+584 Now,I want to show you guys some physics and mathematics.
acceleration is the change in velocity in a given amount of time.\(a=\frac{V-Vo}{t}\)
where V is the final veocity and Vo is the initial velocity.
rearrange,we have \(V=at+Vo\)If the object's accelration is constant,then the graph of the velocity is a linear function.
But I would like to express it in y=mx+b form in my graph.And here is my graph
The area of the triangle is the sum of the two base times the hight divide by two.or A=(b1+b2)*h/2
In here ,we could express it as \(A=(V+Vo)*\Delta t/2\)
the area under the curve(line) of function of velocity is the displacement.
Therefore,\(\Delta S=(V+Vo)*\Delta t/2\)\(a=\frac{V-Vo}{t}\Rightarrow t=\frac{V-Vo}{a}\)
If the object start moving at time 0 seconds
\(\Delta t=T-To=t\)
the displacement of the object from time 0 seonds is \(\Delta S=\frac{V^2-Vo^2}{2a}\)
substitute V=at+Vo into \(\Delta S=(V+Vo)*\Delta t/2\)
we have \(\Delta S=(2Vo+at)*t/2\Rightarrow \Delta S=1/2at^2+Vo*t\)
displacement equal final position subract inital position \(\Delta S=S-So\)
\(S=\Delta S+So\)\(S=1/2at^2+Vo*t+So\)
Confirm my previous equation by using integration
\(V=\frac{ds}{dt}=a*t+Vo \Rightarrow \int \frac{ds}{dt}dt=\int at+Vo dt\Rightarrow S=1/2at^2+Vo*t+C\) for some costant C
In here C is the initial position,so \(S=1/2at^2+Vo*t+So\)
http://www.desmos.com/calculator/ibbxgibq5n (my original graph from demos)
