Answer to Question #135298 in Mechanics | Relativity for Vanel

Question #135298
The position of a particle moving under uniform acceleration is some function of time, t and the acceleration, a. Suppose we write this position as x = ka^m.(t^n) where k is a dimensionless constant.
Determine the values of m and n.
Can this analysis give the value of k? Explain.
1
Expert's answer
2020-09-30T12:29:19-0400

Explanations & Calculations


  • A proper equation is dimensionally correct & it's the same for this function.
  • Consider the dimensions of all the quantities,

"\\qquad\\qquad\n\\begin{aligned}\n\\small [x] &= \\small L \\qquad \\qquad [a] =LT^{-2}\\qquad\\qquad[t]=T\\\\\n\\end{aligned}"

  • According to the requirement stated above,

"\\qquad\\qquad\n\\begin{aligned}\n\\small [x] &= \\small [ka^mt^n] \\\\\n&= \\small [a^mt^n]\\\\\n\\small L&= \\small (LT^{-2})^m \\times(T)^n\\\\\n\\small L^1T^0&= \\small L^{m}T^{(-2m+n)}\\\\\n&.............................................\\\\\n\\small 1&=\\small m\\implies m=1\\\\\n\\small 0 &= \\small -2m+n \\implies n = 2\\\\\n&.............................................\\\\\n\\therefore \\bold{x} & =\\small \\bold{kat^{2}}\n\\end{aligned}"


  • Usually values of dimensionless constants cannot be found using this method. They are found experimentally.
  • For this reason the validity of an equation cannot be fully assessed by this method.
  • For an example "\\small V = U +at" and "\\small V =\\frac{1}{3}U+2at" which are both dimensionally correct while only the first being the valid one.

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