QuestionJune 11, 2025

55. ... Acceleration is related to velocity and time by the following expression: a=v^pt^q Find the powers p and q that make this equation dimensionally consistent.

55. ... Acceleration is related to velocity and time by the following expression: a=v^pt^q Find the powers p and q that make this equation dimensionally consistent.
55. ... Acceleration is related to velocity and time by the following expression: a=v^pt^q Find the powers p and q that make this equation dimensionally consistent.

Solution
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Answer

p = 1, q = -1 Explanation 1. Define Dimensions Velocity v has dimensions [LT^{-1}], time t has dimensions [T], and acceleration a has dimensions [LT^{-2}]. 2. Express Equation Dimensionally Substitute dimensions into the equation: [LT^{-2}] = [LT^{-1}]^p[T]^q. 3. Simplify Dimensional Equation Simplify to get: [L^pT^{-p}][T^q] = [L^1T^{-2}]. 4. Equate Powers of L From L: p = 1. 5. Equate Powers of T From T: -p + q = -2. Substitute p = 1: -1 + q = -2. 6. Solve for q q = -1.

Explanation

1. Define Dimensions<br /> Velocity $v$ has dimensions $[LT^{-1}]$, time $t$ has dimensions $[T]$, and acceleration $a$ has dimensions $[LT^{-2}]$.<br /><br />2. Express Equation Dimensionally<br /> Substitute dimensions into the equation: $[LT^{-2}] = [LT^{-1}]^p[T]^q$.<br /><br />3. Simplify Dimensional Equation<br /> Simplify to get: $[L^pT^{-p}][T^q] = [L^1T^{-2}]$.<br /><br />4. Equate Powers of L<br /> From $L$: $p = 1$.<br /><br />5. Equate Powers of T<br /> From $T$: $-p + q = -2$. Substitute $p = 1$: $-1 + q = -2$.<br /><br />6. Solve for q<br /> $q = -1$.
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