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Chapter 3: Problem 81

A jetliner flies east for \(600.0 \mathrm{km},\) then turns \(30.0^{\circ}\) toward the south and flies another \(300.0 \mathrm{km} .\) (a) How far is the plane from its starting point? (b) In what direction could the jetliner have flown directly to the same destination (in a straight-line path)? (c) If the jetliner flew at a constant speed of \(400.0 \mathrm{km} / \mathrm{h}\), how long did the trip take? (d) Moving at the same speed, how long would the direct flight have taken?

Short Answer

Expert verified
Answer: There is a time difference of approximately 0.067 hours between the two-part trip and the direct flight.

Step by step solution

01

Calculate the components of the two displacement vectors

To begin, we can represent the first displacement (flying east) as a vector, and the second displacement (rotating 30 degrees towards south) as another vector. Let's call these vectors A and B. We can find the components of A and B by using their magnitudes and the given angles. For displacement A (600.0 km east): Ax = 600.0 km (due east) Ay = 0 km (no vertical component) For displacement B (300.0 km at 30 degrees south of east): Bx = 300.0 \times \cos{(30^{\circ})} = 300.0 \times \frac{\sqrt{3}}{2} \approx 259.81 \mathrm{km} \\ By = - 300.0 \times \sin{(30^{\circ})} = - 300.0 \times \frac{1}{2} = - 150.0 \mathrm{km} (south is negative)
02

Find the total displacement of the plane

Now, we can calculate the total displacement vector (C) by adding vectors A and B component-wise: Cx = Ax + Bx = 600.0 + 259.81 = 859.81 \mathrm{km} \\ Cy = Ay + By = 0 - 150.0 = -150.0 \mathrm{km} The magnitude of the total displacement vector C can be found using the Pythagorean theorem: \(C = \sqrt{Cx^2 + Cy^2} = \sqrt{(859.81)^2 + (-150.0)^2} \approx 873.21 \mathrm{km}\) Thus, the plane is 873.21 km away from its starting point.
03

Find the angle of the direct flight path

We can find the angle, θ, of the direct flight path by using the inverse tangent function (arctan): θ = arctan (Cy / Cx) = arctan (-150.0 / 859.81) = -9.69° The angle is negative because south (180°) is less than east (90°). Therefore, the angle of the direct flight path is 9.69° south of the east.
04

Calculate the time of the direct flight

We're given that the jetliner flew at a constant speed of 400.0 km/h. The total time for the trip (part c) can be found by adding the times for each leg: t1 = A / v = 600.0 / 400 = 1.5 \mathrm{h} \\ t2 = B / v = 300.0 / 400 = 0.75 \mathrm{h} t_total = t1 + t2 = 1.5 + 0.75 = 2.25 \mathrm{h} Now we calculate the time to fly the direct path (part d): t_direct = C / v = 873.21 / 400 \approx 2.183 \mathrm{h} So, the jetliner took 2.25 hours to complete the two-part trip, while the direct flight would have taken 2.183 hours.

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Most popular questions from this chapter

A particle has a constant acceleration of \(5.0 \mathrm{m} / \mathrm{s}^{2}\) to the east. At time \(t=0,\) it is \(2.0 \mathrm{m}\) east of the origin and its velocity is \(20 \mathrm{m} / \mathrm{s}\) north. What are the components of its position vector at \(t=2.0 \mathrm{s} ?\)
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