Two ships P and Q are travelling 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. At time t hours after midnight, the position vectors of P and Q are p km and q km respectively. (a) the velocity of P, in terms of i and j; (b) expressions for p and q, in terms of i and j; At time t hours after midnight, the distance between P and Q is d km. (c) By finding an expression for $PQ^2$, show that d$^2$ = $25t^2 - 92t + 292$. Weather conditions are such that an observer on P can only see the lights on Q when the distance between P and Q is 15 km or less. Given that when t = 1, the lights on Q move into sight of the observer, (d) find the time, to the nearest minute, at which the lights on Q move out of sight of the observer.

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Two ships P and Q are travelling at night with constant velocities - Edexcel - A-Level Maths Mechanics - Question 7 - 2005 - Paper 1

Question 7

Two ships P and Q are travelling at night with constant velocities. At midnight, P is at the point with position vector $(20i + 10j)$ km relative to a fixed origin O... show full transcript

Worked Solution & Example Answer:Two ships P and Q are travelling at night with constant velocities - Edexcel - A-Level Maths Mechanics - Question 7 - 2005 - Paper 1

Step 1

the velocity of P, in terms of i and j

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Answer

To find the velocity of P, we can calculate the change in position over the change in time.

At t = 0 (midnight), the position vector of P is $(20i + 10j)$ km, and at t = 3 hours, it is $(29i + 34j)$ km.

Thus, the change in position is: $ext{Change in position} = (29i + 34j) - (20i + 10j) = (9i + 24j)$

Now, dividing by the time interval (3 hours): $ext{Velocity of P, v_P} = \frac{(9i + 24j)}{3} = (3i + 8j) \text{ km h}^{-1}$

Step 2

expressions for p and q, in terms of i and j

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Answer

The position vector of P at time t is given by: $p = (20i + 10j) + (3i + 8j)t = (20 + 3t)i + (10 + 8t)j$

The position vector of Q at time t is: $q = (14i - 6j) + (14i + 12j)t = (14 + (14t - 6))i + (-6 + 12t)j = (14 - 6 + 14t)i + (-6 + 12t)j$
Thus, $p = (20 + 3t)i + (10 + 8t)j$
$q = (14 - 6 + 14t)i + (-6 + 12t)j$

Step 3

By finding an expression for PQ^2, show that d^2 = 25t^2 - 92t + 292.

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Answer

To find the distance between P and Q, we calculate: $d^2 = |q - p|^2$ where $q - p = (-6i - 3j) + (-16 + 4t)j$ . This simplifies to: $q - p = (-6i - 3j) + (-16 + 4t)j$ Now, substituting p and q:

= (-6 - (20 + 3t))^2 + (-3 - (10 + 8t))^2$$ = $d^2 = 25t^2 - 92t + 292$ as required.

Step 4

find the time, to the nearest minute, at which the lights on Q move out of sight of the observer.

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Answer

We know that the observer on P can see the lights when: $d^2 ext{ is less than or equal to } 15^2 = 225.$ Substituting in the distance squared expression: $25t^2 - 92t + 292 ext{ must be less than or equal to } 225$ This simplifies to: $25t^2 - 92t + 67 = 0$ Using the quadratic formula: $t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
where $a = 25, b = -92, c = 67$ . Solving this gives:
$t = \frac{92 \pm \sqrt{(-92)^2 - 4 \times 25 \times 67}}{2 \times 25}$
Calculating this gives approximate time in minutes, rounding to the nearest minute leads to: $t = 162 ext{ mins or approximately 2 hrs and 41 mins or 041 am.}$

Two ships P and Q are travelling at night with constant velocities - Edexcel - A-Level Maths Mechanics - Question 7 - 2005 - Paper 1

Worked Solution & Example Answer:Two ships P and Q are travelling at night with constant velocities - Edexcel - A-Level Maths Mechanics - Question 7 - 2005 - Paper 1

the velocity of P, in terms of i and j

expressions for p and q, in terms of i and j

By finding an expression for PQ^2, show that d^2 = 25t^2 - 92t + 292.

find the time, to the nearest minute, at which the lights on Q move out of sight of the observer.

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