Motion: How to Calculate Speed with Acceleration?

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To calculate speed with constant acceleration, the formula derived is v = 2Δx/t, where Δx is the distance covered in time t. This result stems from the average speed being the mean of the initial and final speeds, which, for an object starting from rest, simplifies to v/2. The distance can be expressed as Δx = (v/2) * t, leading to the conclusion that v = 2Δx/t. The discussion emphasizes that this relationship holds true specifically under conditions of constant acceleration. Understanding this concept is crucial for solving motion problems effectively.
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Hi, I was given the following question:

Q: A body starts moving with an acceleration a=const. After a time t it has covered the distance \Delta x[/itex]. What is its speed at time t?<br /> <br /> Answer: \frac{2\Delta x}{t}<br /> <br /> Can somebody say how to arrive at this answer (both mathematically and intuitionally)? I know that \Delta x/twould be the speed w/o acceleration, so the &quot;2&quot; must come from acceleration, but where does it come from?<br /> <br /> Can the answer be obtained from the formula v=v_0+at^2?<br /> <br /> - Kamataat
 
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x-x_0=\frac{1}{2}(v_0+v_f)\Delta{t},\ \ v_0=0 m/s
 
Thank you!

- Kamataat
 
Another way: for CONSTANT acceleration, the "average" speed is just the average of the starting speed and ending speed: \frac{v_i+v_f}{2}. Since the starting speed is 0, if the ending speed is v, this is \frac{v}{2} so the distance covered in time \Delta t would be \Delta x =\frac{v}{2}\Delta t} so v= 2\frac{\Delta x}{\Delta t}.

CAUTION: This is only true for CONSTANT acceleration!
 
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HallsofIvy said:
Another way: for CONSTANT acceleration, the "average" speed is just the average of the starting speed and ending speed: \frac{v_i+v_f}{2}. Since the starting speed is 0, if the ending speed is v, this is \frac{v}{2} so the distance covered in time \Delta t would be \Delta x =\frac{v}{2}\Delta t} so v= 2\frac{\Delta x}{\Delta t}.

CAUTION: This is only true for CONSTANT acceleration!

Halls, that's what my equation says :) 1/2(v_0 + v_f) = average velocity :smile:
 
yup, i get it now. tnx
 
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