RandomizedProblemQuiz2

course Phy121

6/27 11

Reason out the quantities v0, vf, `dv, vAve, a, `ds and `dt: If an object’s velocity changes at a uniform rate from 10 cm/s to 13 cm/s as it travels 57.5 cm, then what is the average acceleration of the object?Using the equations which govern uniformly accelerated motion determine vf, v0, a, `ds and `dt for an object which accelerates through a distance of 57.5 cm, starting from velocity 10 cm/s and accelerating at .6 cm/s/s.

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

57.5 cm= (10 cm/sec+ 13 cm/sec)/2* `dt

`dt= 5 sec

aAve=`dv/`dt

aAve= (13 cm/sec- 10 cm/sec)/(5 sec)

aAve= 3/5 cm/sec^2

v0= 10 cm/sec

a= 0.6 cm/sec^2

`ds= 57.5 cm

vf^2 = v0^2 + 2 a `ds

vf^2= 10^2+ 2 (0.6)*(57.5)

vf= 13 cm/sec

vf = v0 + a * `dt

13 cm/sec= 10 cm/sec+ 0.6 cm/sec^2* `dt

`dt= 5 sec

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