Solution: By using Newton's First Law of Motion: VFinal = Uinitial + (acceleration x time) Part 1: Let us integrate that the car with an initial velocity of 13.89 m/s comes to a stop at 3 seconds and reaches a final velocity of 0 m/s2 This event is calculated as follows: 0m/s2 = 13.89 m/s + (a x 3s) %3D a = S-13.89 dt -13.89t +c a = -13.89 () + (0 m/s2) [c, = 0 m/s: a = -13.89 t+0 m/s²] 13.89 a = - 3 1 - t= 3 a = -4. 63 m/s2 The distance traveled during this time is given by: s = Vinitial X time + [5 (acceleration) x (time) ] s = (-13.89 t+ 0m/s ) dt (-13.89 t + 0 m/s*) dt --13.89t + s = (13.89 m/s) x (3 s) +[(-4.63 m/s) x (3 s) ] s = 41.67 m - 20.835 m S 20.84 m

Solution: By using Newton's First Law of Motion: VFinal = Uinitial + (acceleration x time) Part 1: Let us integrate that the car with an initial velocity of 13.89 m/s comes to a stop at 3 seconds and reaches a final velocity of 0 m/s2 This event is calculated as follows: 0m/s2 = 13.89 m/s + (a x 3s) %3D a = S-13.89 dt -13.89t +c a = -13.89 () + (0 m/s2) [c, = 0 m/s: a = -13.89 t+0 m/s²] 13.89 a = - 3 1 - t= 3 a = -4. 63 m/s2 The distance traveled during this time is given by: s = Vinitial X time + [5 (acceleration) x (time) ] s = (-13.89 t+ 0m/s ) dt (-13.89 t + 0 m/s*) dt --13.89t + s = (13.89 m/s) x (3 s) +[(-4.63 m/s) x (3 s) ] s = 41.67 m - 20.835 m S 20.84 m

Name: Given the initial velocity as 13.89 m/s, can you perform/integrate the
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Description: Given the initial velocity as 13.89 m/s, can you perform/integrate the solution/function of acceleration and distance based on the typewritten formula / highlighted parts. Thank you Solution:By using Newton's First Law of Motion:Final =Uinitial + (ac

Principles of Physics: A Calculus-Based Text

5th Edition

ISBN:9781133104261

Author:Raymond A. Serway, John W. Jewett

Publisher:Raymond A. Serway, John W. Jewett

Chapter3: Motion In Two Dimensions

Section: Chapter Questions

Problem 18P

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