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 Sep 20th 2017, 08:57 PM #1 Junior Member   Join Date: Sep 2017 Posts: 2 AP Acceleration Problems I have been stuck on these two questions for roughly an hour now. I'm not quite sure which formula to use to find the answer. The questions are: 1. A car starts from rest at position x0=6.2m and accelerates to 11m/s in 4.0s. What is its position at the end of the acceleration? 2. How much time does it take to an object to accelerate from 10.0m/s to 15m/s while moving from x0=8.0m to x=29m? I tried both x = x0 + v0t + 1/2at^2 and v^2=v0^2+ 2a(x-x0) but couldn't the answer. My teacher says the answer is 28.2m on the first question, but I want to know how to get to that. For the first one, it seems like I always come out to like 88m or something similar, but not close to the answer. I don't even know where to start on the second one. Last edited by sussus; Sep 20th 2017 at 09:00 PM.
Sep 20th 2017, 09:11 PM   #2
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 Originally Posted by sussus I have been stuck on these two questions for roughly an hour now. I'm not quite sure which formula to use to find the answer. The questions are: 1. A car starts from rest at position x0=6.2m and accelerates to 11m/s in 4.0s. What is its position at the end of the acceleration? 2. How much time does it take to an object to accelerate from 10.0m/s to 15m/s while moving from x0=8.0m to x=29m? I tried both x = x0 + v0t + 1/2at^2 and v^2=v0^2+ 2a(x-x0) but couldn't the answer.
Why? Do you know the meaning of that equation? If so then substitute the appropriate values and see what you get.

Sep 20th 2017, 09:19 PM   #3
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 Originally Posted by Pmb Why? Do you know the meaning of that equation? If so then substitute the appropriate values and see what you get.
I keep getting 88.

Sep 21st 2017, 01:46 AM   #4
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 Originally Posted by sussus I keep getting 88.
I expect that is because you need to understand the difference between distance travelled and position x.

Try calculating distance travelled first and then adding it to the start position to get the finish position.

Why are you using those particular formulae?
They make things particularly difficult.

 Sep 21st 2017, 08:18 PM #5 Senior Member     Join Date: Aug 2008 Posts: 113 derive another equation from $v_f^2=v_0^2+2a\Delta x$ and $a=\dfrac{v_f-v_0}{\Delta t}$ move $v_0^2$ and substitute for $a$ ... $v_f^2-v_0^2 = 2\left(\dfrac{v_f-v_0}{\Delta t}\right) \cdot \Delta x$ $(v_f+v_0)(\cancel{v_f-v_0})= 2\left(\dfrac{\cancel{v_f-v_0}}{\Delta t}\right) \cdot \Delta x$ solve for $\Delta x$ ... $\Delta x = \dfrac{1}{2}(v_f+v_0) \cdot \Delta t$ $\Delta x = \dfrac{1}{2}(v_f+v_0) \cdot \Delta t$ $x_f - 6.2 = \dfrac{1}{2}(11+0) \cdot 4$ $x_f = 22 + 6.2 = 28.2$ solving the derived equation for $\Delta t$ ... $\Delta t = \dfrac{2 \Delta x}{v_f+v_0}$ can you complete #2 ?
 Sep 24th 2017, 08:23 AM #6 Senior Member   Join Date: Aug 2010 Posts: 372 For (1) I presume you were able to calculate that, since the car "accelerates to 11m/s in 4.0s.", the (average) acceleration is a= 11/4 m/s^2. Since the car "starts from rest", v0= 0 and since it starts "at x0= 6.2m", x0= 6.2. So using your "x = x0 + v0t + 1/2at^2", x= 6.2+ (11/8)t^2. With t= 4s that is x= 6.2+ (11/8)(16)= 6.2+ 22= 28.2 m. I have absolutely no idea how you got "88". Could you show us your working? For (2), "How much time does it take to an object to accelerate from 10.0m/s to 15m/s while moving from x0=8.0m to x=29m?", taking "t" to be the required time and "a" the (average) acceleration, x= x0+ v0t+ at^2= 8+ 10t+ at^2= 29 while v= v0+ at= 10+ at= 15. Solve the two equations 8+ 10t+ at^2= 29, so at^2+ 10t- 21= 0, and 10+ at= 15, so at= 5, for a and t. I would start by solving the second for a= 5/t so that at^2+ 10- 21= 0 becomes (5/t)(t^2)+ 10t- 21= 5t+ 10t- 21= 0 or 15t= 21.

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