Question

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?

Answer 1

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 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*t" " " "(1)`

`s=1/2*a*t^2-u*t" " " "(2)`

By solving `(1)` for `t`:

`0=5+a*t`

`a*t=-5`

`t=-5/a`

Then substituting in `(2)`:

`100=1/2*a*t^2-0*t`

`100=1/2*a*t^2`

`100=1/2*a*(-5/a)^2`

`100=1/2*a*(-5)^2/a^2`

`100=1/2*25/a`

`a=25/(2*100)=0.125 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=(du)/dt` we can transform by using chain rule through Leibniz's notation:

`a=(du)/dt=(du)/dt*(dx)/dx=(dx)/dt*(du)/dx`

Knowing that `u=(dx)/dt` gives us:

`a=u*(du)/dx`

By integrating:

`a*dx=u*du`

`aint_0^100dx=int_5^0udu`

`a*[x]_0^100=[u^2/2]_5^0`

`a*(100-0)=(0^2/2-5^2/2)`

`a=5^2/(2*100)=25/(2*100)=1/(2*4)=0.125 m/s^2`