Why distance equals initial velocity times time Plus acceleration over two times time Squared?

Question

i am a beginner in physics and I do not understand why is the d=vi(t)+(1/2)a(t^2).

Answer 1

For a constant acceleration $a$, $s = s_i + v_it + {{1} \over {2}} a t^2$; $s_i$ is the location at the initial time, $v_i$ is the velocity at the initial time, and $t$ is the total time. Your relationship assumes $s_i$ is zero. I use $s$ instead of $d$ for distance to not confuse distance with the derivative used below.

Acceleration is defined as $a = d^2s/dt^2$ for distance $s$ for one dimensional motion. (For general motion, acceleration is a vector $\vec a = {\vec d}^2s/dt^2$.) This differential equation can be solved for $s$ giving the above result.

Again, this result is for constant acceleration only.

Answer 2

So, this is easier when you learn calculus.

Without calculus, it requires one of many "magical" facts to cover the gap that calculus would do for you. Here is the simplest that I know: that under a circumstance of constant acceleration $a$ over a time $t$ from velocity $v_0$ to velocity $v_1=v_0 + a~t$, the average velocity can be written in two different ways:$$ v_\text{avg} = \frac{d}{t} = \frac{v_0 + v_1}2. $$ That first equals sign is the definition of average velocity, the second equal sign is a relation that only holds for constant acceleration. If for example you had a sudden acceleration “kick” near the start time, then for almost all of the time your velocity would have been $v_1$ and that would be your average velocity. Or if you had had a sudden “kick” from $v_0$ to $v_1$ near the end of the time $t$, then for almost all of the time your velocity would have been $v_0$ and that would be your average velocity. There are lots of possibilities. But for constant acceleration it turns out that the average velocity is also the average of the starting and ending velocities, which makes constant acceleration very simple.

Anyway, if you substitute the above relation for $v_1=v_0+a~t$ into that expression and multiply both sides by $t$ you get the desired equation for $d$. So they are perfectly equivalent. In addition, by eliminating $t$ instead of $v_1$ you can instead prove the work energy theorem by a difference of squares factorization.