An engineer designs a roller coaster so that a car travels horizontally for 202 ft, then climbs 147 ft at an angle of 32.0° above the horizo

Question

An engineer designs a roller coaster so that a car travels horizontally for 202 ft, then climbs 147 ft at an angle of 32.0° above the horizontal. It then moves 147 ft at an angle of 47.0° below the horizontal. If we take the initial horizontal motion of the car to be along the +x-axis, what is the car’s displacement? (Give the magnitude of your answer, in ft, to at least four significant figures and give the direction of your answer in degrees counterclockwise from the +x-axis.)

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Hồng Cúc 3 years 2021-07-15T13:35:21+00:00 1 Answers 384 views 0

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  1. Dau
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    2021-07-15T13:37:01+00:00
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    Answer: The car’s displacement has magnitude 228.3381 ft and direction of 7.46°.

    Explanation: Displacement is defined as the change in position of an object. It is a vector, so it has magnitude and direction.

    To determine the car’s displacement, divide its route into 3 parts:

    1) Horizontally for 202ft;

    2) Climbs 147ft at an angle of 32°;

    3) Moves 147ft at an angle of 47° below the horizontal;

    Now calculate each x- and y- displacement of each part.

    Part (1):

    The car only displaces on x-axis, so displacement is 202ft.

    Part (2):

    The car climbs at an angle with the horizontal, so it creates a right triangle with an angle of 32 with the horizontal line. Then displacement is

    d_{x} = 147cos(32) = 124.6631

    d_{y}=147sin(32) = 77.8981

    Part (3):

    The car goes down forming an angle below the horizontal. This trajectory also form a right triangle with an external angle. Knowing that alternate interior angles are congruent, the angle the right triangle forms with the horizontal line is 47°. So, displacement is

    d_{x}=147cos(47) = -100.2537

    d_{y}=147sin(47) = -107.509

    Displacements in x and y are negative because they point towards the negative side of the reference.

    To calculate total displacement:

    d_{total}=\sqrt{d_{Tx}^{2}+d_{Ty}^{2}}

    Total displacement at each coordinate are:

    d_{Tx} = 202 + 124.6631 – 110.2537 = 226.41

    d_{Ty} = 77.8981 – 107.509 = -29.611

    d_{total}=\sqrt{(226.41)^{2}+(-29.611)^{2}}

    d_{total} = 228.3381

    The car’s displacement is 228.3381 ft.

    For the direction, determine reference angle by:

    \theta = tan^{-1}(\frac{d_{Ty}}{d_{Tx}} )

    \theta = tan^{-1}(\frac{29.611}{226.41} )

    θ = 7.46°

    The direction of the car is at 7.46°.

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