles, and no hope, so can someone please help thank you.

 

[Dick]

Just subtract Q from P.

"I got

P= (20i + 10j)+ ( 3i + 8J)t

and For Q = (14i - 6j) + (12j)t"

I.e. [(20i + 10j)+ ( 3i + 8J)t]-[(14i - 6j) + (12j)t].