course PHY 121
8/11
Using the equations which govern uniformly accelerated motion determine vf, v0, a, s and t for an object which starts at velocity 10 cm /s and accelerates at .333 cm/s/s through a distance of 103.5 cm. v0 = 10cm/s
`ds = 103.5cm
aAve = .333cm/s^2
..if vf^2 = v0^2 + 2 a `ds, then vf^2 = (10cm/s)^2 + [2*.333cm/s^2*103.5cm] = 100cm/s^2 + 68.931 = 168.9 = vf^2 = sqr[168.9] = 12.99 = ~13cm/s = vf
`dt = `ds/vAve. vAve = (13+10)/2 = 11.5cm/s - - (103.5cm)/(11.5cm/s)= 9s
Reason out the quantities v0, vf, v, vAve, a, s and t: If an object’s velocity changes at a uniform rate from 10 cm/s to 13 cm/s as it travels 103.5 cm, then how long does it take to cover the distance?
`ds = 103.5cm
since, vAve = `ds/`dt, we can rearrange the equation to read `dt = `ds/vAve. vAve = 10+13/2 = 11.5. So `dt = 103.5/11.5 = 9s.
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