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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 !  Tags acceleration, basic, physics, problems, speed Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post fmsdel Physics 1 January 31st, 2012 03:18 PM Sunde Algebra 4 August 7th, 2011 01:25 AM r-soy Physics 0 February 7th, 2011 08:53 AM football Physics 2 June 8th, 2010 08:07 AM DLDude Algebra 0 January 17th, 2009 09:18 AM

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