Part C Learning Goal: To analyze the motion of a particle using a translating fr

Question

Part C Learning Goal: To analyze the motion of a particle using a translating frame of reference. Two particles, A and B, are moving along arbitrary paths and are at positions ry and rs from a common origin. The relative position of point B with respect to A is designated by a relative-position vector rBA and is specified with the equation rs =A+IBA Taking the time derivative of the relative-position equation above results in the following equation that After crossing the ocean current, the speedboat continues to approach the cruise ship at U2 = 19.3 m/s and at an angle = 57.0°. The cruise ship sees the speedboat and slows down by 2.5 m/s. At the same time, the speedboat docks with the cruise ship by turning with a radius of curvature of r = 20.0 m and by reducing its speed to match that of the cruise ship once it is parallel to the cruise ship. (Figure 3) What area, and ay, the scalar components of the speedboat's acceleration relative to the cruise ship's when the speedboat is docked parallel to the cruise ship? Assume that the cruise ship's acceleration and the speedboat's transverse acceleration are constant. Express your answers to three significant figures in meters per squared second separated by a comma. » View Available Hint(s) Figure 1 of 3 ( > 1o Axpo vec 6 - 0 2 ? azdy = m/s2 Submit Previous Answers X Incorrect; Try Again (Return to Aecionment Provide feedback 5.18 AM 2017-12...0.26

Explanation / Answer

The cruie ship slows by 2.5 m/s in some time T so it has an accleration

a1 =- v1' - v1/ T= 2.5 m/s/T ( i)

since the speed boat is traveling in a circular path its accleration is

a2 = an + at

when the speedboat and cruise ship are parallel

a2 = - an i - atj

an = v2'^2/ r1 = v1'^2/ r1 = ( 12.5-2.5)^2/20 = 5 m/s^2 i

the change in the speed boat speed

v2'^2 = v2^2 + 2 at( s- so)

s- so = 2pi r1( 90-57/360)

s-so = pi r1/5.45

v2'^2 - v2^2 = 2 at (  pi r1/5.45)

at = v2'^2 - v2^2 ) 5.45/ 2 * pi ( 20 m)

= ( 12.5-2.5)^2 - (19.3)^2 * 5.45/ 40 pi

= -11.81 m/s^2 j

a2 = -5 m/s^2 i - 11.81 m/s^2 j

---------------------------------------------------------

T = v2'- v2/ at

= ( 12.5-2.5)-19.3/-11.81

=0.787 s

a1 = - 2.5/T ( j) = -2.5/0.787 = -3.17 j

a21 = a2- a1

a21 =  5 m/s^2 i - 11.81 m/s^2 j+3.17 j

= 5 m/s^2 i -8.64 m/s^2 j

ax = 5 m/s^2 i

ay = -8.64 m/^2 j

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