course Phy 121
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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 11 cm/s to 15 cm/s as it travels 117 cm, then what is the average acceleration of the object?
vo = 11cm/s
`dv = vf - vo = 15cm/s - 11cm/s = 4cm/s
`ds = 117cm
vAve = (vf + vo)/2 = (15cm/s + 11cm/s)/2 = 13cm/s
`dt = `ds/vAve = 117cm/13cm/s = 9 seconds
aAve = `dv/`dt = (4cm/s)/(9s) = 0.444cm/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 117 cm, starting from velocity 11 cm/s and accelerating at .444 cm/s/s.
Impossible situation, must use fourth equation:
vf^2 = vo^2 + 2(a)(`ds)
vo = 11cm/s
a = 0.444cm/s^2
`ds = 117cm
vf^2 = (11cm/s)^2 + 2(.444cm/s^2)(117cm)
vf^2 = 121cm/s + 103.896
vf^2 = 224.896
TAKE SQUARE ROOT OF BOTH SIDES
vf = 14.996cm/s = 15cm/s
aAve = `dv/`dt
`dt = `dv/aAve = (15cm/s)/(0.444cm/s^2) = 9 seconds"
Very good. Let me know if you have questions.