Asked by Jess
Suppose you are given the following equation, where xf and xi represent positions at two instants of time, vxi is a velocity, a is acceleration, t is an instant of time, and a, b, and c are integers.
xf = xi ta + vxi tb + 1/2 ax tc
For what values of a, b, and c is this equation dimensionally correct?
I know that time = T, velocity = L/T, acceleration = L/T^2 and that xf=xi=L/T. I can't seem to solve it so that things cancel out.
xf = xi ta + vxi tb + 1/2 ax tc
For what values of a, b, and c is this equation dimensionally correct?
I know that time = T, velocity = L/T, acceleration = L/T^2 and that xf=xi=L/T. I can't seem to solve it so that things cancel out.
Answers
Answered by
MathMate
It's probably a duplication of symbols.
If we separate the constant a from acceleration a, it may work out.
α = acceleration
xf = xi ta + vxi tb + 1/2 αx t c
L = (a)L*T + (b)LT^-1*T + (c)LT^-2*L*T
Can you take it from here, assuming the symbols are correct?
If we separate the constant a from acceleration a, it may work out.
α = acceleration
xf = xi ta + vxi tb + 1/2 αx t c
L = (a)L*T + (b)LT^-1*T + (c)LT^-2*L*T
Can you take it from here, assuming the symbols are correct?
Answered by
Zoe
I got close!
I tried to solve it and cancelled things out, to which I got A=0, B=1, but I can't seem to get C right. I thought C would = 0, but apparently that's wrong. Any clues as to where it went bad?
I tried to solve it and cancelled things out, to which I got A=0, B=1, but I can't seem to get C right. I thought C would = 0, but apparently that's wrong. Any clues as to where it went bad?
Answered by
MathMate
Are you supposed to check the dimensions or to find the numerical values?
Without numerical values of all the variables, it would not be possible to check the values of a, b and c.
Without numerical values of all the variables, it would not be possible to check the values of a, b and c.
Answered by
Josh
C=2
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