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.
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?
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.
C=2
To determine the values of a, b, and c for which the equation is dimensionally correct, we need to ensure that the dimensions on both sides of the equation match.
Let's break down the dimensions for each term in the equation:
xf: Represents a position, which has the dimension of length (L).
xi: Also represents a position, with the dimension of length (L).
ta: Represents time (T), and multiplying by a position gives the dimension of (L * T).
vxi: Represents velocity, which has the dimension of length divided by time (L / T).
tb: Represents time (T), and multiplying by velocity gives the dimension of (L / T * T = L / T^2).
ax: Represents acceleration, which has the dimension of length divided by time squared (L / T^2).
tc: Represents time (T), and multiplying by acceleration gives the dimension of (L / T^2 * T = L / T^3).
Putting it all together, we have:
xf = L
xi = L
ta = T
vxi = L / T
tb = L / T^2
ax = L / T^2
tc = L / T^3
Substituting these dimensions into the equation:
L = L * T * a + L / T * b + 1/2 * (L / T^2) * c
Now, we can equate the dimensions to solve for a, b, and c:
For the length dimension:
1 = a + b + c/2
For the time dimension:
0 = -b - c
We now have a system of linear equations, which can be solved by substitution or by forming an augmented matrix and performing row operations.
From the second equation, we can express b = -c, and substitute into the first equation:
1 = a - c + c/2
1 = a - c/2
Simplifying further:
a = 1 + c/2
From this equation, we can see that a can have any value as long as c is an even integer. Since a, b, and c are integers, we can set c = 2k, where k is any integer.
Then we have:
a = 1 + 2k/2 = 1 + k
Therefore, the equation is dimensionally correct for any values of a, b, and c where a = 1 + k, b = -2k, and c = 2k, where k is an integer.
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?