query 4

course phy 231

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004. `Query 4

Physics I

02-02-2008

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18:51:12

Intro Prob 6 given init vel, accel, `dt find final vel, dist

If initial velocity is v0, acceleration is a and time interval is `dt, then in symbols what are the final velocity vf and the displacement `ds?

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RESPONSE -->

a=v0-vf/dt

confidence assessment: 3

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18:51:45

**You would use accel. and `dt to find `dv:

a * `dt = `dv. Adding `dv to initial vel. vo you get final vel.

Then average initial vel. and final vel. to get ave. vel.:

(v0 + vf) / 2 = ave. vel.

You would then multiply ave. vel. and `dt together to get the distance.

For example if a = 3 m/s^2, `dt = 5 s and v0 = 3 m/s:

3 m/s^2 * 5 s = 15 m/s = `dv

15 m/s + 3 m/s = 18 m/s = fin. vel.

(18 m/s + 3 m/s) / 2 = 10.5 m/s = vAve

10.5 m/s * 5 s = 52.5 m = dist.

In more abbreviated form:

a * `dt = `dv

v0 + `dv = vf

(vf + v0) /2 = vAve

vAve * `dt = `ds so

`ds = (vf + v0) / 2 * `dt.

**

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RESPONSE -->

self critique assessment: 0

&#

Your response did not agree with the given solution in all details, and you should therefore have addressed the discrepancy with a full self-critique, detailing the discrepancy and demonstrating exactly what you do and do not understand about the given solution, and if necessary asking specific questions (to which I will respond).

&#

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18:58:41

** Since accel is uniform vAve = (v0 + vf) / 2.

Thus displacement is

`ds = vAve * `dt = (v0 + vf) / 2 * `dt,

which is the first equation of uniformly accelerated motion. **

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RESPONSE -->

was unsure of what the question was asking

self critique assessment: 1

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19:03:34

Describe the flow diagram we obtain for the situation in which we know v0, vf and `dt.

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RESPONSE -->

in the flow diagram, v0, vf and dt would be independent of each other on the graph. so therefore it would be 3 different lines showing the flow of each

confidence assessment: 2

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See my notes here and in previous assignments regarding self-critique.