# Motion under constant acceleration

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 In what follows we will describe the motion of a body moving under constant aceleration a0.     That might f.x. be a body making a free fall.  If we look at the (t,v)-graph for such a motion we end up with a straight line, because the increase in velocity per unit time is a constant. The acceleration a0 is determined as the slope of the line, as we know that v'(t) = a.

From the fact that we are dealing with a linear function of velocity we can write down the velocity equation

v(t)= v0 + a0.t ,

where v0 is the initial velocity.

We wish to determine the function of position s(t) of the motion.

From what we have seen in the preceeding section we know that the covered distance Ds at the time t is equal to the area lying under the (t,v)-graph.

Fortunately the graph of velocity is a straight line, making it possible for us to determine the area with the help of some simple geometrical considerations.

Start!
In order to find out just how far the ball has dropped under the constant of acceleration, we must find an expression of the blue area under the (t,v)-graph.

1. Show the area.

2. Slice the area in two. The total area is the sum of the rectangular area and the triangular area.

3. Put on notation. Now, we know that v(t) concists of two terms, and as the height of the rectangle is v0 , then the height of the triangle must be given by the second term - that is a0.t.

4. Calculate the area.

In other words the covered distance Ds can be calculated as

Ds = ˝.a0.t2 + v0.t

The covered distance Ds only expresses the change in the bodys position.

The final positionen is s(t), and if we denotes the initial positionen s0, we get

Ds = s(t) - s0 =  ˝.a0.t2 + v0.t

or

s(t) =  ˝.a0.t2 + v0.t + s0 .

The graph of position for the motion under constant acceleration is a parabola.

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- Sorry, no translations.

- End -

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