# While sledding down a snowy hill Ed slowed down from 5 m/s to rest in a distance of 100 m. What was Ed's acceleration?

Jan 2, 2016

Since you also have time as an unknown value, you need 2 equations that combine these values. By using the equations of speed and distance for deceleration, the answer is:

$a = 0.125 \frac{m}{s} ^ 2$

#### Explanation:

1st way

This is the simple elementary path. If you are new to motion, you want to go this path.

Provided that the acceleration is constant, we know that:

$u = {u}_{0} + a \cdot t \text{ " " } \left(1\right)$

$s = \frac{1}{2} \cdot a \cdot {t}^{2} - u \cdot t \text{ " " } \left(2\right)$

By solving $\left(1\right)$ for $t$:

$0 = 5 + a \cdot t$

$a \cdot t = - 5$

$t = - \frac{5}{a}$

Then substituting in $\left(2\right)$:

$100 = \frac{1}{2} \cdot a \cdot {t}^{2} - 0 \cdot t$

$100 = \frac{1}{2} \cdot a \cdot {t}^{2}$

$100 = \frac{1}{2} \cdot a \cdot {\left(- \frac{5}{a}\right)}^{2}$

$100 = \frac{1}{2} \cdot a \cdot {\left(- 5\right)}^{2} / {a}^{2}$

$100 = \frac{1}{2} \cdot \frac{25}{a}$

$a = \frac{25}{2 \cdot 100} = 0.125 \frac{m}{s} ^ 2$

2nd way

This path is not for beginners, as it is the calculus path. All it provides is actual proof of the above equations. I am just posting in case you are interested in how it works.

Knowing that $a = \frac{\mathrm{du}}{\mathrm{dt}}$ we can transform by using chain rule through Leibniz's notation:

$a = \frac{\mathrm{du}}{\mathrm{dt}} = \frac{\mathrm{du}}{\mathrm{dt}} \cdot \frac{\mathrm{dx}}{\mathrm{dx}} = \frac{\mathrm{dx}}{\mathrm{dt}} \cdot \frac{\mathrm{du}}{\mathrm{dx}}$

Knowing that $u = \frac{\mathrm{dx}}{\mathrm{dt}}$ gives us:

$a = u \cdot \frac{\mathrm{du}}{\mathrm{dx}}$

By integrating:

$a \cdot \mathrm{dx} = u \cdot \mathrm{du}$

$a {\int}_{0}^{100} \mathrm{dx} = {\int}_{5}^{0} u \mathrm{du}$

$a \cdot {\left[x\right]}_{0}^{100} = {\left[{u}^{2} / 2\right]}_{5}^{0}$

$a \cdot \left(100 - 0\right) = \left({0}^{2} / 2 - {5}^{2} / 2\right)$

$a = {5}^{2} / \left(2 \cdot 100\right) = \frac{25}{2 \cdot 100} = \frac{1}{2 \cdot 4} = 0.125 \frac{m}{s} ^ 2$