How Do You Calculate the Distance Between Two Moving Ships Using Vectors?

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Two ships, P and Q, are moving with constant velocities, starting from different position vectors at midnight. Ship P's position after three hours is given, and its velocity needs to be calculated in terms of vectors. The position vectors for both ships, P and Q, can be expressed in terms of time t. The discussion centers on deriving the squared distance between the two ships, leading to the equation d^2 = 25t^2 – 92t + 29^2. Additionally, participants seek resources for further understanding of vector calculations and distance measurement.
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Two ships P and Q are traveling at night with constant velocities. At midnight, P is at the point with position vector (20i + 10j) km relative to a fixed origin O. At the same time, Q is at the point with position vector (14i – 6j) km. Three hours later, P is at the point with position vector (29i + 34j) km. The ship Q travels with velocity 12j km h–1. At time t hours after midnight, the position vectors of P and Q are p km and q km respectively. Find
(a) the velocity of P, in terms of i and j,

(b) expressions for p and q, in terms of t, i and j.

At time t hours after midnight, the distance between P and Q is d km.
(c) By finding an expression for PQ, show that
d^2 = 25t^2 – 92t + 29^2.
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Hi... another vectors question...
i get the answers to part a and and b... however, for C i don't really know how to do this one... maybe I am lacking the principles to this question. :confused: ... If anyone could walk me through part C.. it would really help...
Thanks :)

PS. Do you know of any good internet source that has extensive tutorials on vectors?
 
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How'd you calculate the distance between two points on a graph? You should apply the same rule here.

~H
 
(y2 -y1) ÷ (x2 -x1)?
 
Using pythag;

d = \sqrt{(x_{2}-x_{1})^2 + (y_{2}-y_{1})^2}

~H
 
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