Find the distance between the following points R(-2, 3), S(3, 15)
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Answer:
a. The distance the train travelled in the first hour is approximately 28.3 miles
b. The location of the train at 5:00 p.m. is 53 miles east, and 46 miles west
c. The location of the train at any given time by the function, f(t) = (-8 + 24·t, -10 + 15·t)
d. The train does not collide with the cyclist when the bike goes over the train tracks
Step-by-step explanation:
a. The given information on the train's motion are;
The location south the train is spotted = 10 miles south and 8 miles west
The time the observer spotted the train = 2:00 pm
The location the train is spotted at 3:00 p.m. = 5 miles north and 16 miles east
Therefore, the difference between the two times the train was spotted, t = 3:00 p.m. - 2:00 p.m. = 1 hour
Making use of the coordinate plane for the two locations the train was spotted, we have;
The initial location of the train = (-10, -8)
The final location of the train = (5, 16)
Therefore the distance the train travelled in the first hour is given by the formula for finding the distance, 'd', between two points, (x₁, y₁) and (x₂, y₂) as follows;
Therefore;
The distance the train travelled in the first hour, d = 3·√89 ≈ 28.3 miles
b. The speed of the train, v = (Distance travelled by the train)/Time
∴ v ≈ 28.3 miles/(1 hour) = 28.3 miles per hour
The speed of the train in the first hour, v ≈ 28.3 mph
The direction of the train, θ, is given by the arctangent of the slope, 'm', of the path of the train;
∴ m = tan(θ) = (5 - (-10))/(16 - (-8)) = 0.625
c. Distance = Velocity × Time
At 5:00 p.m., we have;
The time difference, Δt = 5:00 p.m. - 3:00 p.m. = 2 hours
The distance, d₁ = (28.3 mph × 2 hours = 56.6 miles
Using trigonometry, we have the horizonal distance travelled, 'Δx', in the 2 hours is given as follows;
Δx = d₁ × cos(θ)
∴ Δx = 56.6 × cos(arctan(0.625)) ≈ 48
The increase in the horizontal position of the train, relative to the point (5, 16), Δx ≈ 48 miles
The vertical distance increase in the two hours, Δy is given as follows;
Δy = 56.6 × sin(arctan(0.625)) ≈ 30
The increase in the vertical position of the train, relative to the point (5, 16), Δy ≈ 30 miles miles
Therefore; the location of the train at 5:00 p.m. = ((5 + 48), (16 + 30)) = (53, 46)
The location of the train at 5:00 p.m. = 53 miles east, and 46 miles west
c. The function, 'f', that would give the train's position at time-t is given as follows;
The P = f(28.3·t, θ)
Where;
28.3·t = √(x² + y²)
θ = arctan(y/x)
Parametric equations
y - 5 = 0.625·(x - 16)
∴ y = 0.625·x - 10 + 5
The equation of the train's track is therefore, presented as follows;
y = 0.625·x - 5
d = 28.3·t
The y-component of the velocity, = 3*√89 mph × sin(arctan(0.625)) = 15 mph
Therefore, we have;
y = -10 + 15·t
The x-component of the velocity, vₓ = 3*√89 mph × cos(arctan(0.625)) = 24 mph
Therefore, we have;
x = -8 + 24·t
The location of the train at any given time, 't', f(t) = (-8 + 24·t, -10 + 15·t)
d. The speed of the cyclist next to the observer at 2:00 p.m., v = 10 mph
The distance of the cyclist from the track = The x-intercept = 5/0.625 = 8
The distance of the cyclist from the track = 8 miles
The time it would take the cyclist to react the track, t = 8 miles/10 mph = 0.8 hours
The location of the train in 0.8 hours, is f(0.8) = (-8 + 24×0.8, -10 + 15×0.8)
∴ f(0.8) = (11.2, 2)
At the time the cyclist is at the track along the east-west axis, at the point (8, 0), the train is at the point (11.2, 2) therefore, the train does not collide with the cyclist when the bike goes over the train tracks.