B Why is displacement the integral of acceleration with respect to time? [corrected in thread]

[spaceman0x2a]

TL;DR Summary

Question about s = ut + 1/2(at^2)

Looking at s = ut + 1/2(at^2), the first part makes sense to me, but I am confused about the 1/2(at^2). I can see that this is the integral of acceleration with respect to time, but I don't understand why. Is this simply a coincidence? I know that this is considering a linearly rising acceleration, but it feels like it means something.

 

[spaceman0x2a]

And yes, I know this is similar to my previous post here.

 

[PeroK]

TL;DR Summary: Question about s = ut + 1/2(at^2)

Looking at s = ut + 1/2(at^2), the first part makes sense to me, but I am confused about the 1/2(at^2). I can see that this is the integral of acceleration with respect to time, but I don't understand why. Is this simply a coincidence? I know that this is considering a linearly rising acceleration, but it feels like it means something.

This is only for constant acceleration. The displacement is the (signed) area under a velocity-time graph. For constant acceleration, the graph has two geometric components:

1) The $ut$ represents the area of a rectangle, of height $u$ and length $t$.

2) The $\frac 1 2 at^2$ represents the area of a triangle of length $t$ and height $at$. The area of a triangle being $\frac 1 2$ times base time height.

 

[spaceman0x2a]

This is only for constant acceleration. The displacement is the (signed) area under a velocity-time graph. For constant acceleration, the graph has two geometric components:

1) The $ut$ represents the area of a rectangle, of height $u$ and length $t$.

2) The $\frac 1 2 at^2$ represents the area of a triangle of length $t$ and height $at$. The area of a triangle being $\frac 1 2$ times base time height.

Thanks.

I just went and looked at this further and realized the mistake I was making - I thought it was the integral of acceleration with respect to time, when it was actually velocity ($at$) with respect to time. I feel really stupid now, but I guess this is how you learn.

 

[kuruman]

A simple way to look at this is to consider the displacement as the "area under the velocity vs. time curve" which is really the integral. The velocity under constant acceleration is a straight line $v=u+at$. The plot on the right shows this and the colored area under the curve.

You can see that the total area under the curve is the area of the blue rectangle plus the area of the red triangle. Then in words, the equation $$s=ut+\frac{1}{2}at^2$$ says that the displacement under constant acceleration in time $t$ is the sum of two displacements: (a) the displacement as if the object moved at constant velocity and (b) the additional displacement due to the constant acceleration.