I was wondering how the average velocity is the arithmetic mean of initial velocity and final velocity if acceleration is constant. So, my first part of the question: Can someone give me an algebraic proof of this?
Okay, so I moved on and thought maybe it was more of an experimental result, so I tried to do this: Lets suppose that an object is at rest, so its initial velocity is $0\ \mathrm{m/s}$. And lets say it starts moving with a constant acceleration of $2\ \mathrm{m/s}$ per second. And lets manually calculate the distance it will cover in $5\ \mathrm{s}$. The result is as follows:
- Distance covered in $0\ \mathrm{s} = 0\ \mathrm{m}$.
- Distance covered in $1\ \mathrm{s} = 2\ \mathrm{m}$.
- Distance covered in $2\ \mathrm{s} = 4\ \mathrm{m}$.
- Distance covered in $3\ \mathrm{s} = 6\ \mathrm{m}$.
- Distance covered in $4\ \mathrm{s} = 8\ \mathrm{m}$.
- Distance covered in $5\ \mathrm{s} = 10\ \mathrm{m}$.
So total distance covered = $(0 + 2 + 4 + 6 + 8 + 10)\ \mathrm{m} = 30\ \mathrm{m}$.
But, if I use this equation instead:
$$s = \frac{v + u}{2}t$$ I get total distance covered of $25\ \mathrm{m}$.
Why don't these results agree?