Query 4

course Phy 201

I am still not understanding when to use Vave and knowing when acceleration is uniform since Average Velocity is calulated differently if acceleration is uniform as indicated in epxeriment 4 and random problem 4. Thanks.

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Average velocity is `ds / `dt. If acceleration is not uniform then this is the only definition that applies.

If acceleration is uniform then vAve is also equal to (vf + v0) / 2.

assignment #004

004. `Query 4

Physics I

02-10-2008

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16:09:12

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 -->

Dt=Dv/Aave

confidence assessment: 0

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16:10:01

**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 -->

a * `dt = `dv

v0 + `dv = vf

(vf + v0) /2 = vAve

vAve * `dt = `ds so

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

self critique assessment: 3

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16:10:55

What is the displacement `ds associated with uniform acceleration from velocity v0 to velocity vf in clock time `dt?

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RESPONSE -->

Dv=vf-v0

confidence assessment: 0

If acceleration is uniform then vAve = (vf + v0) / 2.

Since vAve = `ds / `dt, then `ds = vAve * `dt, so for the given situation we have

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

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16:12:47

Query Add any surprises or insights you experienced as a result of this assignment.

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RESPONSE -->

I am still not understanding when to use Vave and knowing when acceleration is uniform since Average Velocity is calulated differently if acceleration is uniform as indicated in epxeriment 4 and random problem 4. Thanks.

confidence assessment: 3

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You're not doing badly. You have asked some very pertinent questions and are making a good effort to understand this.

See my notes and let me know if they don't clarify the situation for you.