Based on the graph below, what was the accelerationof the object between 0.0 s and 5.0 s? Round youranswer to 3 sig figs.16.0 -0.0s, 12.0 ms)12.08.0(5.0 s,0.0 ms)4.00.0++ time (s)0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0a-2.00 m/s/sb-2.40 m/s/s3.20 m/s/sd-2.80 m/s/se2.00 m/s/sf2.80 m/s/s-3.20 m/s/sh2.40 m/s/svelocity (m/s)

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Description: Based on the graph below, what was the accelerationof the object between 0.0 s and 5.0 s? Round youranswer to 3 sig figs.16.0 -0.0s, 12.0 ms)12.08.0(5.0 s,0.0 ms)4.00.0++ time (s)0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0a-2.00 m/s/sb-2.40 m/s/s3.2

Acceleration a=vf-vitf-ti =0-12.0 ms5.0 s-0.0 s =-12.0 ms5.0 s =-2.40 m/s/s

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Based on the graph below, what was the acceleration of the object between 0.0 s and 5.0 s? Round your answer to 3 sig figs. 16.0 0.0s,120 ms) 12.0 (50s, 0.0 m's) - time (s) 4.0 0.0 0.0 0.5 1.0 1.5 20 2.5 3.0 3.5 4.0 4.5 5.0 -2.00 m/s/s -2.40 m/s/s 3.20 m/s/s d. -2.80 m/s/s 2.00 m/s/s f 2.80 m/s/s -3.20 m/s/s 2.40 m/s/s (s/1) fipopa

Based on the graph below, what was the acceleration of the object between 0.0 s and 5.0 s? Round your answer to 3 sig figs. 16.0 0.0s,120 ms) 12.0 (50s, 0.0 m's) - time (s) 4.0 0.0 0.0 0.5 1.0 1.5 20 2.5 3.0 3.5 4.0 4.5 5.0 -2.00 m/s/s -2.40 m/s/s 3.20 m/s/s d. -2.80 m/s/s 2.00 m/s/s f 2.80 m/s/s -3.20 m/s/s 2.40 m/s/s (s/1) fipopa

Glencoe Physics: Principles and Problems, Student Edition

1st Edition

ISBN:9780078807213

Author:Paul W. Zitzewitz

Publisher:Paul W. Zitzewitz

Chapter3: Accelerated Motion

Section: Chapter Questions

Problem 59A

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