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course PHY121
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?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?
vA = `ds / `dt
`dt = `ds / vA
`dt = 117cm / 13cm/s = 9s
aA = `dv / `dt = 4cm/s / 9s = 0.4cm/s2
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.
`ds = 117cm
v0 = 11cm/s
a = 0.444cm/s2
vf2 = v02 + 2 a*ds = (11cm/s)2 + (2 * 0.444cm/s2 * 117cm) = 121cm2/s2 + 102.96cm2/s2 = 224.96cm2/s2
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I know what this means and it's good, but for future clarity you want to use the ^ to indicate expontiation (e.g., vf^2). You are likely going to the trouble to put that 2 in the exponent, which takes longer than writing ^2 and doesn't come through properly in a text-based form.
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vf2 = 224.96cm2/s2, sqrt (vf2) = sqrt (224.96cm2/s2), vf = 15cm/s
`dt = `ds / vA = 117cm / 13cm/s = 9s
&#Good responses. See my notes and let me know if you have questions. &#