
#1
Dec2612, 12:49 PM

P: 273

[tex]d=\left( \left. \frac { { d }x }{ { d }t } \right _{ 0 } \right) t+\left( \left. \frac { { d }^{ 2 }x }{ { d }t^{ 2 } } \right _{ 0 } \right) \frac { t^{ 2 } }{ 2 }[/tex]
is supposed to be the same as writing [tex]d=vt+a\frac { t^{ 2 } }{ 2 }[/tex] The first equation seems to use the definite integral, but there's no upper limit? Furthermore, shouldn't it be the following? [tex]d=\frac { { d }x }{ { d }t } t+\frac { { d^{ 2 } }x }{ { d }t^{ 2 } } \frac { t^{ 2 } }{ 2 }[/tex] 



#2
Dec2612, 01:02 PM

P: 123

I think it is about linear motion with constant acceleration, which is described by the equation:
[tex]\frac{{{d}^{2}}x\left( t \right)}{d{{t}^{2}}}=a[\tex] where a is a constant. Integrate once (by changing the integration variable from t to τ) from τ = 0 to τ = t, to get: 



#3
Dec2612, 01:04 PM

P: 123





#4
Dec2612, 01:15 PM

Mentor
P: 11,219

Can someone explain this equation to me?$$\left. \frac {dx}{dt} \right _{0}$$ means "take the derivative of x(t) with respect to t, then evaluate it at t = 0." That is, it's simply a more complicated way of writing ##v_0##. 



#5
Dec2612, 01:22 PM

P: 123

It is about linear motion with constant acceleration, which is described by the equation:
[tex]\frac{{{d}^{2}}x\left( t \right)}{d{{t}^{2}}}=a[/tex] where "a" is a constant. Integrate once (by changing the integration variable from t to τ) from τ = 0 to τ = t, to get: [tex]\int_{0}^{t}{\frac{{{d}^{2}}x\left( \tau \right)}{d{{\tau }^{2}}}d\tau }=at\Rightarrow \left. \frac{dx\left( \tau \right)}{d\tau } \right_{0}^{t}=\frac{dx\left( t \right)}{dt}{{\left. \frac{dx\left( \tau \right)}{d\tau } \right}_{0}}=at[/tex] Integrate again using the same limits to get: [tex]\int_{0}^{t}{\left[ \frac{dx\left( \tau \right)}{d\tau }{{\left. \frac{dx\left( t \right)}{dt} \right}_{0}} \right]}d\tau =\frac{1}{2}a{{t}^{2}}\Rightarrow \left. x\left( \tau \right) \right_{0}^{t}{{\left. \frac{dx\left( t \right)}{dt} \right}_{0}}t=\frac{1}{2}a{{t}^{2}}\Rightarrow x\left( t \right)x\left( 0 \right)={{\left. \frac{dx\left( t \right)}{dt} \right}_{0}}t+\frac{1}{2}a{{t}^{2}}[/tex] Because of the constancy of the second derivative, you could set: [tex]a={{\left. \frac{{{d}^{2}}x\left( t \right)}{d{{t}^{2}}} \right}_{0}}[/tex] (alternatively, you could start by the equation [itex]\frac{{{d}^{3}}x\left( t \right)}{d{{t}^{3}}}=0[/itex] , integrate three times and get to the same result). Also you can define: [itex]x\left( t \right)x\left( 0 \right)=d\left( t \right)[/itex] so you get the equation: [tex]d\left( t \right)={{\left. \frac{dx\left( t \right)}{dt} \right}_{0}}t+\frac{1}{2}{{\left. \frac{{{d}^{2}}x\left( t \right)}{d{{t}^{2}}} \right}_{0}}{{t}^{2}}[/tex] 



#6
Dec2612, 01:23 PM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,879

Well, let the rest of us know!
The difference between what you give initially and your suggestion is that the acceleration is assumed to always be the acceleration of x at t= 0, thus a constant. 



#7
Dec2612, 01:26 PM

P: 741

The reason is that dx/dt and d^2 x/dx^2 may be changing with time. By placing the subscripts after the parentheses, on is implying that the derivative itself isn't changing in time. The subscript is defined as meaning that the derivative is evaluated only at t=0. 



#8
Dec2612, 02:10 PM

P: 597

Summarizing in more vernacular terms: What is the first derivative of x (position) with respect to time? What is the second derivative of x with respect to time? Velocity and acceleration, respectively. I see no need for integrals. We are evaluating these values at the point where no time has passed. Thus, I think it would be more appropriate to write [itex]d = v_0 t + a_0 \frac{t^2}{2}[/itex] because we are using the initial velocity and initial acceleration. 



#9
Dec2612, 09:51 PM

P: 273

Thanks guys!
@cosmic dust: Why did you change the variable from t to τ? Which one of the two represents time? What does the other represent? 



#10
Dec2712, 12:32 AM

Mentor
P: 11,219




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