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Sorry people, very basic kinematic stuff here.

(1) Velocity: $$v=\frac{d}{t}$$

(2) Acceleraton: $$a=\frac{v_{f}-v_{i}}{t}$$

(3) Re-arrange acceleration: $$v_{f} = v_{i}+at$$

(4) Ok here is my question, my lecturer produces this equation by "combining" (1) and (3): $$d=v_{i}t+\frac{1}{2}at^{2}$$

(5) So now I want to figure out how (4) was formed, I stick to algebra and this is my process/result:

(5.1) If $v=\frac{d}{t}$ then place $\frac{d}{t}$ into the final velocity equation: $$\therefore \left (\frac{d}{t}\right )_{f} = \left (\frac{d}{t}\right )_{i} + at$$ $$\therefore \left ( \left (\frac{d}{t}\right )_{f} \right )\times t= \left (\left (\frac{d}{t}\right )_{i} + at\right ) \times t$$ $$\therefore d=d+\left (at\right )\times t$$ $$\therefore d=d+at^{2}$$ Obviously: $$\left (d=d+at^{2} \right )\neq \left (d=v_{i}t+\frac{1}{2}at^{2}\right )$$

So what have I done wrong?

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closed as off-topic by ACuriousMind, LDC3, Qmechanic Jun 21 '15 at 14:21

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  • 1
    $\begingroup$ Did you learn to write acceleration and velocity in term of derivatives? $\endgroup$ – Bernhard Nov 7 '13 at 8:23
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    $\begingroup$ If you use calculus this result follows easily. Without calculus any derivation of equation 4 is going to be somewhat contrived. $\endgroup$ – John Rennie Nov 7 '13 at 8:37
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    $\begingroup$ @John Rennie It doesn't really need to be contrived. The derivation with calculus says that the displacement should be the area under the velocity vs time curve, and then uses calculus to get that area. One explanation without calculus would say that the displacement should be the area under the velocity vs time curve, and then uses basic geometry (i.e., the formula for the area of a triangle) to get that area. $\endgroup$ – Brian Moths Nov 7 '13 at 15:00
  • $\begingroup$ JohnRennie, Bernhard, This is interesting, I don't think I know how to write velocity and acceleration in term of derivatives. I can do derivatives and integration. Perhaps a link to an easy to understand resource to get me started? So many examples on the internet are too complicated. Thanks :) $\endgroup$ – Joseph Nov 7 '13 at 17:19
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Equation 1 needs to be stated correctly, the way you wrote it ignores the fact that velocity is changing. The correct expression (assuming constant acceleration) is enter image description here

Now substitute your equation 3 into my equation gives:

enter image description here

Which is the expression you are after, the mistake you had was that you didn’t take into account the variable speed into equation 1. For the record this derivation is only valid for constant acceleration. For variable acceleration, as John mentioned you can derive those expressions from calculus. If you assumed constant acceleration in calculus you will end up with the same equations derived here.

Hopefully that was useful

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  • $\begingroup$ What does $v_{(t)}$ mean in the first equation? $$\left ( \frac{v_{(t)}+v_{i}}{2}\right )\times t$$ $\endgroup$ – Joseph Nov 7 '13 at 17:14
  • $\begingroup$ It is the variable velocity as function of time. Or the instantaneous velocity as function of time. "vf" in your equation 3 @Joseph $\endgroup$ – Gotaquestion Nov 7 '13 at 17:28
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You should simply integrate $at$ which gives you the result $\frac{at^2}{2}$ (if you know how to integrate for sure!)

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I know this is an old thread.. but just thought i would chime in on this.. What you did wrong to try and get the equation that your instructor came up with:

  1. $V=d/t$
  2. $V_f= V_i + at$ (the rewritten equation)

to get to the final equation of $d=V_it + at^2$ you have to assume that $V=V_f$. Because $V_i$ and $V_f$ are two separate variable and the first equation is $V$ at a given point in time you can choose which one to replace with $\frac{d}{t}$ but not both. So assuming $V_f=\frac{d}{t}$ and you substitute you get:

$$\frac{d}{t}=V_i + at$$

Then to get $t$ on the right only:

$$t\left(\frac{d}{t}\right)= (V_i + at)t$$

Simplify:

$$d= V_it + at^2$$

Which is the equation you are looking for.

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    $\begingroup$ He is looking for an equation with $\frac{1}{2}at^2$ Your result has just $at^2$. It is not the same. This is not the way to get his desired result $\endgroup$ – Jim Jan 27 '15 at 15:07
  • $\begingroup$ whoops misread that equation.. lets try this again then heh.. $\endgroup$ – Package Jan 28 '15 at 0:08
  • $\begingroup$ $V_avg = \frac {delta d}{delta t}$ is the correct average velocity formula. so we know that if graphed this would be V on the Y axis and T on the X axis. assuming constant acceleration the variables would be $v_i$ and $v_f$ for the Y axis and $t_i$ and $t_f$. you would find a trapezoid shaped area under the line of acceleration and that area would be the displacement. so the area of this trapezoid is $delta d = (\frac {1}{2})(v_i + v_f)t$. because $v_f = v_o + at$ subtitue that equation for $v_f$ and do the algebra. to get $d = v_i t +(\frac{1}{2})at^2$ $\endgroup$ – Package Jan 28 '15 at 0:28

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