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[In this question i and j are horizontal unit vectors due east and due north respectively and position vectors are given relative to a fixed origin O.] Two cars P and Q are moving on straight horizontal roads with constant velocities - Edexcel - A-Level Maths Mechanics - Question 1 - 2016 - Paper 1

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Question 1

[In-this-question-i-and-j-are-horizontal-unit-vectors-due-east-and-due-north-respectively-and-position-vectors-are-given-relative-to-a-fixed-origin-O.]--Two-cars-P-and-Q-are-moving-on-straight-horizontal-roads-with-constant-velocities-Edexcel-A-Level Maths Mechanics-Question 1-2016-Paper 1.png

[In this question i and j are horizontal unit vectors due east and due north respectively and position vectors are given relative to a fixed origin O.] Two cars P a... show full transcript

Worked Solution & Example Answer:[In this question i and j are horizontal unit vectors due east and due north respectively and position vectors are given relative to a fixed origin O.] Two cars P and Q are moving on straight horizontal roads with constant velocities - Edexcel - A-Level Maths Mechanics - Question 1 - 2016 - Paper 1

Step 1

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

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Answer

To find the direction of motion of Q, we can start with its velocity vector:

extVelocityofQ=20i5j ext{Velocity of Q} = 20i - 5j

The direction can be calculated using the tangent of the angle formed by the vector with the positive x-axis (east direction):

anheta=oppositeadjacent=520 an heta = \frac{\text{opposite}}{\text{adjacent}} = \frac{-5}{20}

Thus, we find:

θ=tan1(520)=tan1(14)\theta = \tan^{-1}\left(-\frac{5}{20}\right) = \tan^{-1}\left(-\frac{1}{4}\right)

Calculating this gives:

θ14.04°\theta \approx -14.04°

Since this angle is measured clockwise from the east direction, the bearing is:

extBearing=360°+(14.04°)=345.96°extorapproximately346°. ext{Bearing} = 360° + (-14.04°) = 345.96° ext{ or approximately } 346°.

Step 2

Find an expression for p in terms of t.

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Answer

The position vector p can be expressed as:

p=(400+15t)i+(20t)j.p = (400 + 15t)i + (20t)j.

This shows that p depends linearly on time t.

Step 3

q in terms of t.

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Answer

For the position vector q, we know the initial position is 800j and the velocity of Q is (20i5j)(20i - 5j) m s1^{-1}:

q=(8005t)j+(20t)i.q = (800 - 5t)j + (20t)i.

Thus, q also varies with time t.

Step 4

Find the position vector of Q when Q is due west of P.

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Answer

For Q to be due west of P, their j-components must be equal:

20t=8005t20t = 800 - 5t

Rearranging gives:

\Rightarrow t = \frac{800}{25} = 32 \text{ seconds}.$$ Now substituting t back into the position vector of Q: $$q = (20(32)i + (800 - 5(32))j) = 640i + 640j.$$ Thus, the position vector of Q when it is due west of P is: $$q = 640i + 640j.$$

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