resubmit

course phy 201

this is the rework on the homework you requested.

ran prob 3 v 4course phy 201

What are the acceleration and final velocity of an object whose average velocity, accelerating uniformly from rest for 7.9 seconds, is 59 cm/s? How far does the object travel during the 7.9 seconds?

well since final velocity is double the average velocity so final velocity is 118 cm/sec

acceleration is the change in velocity / the change in elasped time

so 59 cm/sec / 7.9 sec = 7.5cm

For the 7.9 sec interval:

What was the initial velocity?

&&&& if it is accelerating uniformly from rest then v0=0

What was the final velocity?

&&&& meaning the vf=2 * vAve = 2*59cm/s=118cm/s

What therefore is the acceleration?

&&&& so acceleration is the change in velocty/ the change in elapsed time

thus vf-v0 = 118-0=118cm/s

tf-t0 = 7.9-0=7.9s

118cm/s/7.9s=14.9cm/s/s

Good.

to determine the final position is 7.9s * 59cm/s = 466.1 cm/s/s

7.9 is how long we accerlerated and 59 shows HOW we are moving.

Good, except for one detail.

Please respond with a copy of this document, including the questions I've posed. Insert your own answers, or if you have them questions, indicating your insertions by &&&&.

now lets try this again.

Good revision.

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Let me know if you have questions. &#