From the main two kinematic equations describing one-dimensional motion under constant acceleration 1 x = Xo + vot + at? and v = vo + at one can derive other equations that can be useful in solving constant acceleration problems by solving for one term in one equation above and substituting it into the other one to eliminate it. A) Starting from the main two equations eliminate vo to derive x xo = vt- B) Start from the main two equations and eliminate t to derive a(x – xo) = (v² – v,²). 1 — хо) %3D

Physics for Scientists and Engineers: Foundations and Connections
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3. From the main two kinematic equations describing one-dimensional motion under constant
acceleration
1
x = xo + vot +, at
and
v = vo + at
one can derive other equations that can be useful in solving constant acceleration problems by
solving for one term in one equation above and substituting it into the other one to eliminate it.
A) Starting from the main two equations eliminate vo to derive x xo = vt – at².
B) Start from the main two equations and eliminate t to derive
a(x – x,) = (v² – vo²).
Transcribed Image Text:3. From the main two kinematic equations describing one-dimensional motion under constant acceleration 1 x = xo + vot +, at and v = vo + at one can derive other equations that can be useful in solving constant acceleration problems by solving for one term in one equation above and substituting it into the other one to eliminate it. A) Starting from the main two equations eliminate vo to derive x xo = vt – at². B) Start from the main two equations and eliminate t to derive a(x – x,) = (v² – vo²).
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