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Two cars, A and B, travel along two straight roads which intersect at an angle $ heta$ where $ an \theta = \frac{4}{3}$ - Leaving Cert Applied Maths - Question 2 - 2011

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Two cars, A and B, travel along two straight roads which intersect at an angle $ heta$ where $ an \theta = \frac{4}{3}$. Car A is moving towards the intersection at... show full transcript

Worked Solution & Example Answer:Two cars, A and B, travel along two straight roads which intersect at an angle $ heta$ where $ an \theta = \frac{4}{3}$ - Leaving Cert Applied Maths - Question 2 - 2011

Step 1

Find (i) the velocity of A relative to B

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Answer

To find the velocity of A relative to B, we first need to define the velocity vectors for both cars.

Car A's velocity: V⃗A=5i^+0j^  m/s\vec{V}_A = 5 \hat{i} + 0 \hat{j} \; \text{m/s}

Car B's velocity: V⃗B=6i^+8j^  m/s\vec{V}_B = 6 \hat{i} + 8 \hat{j} \; \text{m/s}

Now, the relative velocity of A with respect to B can be calculated as follows: V⃗AB=V⃗A−V⃗B=(5i^+0j^)−(6i^+8j^)=−1i^−8j^\vec{V}_{AB} = \vec{V}_A - \vec{V}_B = (5 \hat{i} + 0 \hat{j}) - (6 \hat{i} + 8 \hat{j}) = -1 \hat{i} - 8 \hat{j}

To find the magnitude of this relative velocity, we compute: ∣V⃗AB∣=(−1)2+(−8)2=1+64=65  m/s|\vec{V}_{AB}| = \sqrt{(-1)^2 + (-8)^2} = \sqrt{1 + 64} = \sqrt{65} \; \text{m/s}

The direction can be found using the tangent function:

ightarrow \theta = \tan^{-1}(8) \; \text{(in degrees)} $$ Thus, the relative velocity of A towards B is approximately \( |\vec{V}_{AB}| \approx 8.18 \; ext{m/s} \).

Step 2

Find (ii) the shortest distance between the cars

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Answer

To find the shortest distance between the two cars, we can use the formula for the perpendicular distance from one line to another. The lines representing the paths of the cars each are 100 extm100 \, ext{m} from the intersection.

The distance traveled by Car A until it reaches the intersection is given by: dA=100⋅sin⁡(θ)  (where θ=tan⁡−1(4/3))d_A = 100 \cdot \sin(\theta) \; \text{(where } \theta = \tan^{-1}(4/3) \text{)}

And for Car B: dB=100⋅sin⁡(ϕ)  (where ϕ=tan⁡−1(10/5))d_B = 100 \cdot \sin(\phi) \; \text{(where } \phi = \tan^{-1}(10/5) \text{)}

Now substituting the values: dA=100⋅sin⁡(tan⁡−1(43))=100⋅45=80 md_A = 100 \cdot \sin \left( \tan^{-1}\left(\frac{4}{3}\right) \right) = 100 \cdot \frac{4}{5} = 80 \, \text{m} dB=100⋅sin⁡(tan⁡−1(105))=100⋅1=100 md_B = 100 \cdot \sin \left( \tan^{-1}\left(\frac{10}{5}\right) \right) = 100 \cdot 1 = 100 \, \text{m}

The shortest distance between the two cars is then: d=∣dA−dB∣=∣80−100∣=20 md = |d_A - d_B| = |80 - 100| = 20 \, \text{m}.

Step 3

Find the range of values of theta if she lands between B and C

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Answer

To find the range of values of heta heta for the boat's path:

  1. We start with the equations for motion:

    • The width of the river is 80 m80 \, m
    • The downstream distance to C is 203 m20 \sqrt{3} \, m from B

  2. We establish: tan⁡β=80203→β=tan⁡−1(80203)\tan \beta = \frac{80}{20 \sqrt{3}} \rightarrow \beta = \tan^{-1}\left(\frac{80}{20 \sqrt{3}}\right)

  3. Using the cosine function to establish limits for heta heta:

cos⁡θ=3.54\cos \theta = \frac{3.5}{4}

  • This gives: θ=cos⁡−1(3.54)≈28.955∘\theta = \cos^{-1}\left(\frac{3.5}{4}\right) \approx 28.955^\circ
  1. Finally, we find: 28.955≤θ≤60  degrees28.955 \leq \theta \leq 60 \; \text{degrees} Thus, the range of values of heta heta for the boat to land between B and C is: 28.955≤θ≤6028.955 \leq \theta \leq 60.

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