Acceleration, velocity and distance

From GeoMod

Imagine we have a ball that is held steady at a certain height and then dropped. How do we figure out where the ball will be at any point in time?

NOTE:

s = distance
v = velocity
a = acceleration
t = time

First, when we release the ball it moves because of gravity. What is gravity? It's the force resulting from the attraction of masses to each other. For our ball it is the force attracting it and the Earth to one another. Of course the Earth is a bit larger than our ball, and indeed anything at the surface of the Earth is mostly affected by the mass of the Earth since its mass dwarfs the mass of pretty much anything at the surface. Since force = mass * acceleration, and the force causing anything to fall at the surface of the Earth is the same for pretty much everything at the surface of the Earth, the acceleration for is the same for any object falling that is near the surface of the Earth. Of course you already know this, and know that the acceleration is approximately 9.8 m/s².

OK. So

• our ball starts with no velocity, therefore our initial velocity (v₁ = 0 m/s);
• it accelerates at the rate of (a = 9.8 m/s²).

So where will be ball be after 1 second? Well we know that

• acceleration is the rate of change of velocity with time (a = dv/dt)
• velocity is the rate of change of distance with time (v = ds/dt)

Therefore to find the position (s) after time (t=1) we need to take our acceleration equation and integrate twice !!!

First integration to get velocity

 a    = dv/dt = g
∫ a dt =   v   = gt + c

if we assume that the initial velocity is v₁ at t = 0 then

c = v1

so our equation becomes

v = gt + v1

Second integration to get position;

 v    = ds/dt = gt + v1
∫ v dt = s = 0.5 g t2 + v1 t + c

To get the new constant we say that the initial position of the ball is s₁ at t = 0, which results in

c = s1

Therefore our final equation for the position of the falling ball is:

s = 0.5 g t2 + v1 t + s1

And this is the analytical solution to finding the position of the falling ball. Of course there are some limitation to the use of this equation but it can be useful.