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course Phy 231
3/3 1
Reason out the quantities v0, vf, Dv, vAve, a, Ds and Dt: If an object’s velocity changes at a uniform rate from 12 cm/s to 15 cm/s as it travels 94.5 cm, then what is the average acceleration of the object?&&&&
V0 = 12 cm/s, vf = 15 cm/s 'ds = 94.5 cm, 'dv =?, vAve = ?, 'dt = ?
dv = vf - v0 = 15 cm/s - 12cm/s = 3 cm/s
vAve = (vf + v0) /2 = (15 cm/s + 12 cm/s) / 2 = 13.5 cm/s
'dt = 'ds / vAve = 94.5 cm / (13.5 cm/s) = 7 seconds
a = 'dv / 'dt = 3cm/s / 7 seconds = 0.428 cm /s^2
check
'ds = vAve * 'dt = 13.5 cm /s * 7 s = 94.5 cm
vf = vo + a * 'dt = 12 cm/s + 0.439 cm / s^2 * 7 s = 15
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Using the equations which govern uniformly accelerated motion determine vf, v0, a, Ds and Dt for an object which accelerates through a distance of 94.5 cm, starting from velocity 12 cm/s and accelerating at .428 cm/s/s.
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'ds = 94.5 cm, v0 = 12 cm/s, a = .428 cm/s/s,
vf = ?,
vf^2 = v0^2 + 2 * a * ds
vf = (v0^2 +2 * a * ds)^.5 = 15 cm/s
'dt =?
'ds = vave * 'dt
'dt = 'ds / vave = 94.5 cm / (( 12 + 15)/2) = 7 sec
vf = 15 cm/s, 'dt = 7 sec
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