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[In this question, the unit vectors i and j are due east and due north respectively - Edexcel - A-Level Maths Mechanics - Question 7 - 2012 - Paper 1

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[In this question, the unit vectors i and j are due east and due north respectively. Position vectors are relative to a fixed origin O.] A boat P is moving with c... show full transcript

Worked Solution & Example Answer:[In this question, the unit vectors i and j are due east and due north respectively - Edexcel - A-Level Maths Mechanics - Question 7 - 2012 - Paper 1

Step 1

Calculate the speed of P.

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Answer

To find the speed of boat P, we use the formula for speed given by the magnitude of the velocity vector:
Speed=(4)2+(8)2 km h1\text{Speed} = \sqrt{(-4)^2 + (8)^2} \text{ km h}^{-1}
This simplifies to:
Speed=16+64=80 km h1\text{Speed} = \sqrt{16 + 64} = \sqrt{80} \text{ km h}^{-1}
Thus, the speed of P is approximately 8.94 km h⁻¹.

Step 2

Write down p in terms of t.

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Answer

The position vector p of boat P as it moves over time can be expressed as:
p = (2i - 8j) + t(-4i + 8j)
This gives:
p = [2 - 4t]i + [-8 + 8t]j.

Step 3

Find the value of t when P is due west of Q.

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Answer

Boat P is due west of boat Q when their respective j-components are equal.
From the position vector of Q:
q = (18 - 6)i + (12 - 8)j
This can be simplified to:
q = 12i + 4j.
Setting the j-components equal gives:
8+8t=4-8 + 8t = 4
Solving for t:
8t=128t = 12
t=128=32=1.5 hourst = \frac{12}{8} = \frac{3}{2} = 1.5 \text{ hours}.

Step 4

Find the distance between P and Q when P is due west of Q.

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Answer

Using the calculated value of t from part (c), we substitute t = 1.5 into the position vector for P:
p = [2 - 4(1.5)]i + [-8 + 8(1.5)]j
This simplifies to:
p = [-4]i + [4]j.
Now we can find the position of Q at t = 1.5:
q = 12i + 4j.
To find the distance between P and Q:
Distance=((412)2+(44)2)=(16)2=16 km.\text{Distance} = \sqrt{((-4 - 12)^2 + (4 - 4)^2)} = \sqrt{(-16)^2} = 16 \text{ km}.

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