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Mamont248 [21]
3 years ago
15

Adhira loves to ride her bike around the neighborhood. She starts riding 1.2 miles at 30° S of E. Then, she rides another 2.0 mi

les at 20° W of S. Next, she rides 1.6 miles at 30° S of W. And finally, she rides 2.6 miles at 15° W of N. What is her final displacement from her starting point?
Physics
1 answer:
zepelin [54]3 years ago
4 0

Answer:

D = 1.8677 miles , θ = 24.28º at South of West

Explanation:

This is an exercise in adding vectors, the easiest way to solve them is to decompose the vectors and add each component algebraically. Let's use trigonometry

first displacement. d = 1.2 miles to 30º south of East

     cos ( 360-30) = cos (-30) = x₁ / d

     sin (-30) = y₁ / d

     x₁ = d cos (-30)

     y₁ = d sin (-30)

     x₁ = 1.2 cos (-30) = 1,039 miles

     y₁ = 1.2 sin (-30) = -0.6 miles

second shift. d = 2.0 miles to 20º West of South

       cos (270-20) = x₂ / d

       cos (250) = y₂ / d

       x₂ = 2.0 cos 250 = -0.684 miles

       y₂ = 2.0 sin250 = -1.879 miles

Third displacement. d = 1.6 miles to 30º South of West

       cos (180 + 30) = x₃ / d

       sin (210) = y₃ / d

       x₃ = 1.6 cos 210 = -1.3856 miles

       y₃ = 1.6 sin 210 = -0.8 miles

Fourth displacement. d = 2.6 miles to 15º West of North

       cos (90 + 15) = x₄ / d

       sin (105) = y₄ / d

       x₄ = 2.6 cos 105 = -0.6729 miles

       y₄ = 2.6 sin 105 = 2,511 miles

having all the components we add

x-axis  (West-East direction)

       X = x₁ + x₂ + x₃ + x₄

       X = 1.039 -0.684 - 1.3846 - 0.6729

       X = -1.7025 miles

   

       Y = y₁ + y₂ + y₃ + y₄

       Y = -0.6 -1.879 -0.8 +2.511

       Y = -0.768

The modulus of this displacement is we use the Pythagorean theorem

      D = √ (X² + Y²)

      D = √ (1.7025² + 0.768²)

      D = 1.8677 miles

let's use trigonometry to find the direction

       tan θ = Y / X

       θ = tan⁻¹ Y / x

       θ = tan⁻¹ (0.768 / 1.7025)

       θ = 24.28º

as the two components are negative this angle is in the third quadrant

therefore in cardinal direction form is

         θ = 24.28º at South of West

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<h3>What is velocity?</h3>

The change of distance with respect to time is defined as speed. Speed is a scalar quantity. It is a time-based component. Its unit is m/sec.

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A seagull flies at a velocity,\rm v_{SA}  = 9 \ m/sec

The time the bird takes,t=18.0 min

The distance traveled relative to the earth = 6.00 km

a)

The seagull's relative velocity with reference to the ground as;

\rm v_{sg} = \frac{6.00 \times 10^3 \ m }{(20 min) \times \frac{60 s }{1 \ min}} \\\\ v_{sg}= 5.00 \ m/sec

Air velocity with reference to the ground is;

\rm v_{AG}= v_{SG}-v_{SA} \\\\ v_{AG} = 5.00 \ m/sec - 9.00 \ m/sec \\\\ v_{AG} = -4.00 \ m/sec

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If the bird turns around and flies with the wind, The time will he take to return 6.00 km is;

\rm v_{SG}=v_{SA}+v_{AG} \\\\ v_{SG}=-900 \ m/sec +(-4.00 \ m/sec) \\\\ v_{SG}= -13.00 \ m/sec

The time the bird takes;

\rm t = \frac{x_{SG}}{v_{SG}} \\\\ t = \frac{6.00 \times 10^3 \ m }{13.00 \ m/sec } \\\\ t = 462 m/sec \\\\ t = 7  \ min \  and  \ 42  \ sec

c)\

The total round-trip time compared to what it would be with no wind. is;

\rm  t = 20 \ min( \frac{60 \ sec }{1 \ min} )+ 462 \ sec \\\\ t = 1200 \ sec +6 462 \ ec \\\\ t= 1662 \ sec

The time for the round trip is;

\rm  t = \frac{12 \times 10^ 3 }{ 9 \ m/sec }  \\\\ t  = 1333.33 \ sec

Hence the wind's speed, the time bird would need to fly back the total round-trip time will be  4 m/s, 7 minutes and 42 seconds and 1333.33 sec.

To learn more about the velocity, refer to the link: brainly.com/question/862972.

#SPJ1

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