Class l Help 04-Kinematics in 2 Dimensions Begin Date: 9/6/2018 11:00:00 AM-Due

Question

Class l Help 04-Kinematics in 2 Dimensions Begin Date: 9/6/2018 11:00:00 AM-Due Date: 9/11/2018 10:50:00 AM End Date: 9/14/2018 10:50:00 AM (16%) Problem ll: A particle moves in the xy plane with a constant acceleration of a-(2.691 + 9.75 j) ms. At time t-o seconds, the particle is located at (1.51+4.63 j) m and has a velocity of (7.29 1 - 3.05) m/s. 33% Part (a) How fast in m s is the particle traveleing at t-4.25 seconds? Grade Summa Deductions Potential Speed- 0% 100% tan0 cosO cotan0asin) acos0 atan0 acotan sinh0 cosh0 tanh cotanh0 sin0 Submissions Attempts remaining: ó 26 per attempt) detailed view END Degrees Radians CLEAR Submit I give up! Hints: for a 0% deduction. Hints remaining:0 Feedback: 0% deduction per feedback. 33% Part (b) what direction is the particle traveling at t = 425 seconds? Give the angle in degrees from the positive x direction. 33% Part (c) How far in meters is the particle from the xy origin at 4.25 seconds? All content 8 2018 Expert TA,

Explanation / Answer

here,

accelration , a = (2.69 i + 9.75 j ) m/s^2

at t = 0 s ,

r0 = ( 1.51 i + 4.63 j) m

v0 = 7.29 i m - 3.05 j m

a)

at t = 4.25 s

final speed , v = v0 + a * t

v = 7.29 i m - 3.05 j + (2.69 i + 9.75 j ) * 4.25 m/s

v = 18.7 i m/s + 38.38 j m/s

the speed of particle , |v| =sqrt(18.7^2 + 38.38^2)

|v| = 42.7 m/s

b)

the direction of particle's speed , theta = arctan(38.38/18.7)

theta = 64 degree counterclockwise from +x axis

c)

the distance travelled , s - r0 = v0 * t + 0.5 * a * t^2

s - ( 1.51 i + 4.63 j) = (7.29 i m - 3.05 j) * 4.25 + 0.5 * (2.69 i + 9.75 j ) * 4.25^2

s = 56.8 i m + 79.7 j m

the distance travelled , |s| = sqrt(56.8^2 + 79.7^2)

|s| = 97.9 m

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