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Two cars, A and B, travel along two straight roads which intersect at an angle of 135° - Leaving Cert Applied Maths - Question 2 - 2015

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Two cars, A and B, travel along two straight roads which intersect at an angle of 135°. Car A is moving towards the intersection at a uniform speed of 60 km h⁻¹. C... show full transcript

Worked Solution & Example Answer:Two cars, A and B, travel along two straight roads which intersect at an angle of 135° - Leaving Cert Applied Maths - Question 2 - 2015

Step 1

Find the magnitude and direction of the velocity of B relative to A.

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Answer

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

  1. The velocity of car A, extbfvA=60extkm/hi^ extbf{v}_A = 60 ext{ km/h} \hat{i}.

  2. For car B, which passes the intersection 2 minutes later, its velocity can be expressed as:

    • In 2 minutes (1/30 hours), car A travels extDistance=60×130=2extkm ext{Distance} = 60 \times \frac{1}{30} = 2 ext{ km}.
    • Hence, car B travels towards the intersection with velocity extbfvB=45 km/hextatanangleof135°. extbf{v}_B = 45 \text{ km/h} ext{ at an angle of } 135°.
    • Converting this to components gives: vB=45cos(135°)i^+45sin(135°)j^31.82i^+31.82j^.\textbf{v}_B = 45 \cos(135°) \hat{i} + 45 \sin(135°) \hat{j} \approx -31.82 \hat{i} + 31.82 \hat{j}.
  3. The relative velocity of B to A is given by:
    vB relative to A=vBvA=(31.82i^+31.82j^)(60i^)=91.82i^+31.82j^.\textbf{v}_{B \text{ relative to } A} = \textbf{v}_B - \textbf{v}_A = (-31.82 \hat{i} + 31.82 \hat{j}) - (60 \hat{i}) = -91.82 \hat{i} + 31.82 \hat{j}.

  4. The magnitude of the relative velocity is: vB relative to A=(91.82)2+(31.82)297.2 km/h.|\textbf{v}_{B \text{ relative to } A}| = \sqrt{(-91.82)^2 + (31.82)^2} \approx 97.2 \text{ km/h}.

  5. To find the direction, use: α=tan1(31.8291.82)19.11°.\alpha = \tan^{-1} \left( \frac{31.82}{-91.82} \right) \approx 19.11°. Thus, the velocity of B relative to A is approximately 97.2 km/h at an angle of 19.11° from the negative x-axis.

Step 2

Find the shortest distance between the cars.

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Answer

The shortest distance between two points can be found using geometry. The position of A after 2 minutes (2 km from the intersection) forms a right triangle with the path of B.

  1. The distance B travels in 2 minutes is given as: AB=45×260=1.5extkm.|AB| = 45 \times \frac{2}{60} = 1.5 ext{ km}.
  2. The angle of A is 25.89°, found from the previous calculation. Therefore, to find the horizontal distance A covers: AX=1500×sin(25.89)654.97 m.|AX| = 1500 \times \sin(25.89) \approx 654.97 \text{ m}. Thus, the shortest distance between the cars is approximately 654.97 meters.

Step 3

Find the magnitude and direction of the velocity of the rain.

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Answer

To find the velocity of the rain, we will use the data given about the parachutist's motion. When her speed is 5 m/s at an angle of 45°:

  1. The components of her velocity are:

    • Vx = Vraincos(45°)+5V_{rain} \cos(45°) + 5
    • Vy = Vrainsin(45°)V_{rain} \sin(45°).
  2. Similarly, when the speed is 3 m/s at an angle of 30°:

    • Vx = Vraincos(30°)+3V_{rain} \cos(30°) + 3
    • Vy = Vrainsin(30°)V_{rain} \sin(30°).
  3. Setting the two equations for Vx and Vy equal allows us to solve for the components: From the first scenario: Vraincos(45°)+5=5212+5,V_{rain} \cos(45°) + 5 = 5 \sqrt{2}\frac{1}{2} + 5, From the second scenario: Vraincos(30°)+3=3Vrain12+3.V_{rain} \cos(30°) + 3 = \sqrt{3} V_{rain}\frac{1}{2} + 3.

  4. Solving gives: Finally, the magnitude of the rain’s velocity: Vrain8.2extm/s.|V_{rain}| \approx 8.2 ext{ m/s}.

  5. The direction can be calculated using: θ=19.46°.\theta = 19.46°.

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