Why is the acceleration value halved in distance calculations?

[eax]

Lets take an example of someone traveling 10m/s^2 the person is traveling for 3seconds how far did he travel. Easy 10*3^2*0.5 = 45. Now with units it would look something like this

10m/s^2 * 3s^2 * 0.5

The 3s^2 makes the s^2 part irrevelant for 10m/s^2 thus becomes 90m. But why is it halved? All I know is that "it is the area under the the graph" so that doesn't proove enough. Why is it "the area under the graph"?

Thanks in advance!

 

[dextercioby]

Because of the definition.The definition involves a definite integral which happens to be the area under the graph of the velocity.But velocity is constant which means its graph is a straight line.Therefore the area is very easy to compute,because it's the area of a right triangle...(and that involves the factor 1/2)

Daniel.

 

[Doc Al]

Another way to see where that 1/2 comes from is to think of it this way:
Distance = (Ave Speed)*(Time)
Ave Speed = (initial speed + final speed)/2 [since the acceleration is uniform]
Ave Speed = (final speed)/2 [since it starts from rest]
final speed = at [since acceleration is uniform]
So... D = (at)/2 (t) = 1/2 a t^2.

 

[VietDao29]

Hi,
First, draw a graph of $v = v_{0} + at$ with the axis Ox is t and axis Oy is v.
And think that, if within a $\Delta t$ time very very small, the speed does not change much. So if $\Delta t \rightarrow 0$, the speed will change less and less. So within that very small amount of time, the distance an object can go is: $v_{t}\Delta t$. That distance can be represented by the area of a small rectangle with the height of $v_{t}$, and the width of $\Delta t$. So in a lot of $\Delta t$ time, which sum up to be the amount of time the object goes, the distance the object will go is the area under the graph.
Hope it help,
Viet Dao,

 

[misogynisticfeminist]

If you see a coefficient which is the reciprocal of a variable's power, you have good reason to suspect integration.