My Math Forum Basic physics problems on speed and acceleration

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 December 17th, 2014, 02:12 PM #1 Senior Member     Joined: Oct 2014 From: EU Posts: 224 Thanks: 26 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.
December 17th, 2014, 03:18 PM   #2
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Quote:
 Originally Posted by szz 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.

 December 17th, 2014, 10:12 PM #3 Senior Member     Joined: Oct 2014 From: EU Posts: 224 Thanks: 26 Math Focus: Calculus Thank you skeeter !

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