course phy121
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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?v0=
vf=13 cm/s
`ds=57.5 cm
`dv = 3cm/s
vAve = (v0+vf)/2 = 11.5cm/s
vAve = `ds/`dt, since we have two of these variables now, we can solve for `dt.
57.5cm/11.5cm/s = 5 seconds
with this we can find the acceleration, 3cm/s/5s = 0.6 cm/s^2
OR
`ds=(vo+vf)/2&`dt ---> 57.5 cm = 11.5cm/s*`dt, therefore `dt = 5s
if `dt = 5s, then a = `dv/`dt or 3cm/s/5s = 0.6 cm/s^2
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.
v0 = 10cm/s
a = .6 cm/s^2
`ds = 57.5 cm
vf^2 = v0^2 + 2a`ds
vf^2 = 100cm^2/s^2 + 69 cm^2/s^2
vf = 10 cm/s + 13 cm/s
vf = 23 cm/s
Then using vf = v0+a*`dt, we get 23cm/s = 10cm/s + 0.6cm/s^2*`dt
21.6s = `dt
This would give a vAve of 57.5/21.6 = 2.66 cm/s
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