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Two cars, A and B, travel along two straight roads which intersect at right angles - Leaving Cert Applied Maths - Question 2 - 2009
Question 2
Two cars, A and B, travel along two straight roads which intersect at right angles. A is travelling east at 15 m/s. B is travelling north at 20 m/s. At a certain in... show full transcript
Worked Solution & Example Answer:Two cars, A and B, travel along two straight roads which intersect at right angles - Leaving Cert Applied Maths - Question 2 - 2009
Step 1
(i) the shortest distance between the cars
Answer
To find the shortest distance between the two cars A and B, we can use the following approach. At the time they are both 800 m from the intersection, let:
- Car A's velocity vector: ( \vec{v_A} = 15 \hat{i} + 0 \hat{j} )
- Car B's velocity vector: ( \vec{v_B} = 0 \hat{i} + 20 \hat{j} )
The relative position of the cars:
- Car A is moving East while Car B is moving North.
The direction angle at which they are approaching each other can be calculated:
Slope, ( m = -\frac{15}{20} = -\frac{3}{4} ) which gives a direction of approximately 36.87°.
We calculate the shortest distance using the formula:
[ d = \frac{\text{Distance}}{\sqrt{m^2 + 1}} = \frac{800}{\sqrt{\left(-\frac{3}{4}\right)^2 + 1}} ] [ d = \frac{800}{\sqrt{\frac{9}{16} + 1}} = \frac{800}{\sqrt{\frac{25}{16}}} = \frac{800 \times 4}{5} = 640 \text{ m} ]
Step 2
(ii) the distance each car is from the intersection when they are nearest to each other
Answer
To find the distance each car is from the intersection when they are nearest to each other, we need to calculate the time taken based on their velocities. Let's denote the time taken by car B to reach the nearest point:
[ \text{Time taken by A} = \frac{800 - 200y}{v_A} = \frac{800 - 40t}{15}] [ \text{Time taken by B} = \frac{800 - 600}{v_B} = \frac{200}{20}]
For this setup:
- For Car A, substituting values yields (t_A \approx 8 \text{ s}) and for Car B (t_B \approx 10 \text{ s}).
From these times, we find:
- Distance of A from intersection when nearest: ( 800 - 600 = 200)
- Distance of B: ( 800 - 896 = -96 \implies 96 \text{ m past the intersection}). Thus:
- Car A is 200 m from the intersection, while Car B is 96 m past the intersection.
Step 3
Find, in terms of u and T, the time to fly from P to Q
Answer
For the aeroplane's journey from point P to Q, we denote:
- ( u ) = speed of the aeroplane in still air (km/h)
- ( s ) = distance from P to Q
- The angle of ascent resolves to: [ \sin 45 = \frac{4}{u} \rightarrow \text{Thus, } u \sin 45 = 4 \text{ or } \cos 45 = \frac{u}{\sqrt{u^2 - 16}}. ]
By substitution, we obtain: [ ext{Time} = \frac{s}{u} = \frac{4}{\sqrt{u^2 - 16}} T.]
Therefore, in terms of u and T, the planar vector calculations yield directional flight time, allowing recovery time to be computed effectively.