Two cars, A and B, travel along two straight roads which intersect at an angle of 135° - Leaving Cert Applied Maths - Question 2 - 2015
Question 2
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.
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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.
The velocity of car A, extbfvA=60extkm/hi^.
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=60×301=2extkm.
Hence, car B travels towards the intersection with velocity
extbfvB=45 km/hextatanangleof135°.
Converting this to components gives:
vB=45cos(135°)i^+45sin(135°)j^≈−31.82i^+31.82j^.
The relative velocity of B to A is given by: vB relative to A=vB−vA=(−31.82i^+31.82j^)−(60i^)=−91.82i^+31.82j^.
The magnitude of the relative velocity is:
∣vB relative to A∣=(−91.82)2+(31.82)2≈97.2 km/h.
To find the direction, use:
α=tan−1(−91.8231.82)≈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.
The distance B travels in 2 minutes is given as:
∣AB∣=45×602=1.5extkm.
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.
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°:
The components of her velocity are:
Vx = Vraincos(45°)+5
Vy = Vrainsin(45°).
Similarly, when the speed is 3 m/s at an angle of 30°:
Vx = Vraincos(30°)+3
Vy = Vrainsin(30°).
Setting the two equations for Vx and Vy equal allows us to solve for the components:
From the first scenario:
Vraincos(45°)+5=5221+5,
From the second scenario:
Vraincos(30°)+3=3Vrain21+3.
Solving gives:
Finally, the magnitude of the rain’s velocity:
∣Vrain∣≈8.2extm/s.
The direction can be calculated using:
θ=19.46°.
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