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Phy 121
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** CQ_1_07.2_labelMessages **
Determine the acceleration of an object whose velocity is initially 15 cm/s and which accelerates uniformly through a distance of 51 cm in 4.1 seconds.
I know:
v0 = 15cm/s
`ds = 51cm
`dt = 4.1s
WTF:
vf = 9.88cm/s
a = -1.25cm/s^2
`ds = (v0 + vf) / 2 * `dt
(`ds / `dt)* 2 - v0 = vf
(51cm / 4.1s) * 2 - 15cm/s = 9.88cm/s
vf = v0 + a `dt
(vf - v0) / `dt = a
(9.88cm/s - 15cm/s) / 4.1s = -1.25cm/s^2
** **
30 minutes
&#Very good responses. Let me know if you have questions. &#