
#1
Dec112, 09:04 PM

P: 41

I'm trying to understand this formula but I'm stuck. I've looked up how to obtain the initial velocity, the acceleration, and time. But all I see are formulas, I don't know how they got it.
Can someone guide me through this please? D = V_{i}t + 1/2at^{2} V_{i} = (d  1/2at^{2}) / t a = 2 x (d  V_{i}t) / t^{2} t = V_{i} / a [itex]\pm[/itex] (√[\(V_{i})^{2} + 2ad]) / a http://http://answers.yahoo.com/ques...1164821AAHxOMD Here's my reference. 



#2
Dec112, 09:31 PM

P: 429

Well, do you know calculus? If so, consider integration of some constant acceleration a. Then the rest is algebra. For example, your V_i = (d  1/2at^2)/t is just the first equation but solved for V_i.




#3
Dec112, 09:46 PM

P: 41





#4
Dec112, 10:21 PM

P: 429

D = Vit + 1/2at^2 HELP!
Well, yes and no. Once you find one equation, you can find the others by algebra. However, I don't know a way of finding one without, at the very least, using ideas from Calculus. *
*At the same time, I've never tried to find such a way, so maybe there exist a way that i'm not aware of. 



#5
Dec212, 01:55 AM

P: 41

So these are the right formulas? But where did they come from? What do they mean?




#6
Dec212, 02:20 AM

P: 429

Well...! I think to answer these questions, you would have to pay me and be in a physics class, because, to give a good answer would require a very long post, and, I'm, to be frank, extremely lazy.
So have fun http://www.lightandmatter.com/lm/ 



#7
Dec212, 03:19 AM

P: 252

If you start with some speed and have a constant acceleration a, after time t your speed will be at. Since the acceleration was constant, the average speed is just the average of the starting and finishing speeds. Then the distance is just the average speed times time. Try to work it out, find the average speed times t and see what you get.




#8
Dec212, 07:43 AM

Math
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PF Gold
P: 38,900

Because you are talking about "instantaneous" speed and acceleration, and those are concepts that cannot even be defined without appealing to some kind of Calculus or limit process, you are not going to be able to derive these equations without Calculus. As chigel says, with constant acceleration, you can average speed over an integral to be the arithetic average of the beginning and ending speeds, but showing that requires Calculus (specifically, the "mean value theorem").




#9
Dec212, 12:32 PM

P: 41

Well I mean average velocity or acceleration. How can I find these variables using algebra if I already have the first formula (s = ut X 1/2at^2)?




#10
Dec812, 10:42 AM

P: 295

If you are referring to weighted averages, the definition of the weighted average of a data set d is given by [itex]\displaystyle \frac{\sum w_k d_k}{\sum w_k}[/itex] where w denotes the weight of each data. Taking the data set to be the velocity and the weight to be the time, one can solve a question like this:
"A car moves x distance with velocity 20 and 4x distance with velocity 40. What is the average velocity of the car?" To solve this, we simply note that the car travelled t time with velocity 20 and 2t time with velocity 40, and hence we have the average as [itex]\displaystyle \frac{20t+80t}{t+2t}=\frac{100}{3}\approx 33.3[/itex]. However, the derivation of the weighted average requires the mean value theorem, so I do not know how useful you might find this to be. 



#11
Dec812, 02:51 PM

HW Helper
P: 6,931

This can be done without calculus as long as you accept that with constant acceleration the average velocity Va = is 1/2 (Vi + Vf), where Vi is initial velocity and Vf is final velocity.
Va = 1/2 (Vi + Vf) Vf = Vi + a t Va = 1/2 (Vi + (Vi + a t)) = Vi + 1/2 a t D = Di + Va t = Di + (Vi + 1/2 a t) t = Di + Vi t + 1/2 a t^{2} 


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