# Can someone explain this equation to me?

1. Dec 26, 2012

### tahayassen

$$d=\left( \left. \frac { { d }x }{ { d }t } \right| _{ 0 } \right) t+\left( \left. \frac { { d }^{ 2 }x }{ { d }t^{ 2 } } \right| _{ 0 } \right) \frac { t^{ 2 } }{ 2 }$$

is supposed to be the same as writing

$$d=vt+a\frac { t^{ 2 } }{ 2 }$$

The first equation seems to use the definite integral, but there's no upper limit?

Furthermore, shouldn't it be the following?

$$d=\frac { { d }x }{ { d }t } t+\frac { { d^{ 2 } }x }{ { d }t^{ 2 } } \frac { t^{ 2 } }{ 2 }$$

2. Dec 26, 2012

### cosmic dust

I think it is about linear motion with constant acceleration, which is described by the equation:
$$\frac{{{d}^{2}}x\left( t \right)}{d{{t}^{2}}}=a[\tex] where a is a constant. Integrate once (by changing the integration variable from t to τ) from τ = 0 to τ = t, to get: 3. Dec 26, 2012 ### cosmic dust Ingore this, I posted it by mistake, sorry. I'm texting the correct answer now, 4. Dec 26, 2012 ### jtbell ### Staff: Mentor Those aren't integrals. The notation \left. \frac {dx}{dt} \right| _{0} means "take the derivative of x(t) with respect to t, then evaluate it at t = 0." That is, it's simply a more complicated way of writing $v_0$. No, because dx/dt is the function v(t), and d2x/dt2 is the function a(t). Last edited: Dec 26, 2012 5. Dec 26, 2012 ### cosmic dust It is about linear motion with constant acceleration, which is described by the equation: [tex]\frac{{{d}^{2}}x\left( t \right)}{d{{t}^{2}}}=a$$
where "a" is a constant. Integrate once (by changing the integration variable from t to τ) from τ = 0 to τ = t, to get:
$$\int_{0}^{t}{\frac{{{d}^{2}}x\left( \tau \right)}{d{{\tau }^{2}}}d\tau }=at\Rightarrow \left. \frac{dx\left( \tau \right)}{d\tau } \right|_{0}^{t}=\frac{dx\left( t \right)}{dt}-{{\left. \frac{dx\left( \tau \right)}{d\tau } \right|}_{0}}=at$$
Integrate again using the same limits to get:
$$\int_{0}^{t}{\left[ \frac{dx\left( \tau \right)}{d\tau }-{{\left. \frac{dx\left( t \right)}{dt} \right|}_{0}} \right]}d\tau =\frac{1}{2}a{{t}^{2}}\Rightarrow \left. x\left( \tau \right) \right|_{0}^{t}-{{\left. \frac{dx\left( t \right)}{dt} \right|}_{0}}t=\frac{1}{2}a{{t}^{2}}\Rightarrow x\left( t \right)-x\left( 0 \right)={{\left. \frac{dx\left( t \right)}{dt} \right|}_{0}}t+\frac{1}{2}a{{t}^{2}}$$
Because of the constancy of the second derivative, you could set:
$$a={{\left. \frac{{{d}^{2}}x\left( t \right)}{d{{t}^{2}}} \right|}_{0}}$$
(alternatively, you could start by the equation $\frac{{{d}^{3}}x\left( t \right)}{d{{t}^{3}}}=0$ , integrate three times and get to the same result). Also you can define: $x\left( t \right)-x\left( 0 \right)=d\left( t \right)$ so you get the equation:
$$d\left( t \right)={{\left. \frac{dx\left( t \right)}{dt} \right|}_{0}}t+\frac{1}{2}{{\left. \frac{{{d}^{2}}x\left( t \right)}{d{{t}^{2}}} \right|}_{0}}{{t}^{2}}$$

6. Dec 26, 2012

### HallsofIvy

Well, let the rest of us know!

tahayassen, there is no integral in the equation. That is saying that d (probably "distance") is equal to the "first derivative evaluated at t= 0" times t plus "the second derivative evaluated at t= 0" times t squared. That would be the distance covered in time t with initial speed the speed at time t= 0 and constant acceleration equal to the acceleration at t= 0.

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. Dec 26, 2012

### Darwin123

No.
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. Dec 26, 2012

### Mandelbroth

Kind of skimmed through the article, but I didn't see anyone put this very plainly because apparently I have problems with reading (edit in red).

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 $d = v_0 t + a_0 \frac{t^2}{2}$ because we are using the initial velocity and initial acceleration.

Last edited: Dec 26, 2012
9. Dec 26, 2012

### tahayassen

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. Dec 27, 2012

### Staff: Mentor

To avoid confusion between the integration variable and the upper limit which you substitute during the final step of the integration process. Other common notations are to use t' for the integration variable and t for the upper limit, or to use t for the integration variable and tf for the upper limit (in that case you often see ti or t0 for the lower limit, which can be nonzero in general).

Both of them represent times, but different ones. In cosmic dust's notation, τ is the "generic" time at any instant during the object's motion. t is the time at the end of the time period that we're integrating over, which extends from τ = 0 to τ = t.