#1
Jul1004, 02:01 PM

P: n/a

[tex]d=v_0t+\frac{1}{2}at^2[/tex]
Why is it [tex]\frac{1}{2}at^2[/tex]? Why not [tex]at^2[/tex] instead? I know that the reason is probably pretty obvious, but I'm having trouble finding it. Thanks for your help. 



#2
Jul1004, 02:57 PM

Sci Advisor
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PF Gold
P: 4,108

Integrate [tex]v=v_0 + a t[/tex] with respect to [tex]t [/tex].




#3
Jul1004, 04:50 PM

Emeritus
PF Gold
P: 8,147

Since a is constant, you can graph v in the t,v plane as a straight line slanting up from t=0, v=v_{0} to some generic point t=t, v=at. Drop a vertical from there and run a horizontal line from the initial point; you have a right triangle and the area of that triangle is the distance made good. But what is the area of a triangle? It is the base (elapsed time=t) times 1/2 the altitude (change in v). But the change in v is at  v_{0}, therefore the distance is given by your formula. Draw the picture. 



#4
Jul1104, 03:15 AM

P: 253

How is this equation derived?
I think its a simple substitution using another constant acceleration equation.
[tex]v=v_0 + a t[/tex] [tex]V^2 = V_0^2 + 2ad[/tex] [tex]V=\sqrt{V_0^2 + 2ad}[/tex] sub. into the first equation [tex]\sqrt{V_0^2 + 2ad} = V_0 + at[/tex] Square both sides [tex]V_0^2 + 2ad = V_0^2 + 2V_0at + a^2t^2[/tex] simplify and divide both sides by 2a [tex]d=v_0t+\frac{1}{2}at^2[/tex] 



#5
Jul1104, 04:17 AM

P: 540

a is the acceleration, which is the change of the velocity per second. You started with velocity [itex]v_{start} = v_0[/itex], so at the end after [itex]t[/itex] seconds the velocity is:
[tex]v_{end} = v_0 + at[/tex] The average velocity during these [itex]t[/itex] seconds was: [tex]\frac{v_{start} + v_{end}}{2} = \frac{v_0 + v_0 + at}{2} = \frac{2v_0 + at}{2} = v_0 + \frac{1}{2}at[/tex] To get the distance travelled during this time you have to multiply this by the time: [tex]d=(v_0 + \frac{1}{2}at)t = v_0t + \frac{1}{2}at^2[/tex] 



#6
Jul1604, 05:01 PM

P: 698

I tried proving this. If you integrate, it is easy. But I tried it the old fashioned way and Im getting the same problem as Esran.
What I did was I had 2 equations to work with. a = (v  vo) / t, and v = d/t a = ((d/t)  vo)/t at = (d/t)  vo at + vo = dt at^2 + vot = d where does the 1/2 come in? 



#7
Jul1604, 05:40 PM

Mentor
P: 28,841

Zz. 



#8
Jul1704, 05:51 PM

P: 698

ohh, k , thanx




#9
Jul1704, 07:48 PM

Sci Advisor
HW Helper
P: 2,275

The reason behind it is this. Acceleration is fow fast your velocity is increasing. If you started out from rest and accelerated at 10 m/s^2, you would be travelling 10 m/s after one second. But you didn't travel 10 m/s for the entire second. You were travelling 0 m/s at time 0, 1 m/s after .1 seconds, 2 m/s after .2 seconds, etc. Your average velocity was 5 m/s (1/2 your acceleration). This continues on even after the first second, except now you're starting at 10 m/s and reach 20 m/s by the end of the 2nd second. In other words, your average velocity for the 2nd second is 15 m/s. You total distance, so far, is 20 meters (5 meters the first second and 15 meters the 2nd second). You can keep going on second by second like this. The 1/2 a accounts for the nonconstant velocity while the t^2 accounts for the accumulating distance from second to second. 



#10
Jul1704, 08:25 PM

P: 608

Yeah back when I was a freshmen and just started physics I wondered what was up with that equation. Since I learned a little calculus I understand.




#11
Jul1804, 02:14 AM

P: 1,322

I vote BobG has having the best answer.




#12
Jul1904, 12:23 AM

P: 253

its good, but i like gerben's better




#13
Jul1904, 04:10 PM

Sci Advisor
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PF Gold
P: 4,108

[tex] \begin{align*} v_{avg} &\equiv\frac{x_fx_0}{t_ft_0}\\ &=\frac{\left[x_0+v_0t+\frac{1}{2}at^2\right]x_0}{\left[ t \right]}\\&=v_0+\frac{1}{2}at \\ &= v_0+\frac{1}{2}\left( v_fv_0 \right) = \frac{1}{2}\left(v_f+v_0 \right) \end{align *} [/tex] In some texts, the above expression for average velocity as onehalf the sum of initial and final velocities is justified by drawing a velocityvstime graph and essentially arguing that the area under the red curve is equal to the area under the green curve. Of course, the area under the red curve is the displacement gained over the interval: the lower rectangle has area [tex]v_0 t[/tex] and the triangle (with height [itex]v_fv_0=at[/itex]) has area [itex]\frac{1}{2} (at) t[/itex]: [tex] \begin{picture}(100,100)(0,0) \put(0,0){\vector(1,0){100}} \put(0,0){\vector(0,1){100}} \put(0,30){\textcolor{red}{\line(3,1){100}}} \put(0,30){\textcolor{red}{v{\scriptsize \textcolor{red}{0} }}} \put(0,47){\textcolor{green}{\line(1,0){100}}} \put(0,50){\textcolor{green}v{\scriptsize \textcolor{green}{avg}}} \put(0,30){\line(1,0){100}} \put(100,64){\line(0,1){64}} \put(0,64){\textcolor{red}{v{\scriptsize \textcolor{red}{f} }}} \put(95,0){t} \end{picture} [/tex] 


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