If a ball is rolling down an inclined ramp straight, will the change in position be squared for every passing second or would it just be multiplied by two? I'm not sure if this makes sense so here's what I mean in an example for the change in position will be squared for every passing second : 1 Second = 4 cm, 2 seconds = 16 cm, 3 seconds = 256 cm... etc

so

dx/dt = a t
dx = a t dt
at t = 0, dx = 0
at t = 1, dx = a
at t = 2, dx = 2a
at t = 3, dx = 3a
at t = 4, dx = 4a etc
in other words you are looking at v(t), the velocity, which is linear with time (the CHANGE in position, not the position)
v = a t

the POSITION itself will be quadratic in time of course. When you integrate x = k t you get x = (1/2)kt^2
POSITION is then

Thank you! And just to be sure I really understand what you're saying would this example be correct?

0 = 0
1 = 4 = a
2= 2(4) = 8
3 = 3(4) = 12
4 = 4(4) = 16
5 = 5(4) = 20

yes,your table is for a = dv/dt = 4

so dv = 4 * dt
dv/dt = a = 4
v = a t = 4 t
x = (1/2) a t^2
t a v x
0 4 0 0
1 4 4 2
2 4 8 8
3 4 12 18
4 4 16 32
5 4 20 50
It is that v column you want, how far it moves PER SECOND
not how far it moved total.

To understand the change in position of a ball rolling down an inclined ramp, we need to consider the concept of acceleration due to gravity and how it affects the ball's motion.

When an object is freely falling under the influence of gravity, it experiences constant acceleration. In the case of a ball rolling down an inclined ramp, the component of gravity along the ramp surface causes the ball to accelerate.

The mathematical relationship between the distance traveled and time for an object with constant acceleration is given by the equation:

d = 0.5 * a * t^2

where:
- d represents the distance traveled,
- a is the acceleration,
- t is the time elapsed.

In this case, since the ball is rolling down an inclined ramp, the acceleration (a) is equal to the acceleration due to gravity (g) multiplied by the sine of the ramp's angle (θ). Therefore:

a = g * sin(θ)

Here, g is the acceleration due to gravity (approximately 9.8 m/s^2 on Earth), and θ is the angle of the inclined ramp.

Now, let's take your example to determine the change in position of the ball for each passing second. Suppose the initial distance (d) covered by the ball in the first second is 4 cm.

Using the equation mentioned earlier (d = 0.5 * a * t^2), we can rearrange it to solve for a:

a = 2 * d / t^2

Substituting the given values, we get:

a = 2 * 4 cm / (1 s)^2 = 8 cm/s^2

Now, for the second second, we can calculate the distance traveled (d) by using the equation:

d = 0.5 * a * t^2

Substituting the values:

d = 0.5 * 8 cm/s^2 * (2 s)^2 = 0.5 * 8 cm/s^2 * 4 s^2 = 16 cm

As you can see, the distance traveled in the second second is indeed 16 cm. However, it is not directly related to the previous distance traveled (4 cm). Each time, the distance is calculated using the time squared, as per the mathematical equation for constant acceleration.

So, in summary, the change in position of the ball rolling down an inclined ramp straight will be governed by the formula d = 0.5 * a * t^2, where the distance (d) is directly related to the time squared (t^2) due to constant acceleration.