Two ships, P and Q, are moving with constant velocities. The velocity of P is $(9i - 2j)$ km h$^{-1}$ and the velocity of Q is $(4i + 8j)$ km h$^{-1}$. (a) Find the direction of motion of P, giving your answer as a bearing to the nearest degree. When $t = 0$, the position vector of P is $(9i + 10j)$ km and the position vector of Q is $(i + 4j)$ km. At time $t$ hours, the position vectors of P and Q are p km and q km respectively. (b) Find an expression for (i) p in terms of t, (ii) q in terms of t.\n (c) Hence show that, at time t hours, $QP^2 = (8 + 5t) + (6 - 10t)j$. (d) Find the values of t when the ships are 10 km apart.

A-Level Edexcel Maths Mechanics Constant Acceleration - 2D: Two ships, P and Q, are moving w

Two ships, P and Q, are moving with constant velocities - Edexcel - A-Level Maths Mechanics - Question 7 - 2017 - Paper 1

Question 7

Two ships, P and Q, are moving with constant velocities. The velocity of P is $(9i - 2j)$ km h$^{-1}$ and the velocity of Q is $(4i + 8j)$ km h$^{-1}$. (a) Find the... show full transcript

Worked Solution & Example Answer:Two ships, P and Q, are moving with constant velocities - Edexcel - A-Level Maths Mechanics - Question 7 - 2017 - Paper 1

Step 1

Find the direction of motion of P, giving your answer as a bearing to the nearest degree.

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Answer

The velocity of P is given by the vector $(9i - 2j)$ km h $^{-1}$ . To find the direction, we first calculate the angle $heta$ using the tangent function:

$an( heta) = \frac{-2}{9}$

Calculating $heta$ , we find:

$\theta = \tan^{-1}\left(-\frac{2}{9}\right)\approx -12.5^{\circ}$

This angle is measured from the positive x-axis (east). To express this as a bearing, we convert it to a positive angle:

Bearing = $90^{\circ} - 12.5^{\circ} = 77.5^{\circ}$ .

Thus, the bearing is approximately $103^{\circ}$ .

Step 2

Find an expression for p in terms of t.

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Answer

$p = (9i + 10j) + t(9i - 2j) = (9 + 9t)i + (10 - 2t)j.$

Step 3

Find an expression for q in terms of t.

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Answer

$q = (i + 4j) + t(4i + 8j) = (1 + 4t)i + (4 + 8t)j.$

Step 4

Hence show that, at time t hours, QP² = (8 + 5t) + (6 - 10t)j.

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Answer

The distance squared between P and Q is given by:

$QP^2 = (p - q) \cdot (p - q)$

Substituting for p and q:

$QP^2 = \left((9 + 9t) - (1 + 4t)\right)^2 + \left((10 - 2t) - (4 + 8t)\right)^2$

Simplifying:

$QP^2 = (8 + 5t)^2 + (6 - 10t)^2.$

Thus, $QP^2 = (8 + 5t) + (6 - 10t)j$ .

Step 5

Find the values of t when the ships are 10 km apart.

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Answer

To find when the ships are 10 km apart, we set:

$QP^2 = 100$

Using the expression for $QP^2$ , we have:

$(8 + 5t)^2 + (6 - 10t)^2 = 100$

Expanding:

$(64 + 80t + 25t^2) + (36 - 120t + 100t^2) = 100$

Combining like terms gives:

$125t^2 - 40t + 0 = 0$

Factoring out results in:

$t(25t - 8) = 0$

Thus, $t = 0$ or $t = 0.32$ .

Two ships, P and Q, are moving with constant velocities - Edexcel - A-Level Maths Mechanics - Question 7 - 2017 - Paper 1

Worked Solution & Example Answer:Two ships, P and Q, are moving with constant velocities - Edexcel - A-Level Maths Mechanics - Question 7 - 2017 - Paper 1

Find the direction of motion of P, giving your answer as a bearing to the nearest degree.

Find an expression for p in terms of t.

Find an expression for q in terms of t.

Hence show that, at time t hours, QP² = (8 + 5t) + (6 - 10t)j.

Find the values of t when the ships are 10 km apart.

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