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course PHY 241
545pm 10/23/2011
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?
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.
v_0 = 11 cm/s
v_f = 15 cm/s
v_ave = (15 cm/s + 11 cm/s) / 2
= 13 cm/s
'dv = (v_f - v_0)
= (15 cm/s - 11 cm/s)
= 4 cm/s
'ds = 117 cm
v_ave = 'ds/'dt
'dt = 'ds / v_ave
= 117 cm / [(15 cm/s + 11 cm/s) /2]
= 9 s
a_ave = 'dv/'dt
= (4 cm/s) / (9 s)
= .44 cm/s^2
vf = v0 + a * `dt
= 11 cm/s + (.444 cm/s^2 * 117 cm)
= 62.948 cm/s
= 63 cm/s
v_0 = 11 cm/s
a = .444 cm/s^2
'ds = 117 cm
'ds = (v0 + vf) / 2 * `dt
'ds/'dt = (v_0 + v_f) / 2
'dt/'ds = 2 / (v_0 + v_f)
'dt = (2 * 'ds) / (v_0 + v_f)
= 234 cm / (74 cm/s)
= 3.2 s
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Self-critique (if necessary):
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Self-critique rating:
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@& You're using the definitions rather that the equations, but you know how to do this either way, so it's OK.*@