Kate crosses a road, of constant width 7 m, in order to take a photograph of a marathon runner, John, approaching at 3 m s⁻¹ - Edexcel - A-Level Maths Pure - Question 2 - 2013 - Paper 8
Question 2
Kate crosses a road, of constant width 7 m, in order to take a photograph of a marathon runner, John, approaching at 3 m s⁻¹.
Kate is 24 m ahead of John when she sta... show full transcript
Worked Solution & Example Answer:Kate crosses a road, of constant width 7 m, in order to take a photograph of a marathon runner, John, approaching at 3 m s⁻¹ - Edexcel - A-Level Maths Pure - Question 2 - 2013 - Paper 8
Step 1
Express 24 sin θ + 7 cos θ in the form R cos(θ - α), where R and α are constants and where R > 0 and 0 < α < 90°, giving the value of α to 2 decimal places.
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Answer
To express it in the required form, we first find R:
R=(7)2+(24)2=25=25
Next, we find α using the tangent function:
tan(α)=724
Thus,
α=arctan(724)≈73.74°
Therefore, we can express it as:
24sinθ+7cosθ=25cos(θ−73.74°).
Step 2
Given that θ varies, find the minimum value of V.
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Answer
To find the minimum value of V, we need to maximize the denominator. The expression to maximize is:
24sinθ+7cosθ
Using the value of R found earlier, we can rearrange:
The maximum is when:
24sinθ+7cosθ=R
Thus, the maximum value becomes:
Vmin=R21=2521=0.84
Step 3
Given that Kate's speed has the value found in part (b), find the distance AB.
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Answer
Using the trig equation:
θ=0.84
Letting x be the distance AB, we can express it as:
x=sin(θ)7
Substituting:
x=sin(0.84)7≈7.29
Thus, the distance AB is 7.29 m.
Step 4
Given instead that Kate's speed is 1.68 m s⁻¹, find the two possible values of the angle θ, given that 0 < θ < 150°.
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Answer
We first substitute Kate's speed into the equation for V:
1.68=24sinθ+7cosθ21
Rearranging gives:
24sinθ+7cosθ=1.6821=12.5
Thus, the equation simplifies to:
Rcos(θ−α)=12.5
We find α using the same method as in part (a).
Solving for θ gives two angles within the specified range (0 < θ < 150°).
