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An unsuspecting bird is coasting along in an easterly direction at 4.00 mph when a strong wind from the south imparts a constant acceleration of 0.200 m/s2. If the acceleration from the wind lasts for 2.70 s, find the magnitude, r, and direction, θ, of the bird's displacement during this time period. (HINT: assume the bird is originally travelling in the x direction and there are 1609 m in 1 mile.) a) 3.59 m, and 8.59º
b) 3.59 m, and 17.2º
c) 4.88 m and 8.59º
d) 4.88 m and 17.2º

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Answer:

The correct answer is option 'c'. [tex](4.88m,8.59^{o})[/tex]

Explanation:

Since for motion along orthogonal axis are independent thus we conclude that the motion of the bird is unchanged along the easterly direction.

Speed in easterly direction is 4.00 mph which is converted to meters per second as

[tex]4.00mph=\frac{4\times 1609}{3600}=1.79m/s[/tex]

Thus the distance covered in x-direction for 2.7 seconds equals

[tex]x=2.7\times 1.79=4.833m[/tex]

Now in for the motion in southern direction is uniformly accelerated thus the distance covered in 2.7 seconds can be evaluated using second equation of kinematics as

[tex]y=ut+\frac{1}{2}at^{2}\\\\s=\frac{1}{2}\times 0.2\times (2.7)^{2}\\\\\therefore y=0.729m[/tex]

This the displacement is given by

[tex]r=\sqrt{x^{2}+y^{2}}\\\\r=\sqrt{4.833^{2}+0.729^{2}}\\\\\therefore r=4.88[/tex]

The angle is given by

[tex]\theta =tan^{-1}\frac{y}{x}\\\\\theta =tan^{-1}\frac{0.729}{4.833}\\\\\theta =8.59^{o}[/tex]

snehashish65

The required value of displacement (r) and the angle are 4.88 m and 8.59 Degree respectively.

Given data:

The speed of bird along the east direction is, [tex]v=4.00 \;\rm mph = 4.00 \times 0.447=1.788 \;\rm m/s[/tex].

The magnitude of acceleration of bird is, [tex]a =0.200 \;\rm m/s^{2}[/tex].

The time interval is, t = 2.70 s.

In the given problem, the motion of the bird is unchanged along the easterly direction. So, the distance covered in x-direction for 2.7 seconds is,

[tex]d = v \times t\\\\d = 1.788 \times 2.70\\\\d= 4.83 \;\rm m[/tex]

Since, the motion in southern direction is uniformly accelerated thus the distance covered in 2.7 seconds can be evaluated using second equation of kinematics as,

[tex]h = ut +\dfrac{1}{2}at^{2}\\\\h = 0 \times t +\dfrac{1}{2} \times 0.200 \times 2.70^{2}\\\\h = 0.729 \;\rm m[/tex]

Then the displacement is given by,

[tex]r=\sqrt{d^{2}+h^{2}}[/tex]

Solving as,

[tex]r=\sqrt{4.83^{2}+0.729^{2}}\\\\r = 4.88 \;\rm m[/tex]

And the angle is given by,

[tex]\theta =tan^{-1}(h/d)\\\\\theta =tan^{-1}(0.729/4.83)\\\\\theta=8.59 ^\circ[/tex]

Thus, the required value of displacement (r) and the angle are 4.88 m and 8.59 Degree respectively.

Learn more about the displacement here:

https://brainly.com/question/12874363

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