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$$ \begin{array}{l}{\text { (II) Three students derive the following cquations in which }} \\ {x \text { refers to distance traveled, } v \text { the specd, } a \text { the acceleration }} \\ {\left(m / s^{2}\right), t \text { the time, and the subscript zero }(a) \text { means a quantity }}\end{array} $$$$\begin{array}{l}{\text { at time } t=0 : \text { (a) } x=u t^{2}+2 a t,(b) x=v_{0} t+\frac{1}{2} a t^{2}, \text { and }} \\ {(c) \quad x=v_{0} t+2 a t^{2} \text { . Which of these could possibly be }} \\ {\text { correct according to a dimensional check? }}\end{array} $$

A) The equation is incorrect.B) Equation can be correct.C) Equation can be correct.

Physics 101 Mechanics

Chapter 1

Introduction, Measurement, Estimating

Physics Basics

Cornell University

University of Michigan - Ann Arbor

Hope College

Lectures

04:16

In mathematics, a proof is…

04:48

In mathematics, algebra is…

02:41

$$\begin{array} { l } { \f…

06:09

$t x^{\prime \prime}-(t+1)…

01:44

$\left(t^{2}-t-2\right) x^…

02:21

$\begin{array}{l}{\text { …

02:58

$$ 2 t x d x+\left(t^{2}-x…

02:56

$$\left(t^{2}-t-2\right)^{…

04:21

$$\frac{a x}{d t}+t x^{3}+…

01:43

$t y^{\prime \prime}+(1-2 …

04:37

$$x-d^{2} y / d t^{2}=t+1$…

02:00

Solve. $$\frac{1}{2} a…

00:51

$$\begin{array} { l } { x …

12:53

$y^{\prime \prime \prime}+…

03:16

$\begin{array}{c}{\text { …

04:53

$t^{2} y^{\prime \prime}-2…

02:55

$w^{\prime \prime}+w=u(t-2…

11:10

$x^{\prime \prime}-x^{\pri…

05:15

$$\begin{array}{l}{x^{\pri…

07:13

$$x^{\prime \prime}(t)-4 x…

04:29

$$\begin{array}{l}{\te…

So we have three possible equations for X. We have X equals the T squared plus 2 80 We have X equals the initial T plus 1/2 of a T squared and we have X equals the initial T plus two a. T squared. So the 1st 1 ex would have units of meters and this would be equal two meters per second time's second squared plus two. Let's get rid of the coefficient, so just be plus meters per second squared times seconds. So here, uh, this would be meters equaling meters seconds plus meters per second. So this obviously cannot to be the solution because dimensionally this would not make sense. So this cannot be correct. For the second equation, we have again meters for x and then this would be meters per second time seconds plus lose the coefficients. We have meters per second squared times, second squared and this would be meters with equal meters plus meters. So here this may be correct and then for the last one was a, uh b see For the last one, we have again meters for the units of X, this will equal meters per second time seconds plus and then we'll lose the coefficient once again. Meters per second squared times, second squared and we have meters equals meters plus meters which of course, also may be correct. So the first this last two equations may be correct to do two dimensional analysis. However, the 1st 1 we know cannot be correct because dimensionally it does not make sense. That is the end of the solution. Thank you for watching.

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