Physics Help Forum Find total time taken?

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 Aug 30th 2008, 09:57 AM #1 Junior Member   Join Date: Aug 2008 Posts: 10 Find total time taken? The speed of a train increases at a constant rate α = 1 m/s^2 from 0 to v = 10 m/s and then remains constant for a time interval and finally decreases to zero at a constant rate β = 2m/s^2. If l = 100 m is the total distance covered then total time taken is? A) 17.5 s B) 2.0 s C) 13.2 s D) None of these __________________ 先生
 Aug 30th 2008, 08:42 PM #2 Senior Member   Join Date: Apr 2008 Location: HK Posts: 886 I guess the answer would not be certain until distace travelled with constant velocity 10m/s is known
Aug 31st 2008, 01:42 AM   #3
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Hello,
 Originally Posted by sensei The speed of a train increases at a constant rate α = 1 m/s^2 from 0 to v = 10 m/s and then remains constant for a time interval and finally decreases to zero at a constant rate β = 2m/s^2. If l = 100 m is the total distance covered then total time taken is?
I've sketched the velocity of the train against time (see attachment). Using the information you're given you can find the value of $\displaystyle \Delta t_1$ and $\displaystyle \Delta t_3$, can't you ? Then simply say that the total distance covered is $\displaystyle 100\,\mathrm{meters} = \int\limits_{0}^{\Delta t_1+\Delta t_2+\Delta t_3}|v(t)|\,\mathrm{d}t$, solve this for $\displaystyle \Delta t_2$ and you'll get the total time taken which is $\displaystyle \Delta t_1+\Delta t_2+\Delta t_3$.
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Aug 31st 2008, 10:13 AM   #4

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 Originally Posted by sensei The speed of a train increases at a constant rate α = 1 m/s^2 from 0 to v = 10 m/s and then remains constant for a time interval and finally decreases to zero at a constant rate β = 2m/s^2. If l = 100 m is the total distance covered then total time taken is? A) 17.5 s B) 2.0 s C) 13.2 s D) None of these
Break the problem into three time frames.

For the first acceleration period you know that
$\displaystyle v = at_1$
so you can find $\displaystyle t_1$. We will also need to know $\displaystyle x_1$, so
$\displaystyle v^2 = 2ax_1$.

For the constant speed period you know that:
$\displaystyle x_2 = vt_2$
We don't yet know x_2, so we are stuck on this for a moment.

For the second accleration (deceleration) period we have
$\displaystyle v = -at_3$
(The acceleration is negative here, so the negative signs cancel.) So we can find $\displaystyle t_3$.
Again,
$\displaystyle v^2 = -2ax_3$
so we can find $\displaystyle x_3$ from this.

Now, $\displaystyle x_2 = 100 - x_1 - x_3$ so you can now find $\displaystyle x_2$ and thus $\displaystyle t_2$. Now just add your times.

-Dan
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