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December 17th, 2014, 02:12 PM   #1
szz
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Math Focus: Calculus
Basic physics problems on speed and acceleration

I have to solve a problem for which I don't have the final solution.
I would like to know if my resolution is correct, and understand where I am wrong, if so.

Statement:

An object is moving with a velocity:

$\displaystyle v(t) = 200t\text{e}^{-t} - 10t^2 -40\quad{\text{km/h}}$

where $\displaystyle t$ is the time given in hours.

Find:

a ) The instantaneous position (given in km) $\displaystyle x(t)$ of the object if at the time $\displaystyle t(0)$ the position is $\displaystyle x = 0$.

b ) The instantaneous acceleration of the object.

c ) The average velocity between $\displaystyle t_1 = 0$ and $\displaystyle t_2 = 3$

Given my solution, I am not sure about answers a and c.

So I proceed as follows:

a ) since the velocity is the rate of change of an object's displacement over the time :

$\displaystyle v = {\text{d}x \over \text{d}t}$

And I see this as a basic differential equation.
So the position is given by the displacement at a certain time $\displaystyle t$. So:

$\displaystyle
\begin{aligned}
& x = \int v\,\text{d}t \\
& x = \int 200t\text{e}^{-t} - 10t^2 -40 \,\text{d}t = 200(-\text{e}^{-t}(1 + t)) - {10t^3 \over 3} - 40t
\end{aligned}
$


But

$\displaystyle x(0) = -200 \,\text{km} \ne 0$

So I am not sure about this result...

b ) Since the acceleration is the rate of change of the velocity:

$\displaystyle a = {\text{d}v \over \text{d}t} = {\text{d}^2x \over \text{d}t^2}$

And I am quiet sure this is correct.

c ) As I know, the average velocity between is given by :

$\displaystyle v_{\text{avg}} = {\Delta x \over \Delta t} = {x(t_2) - x(t_2) \over t_2 - t_1} = {x(3) - x(0) \over 3} = -39.95741 \,\text{km/h} $

I am a little bit confused on the point a and the negative result of answer c..

Thank you in advance.
szz

Last edited by szz; December 17th, 2014 at 02:17 PM.
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December 17th, 2014, 03:18 PM   #2
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Quote:
Originally Posted by szz View Post
Statement:

An object is moving with a velocity:

$\displaystyle v(t) = 200t\text{e}^{-t} - 10t^2 -40\quad{\text{km/h}}$

where $\displaystyle t$ is the time given in hours.

Find:

a ) The instantaneous position (given in km) $\displaystyle x(t)$ of the object if at the time $\displaystyle t(0)$ the position is $\displaystyle x = 0$.

b ) The instantaneous acceleration of the object.

c ) The average velocity between $\displaystyle t_1 = 0$ and $\displaystyle t_2 = 3$

So the position is given by the displacement at a certain time $\displaystyle t$. So:

$\displaystyle
\begin{aligned}
& x = \int v\,\text{d}t \\
& x = \int 200t\text{e}^{-t} - 10t^2 -40 \,\text{d}t = 200(-\text{e}^{-t}(1 + t)) - {10t^3 \over 3} - 40t
\end{aligned}
$


But

$\displaystyle x(0) = -200 \,\text{km} \ne 0$ ...
you forgot the constant of integration ...

$\displaystyle x(t) = -200e^{-t}(1 + t) - {10t^3 \over 3} - 40t + C$

initial condition is $\displaystyle x(0) = 0$

$\displaystyle 0 = -200e^{0}(1 + 0) - {10(0)^3 \over 3} - 40(0) + C$

$\displaystyle 0 = -200 + C$

$\displaystyle C = 200$

$\displaystyle x(t) = -200e^{-t}(1 + t) - {10t^3 \over 3} - 40t + 200$

also, it's OK to have a negative velocity ... it merely indicates a direction opposite to a positive velocity.
Thanks from szz
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December 17th, 2014, 10:12 PM   #3
szz
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Math Focus: Calculus
Thank you skeeter !

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