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I am having a head "scratcher moment" with the equations for linear motion.
Im having a real problem getting the two equations for Vf to agree.
When dt is known, the first equation typically given works fine to solve for Vf:
Vf = Vi + A(t)
However,when time is not known the equation given is:
(Vf)^2 = ((Vi)^2) + 2A(X)
Vf = sqrt( ((Vi)^2) + 2A(X) )
where Vf = final velocity, Vi = initial velocity, A = acceleration, X = displacement (Xf-Xi), and t = time in seconds.
Unfortunately for me, the two equations keep yielding a different answer.
If I plug the following data into the first equation
Vi = 0
Xf = 0
Xi = 10
A = -10
and allow that (dt) is known to be 1s, I can easily solve correctly for the rest of the variables:
Vf = -10
Vavg = -5
dV = -10
dX = -10
BUT, if I plug the same data into the second equation and treat dt as unknown, the yield is
Vf = -14.14
Vavg = -7.07
dV = -14.14
dX = -10
because 2A(dX) = 200.
As far as I can tell Im using the common formulas and the formulas are correctly interpreted. Obviously, I am either using the equations in the wrong context, have the equations wrong, or I am making a math error. Yes, Im certain that Im doing SOMETHING wrong, but, for some odd reason my physics teacher just laughed me off nervously when I asked about the discrepancy I am no doubt causing myself and gave no further explanation.
Thanks for your time.
- CJ (age 37)
Paradise Valley Community College, Phoenix, AZ
If V_avg= -5m/s and dt = 1s then dX= -5m, not -10 m.
Yup, you've "overconstrained" the problem. You have too many known
quantities and the values are inconsistent with the assumption of
constant acceleration, which is implicit in the formulas you quote.
is enough to fully specify the constant acceleration motion.
Further saying that dt=1 sec is too much because the above constrains
dt to be sqrt(2) sec. You have the freedom of changing any one of those
five numbers to fix up the inconsistency.
(published on 10/22/2007)
Follow-up on this answer.