# Why is there a $\frac{1}{2}$ in the kinematic equation? [duplicate]

In a few of the kinematic equations there is a $2$ or a $0.5$ coefficient. Why is this?

For example the kinematic equation for distance is:

$$\text{previous velocity} * \text{time} + \frac{1}{2} * \text{acceleration} * \text{time}^2$$

But why the $\frac{1}{2}$?

If I use the equation for acceleration to get to that equation I don't have a $\frac{1}{2}$? Here:

$$\frac{\Delta v}{t} = a | \cdot t$$

$$v_{new} - v_{old} = a \cdot t | + v_{old}$$

$$\frac{s}{t} = a \cdot t + v_{old} | \cdot t$$

$$s = a \cdot t^2 + v_{old} \cdot t$$

Are my calculations wrong? If so, could someone please show me where I went wrong or explain to me how the $\frac{1}{2}$ comes into play?

## marked as duplicate by Jon Custer, M. Enns, Qmechanic♦Aug 18 '17 at 17:09

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## 2 Answers

The step from the second line to the third line is wrong: V = s/t only for uniform motion, otherwise you should take the mean value whice is where the factor 1/2 comes from.

You should have $\dfrac{s}{t}=v_{\text{mean}}=\dfrac{v_{\text{old}}+v_{\text{new}}}{2}=v_{\text{old}}+\dfrac{at}{2}$.