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course Phy 121
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 46 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 46 cm, starting from velocity 10 cm/s and accelerating at .75 cm/s/s.
Ds = (vf + v0)/2 * Dt, or, Ds / (vf + v0)/2 = Dt
46cm / (13cm/s + 10cm/s) / 2 = 4s, and since we're told it's uniform we know that (vf + v0)/2 = vAve, or (23cm/s) / 2 = 11.5cm/s
v0 = 10cm/s,
vf = 13cm/s
Dv = 13cm/s - 10cm/s = 3cm/s
vAve = Ds / Dt = 11.5cm/s = 46cm / 4s
a = 3cm/s / 4s = 0.75cm/s^2
Ds = 46cm
vf^2 = v0^2 + 2 a Ds
169cm^2/s^2 = 100cm^2/s^2 + 2(a)(46cm)
169cm^2/s^2 - 100cm^2/s^2 = 2(a)(46cm)
69cm^2/s^2 / 46cm = 2a
1.5cm/s^2 / 2 = a
a = 0.75cm/s^2
To test my numbers,
Ds = v0*Dt + .5(a)(4)^2 = 46cm
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