Q2/Deceleration of a particle is based on relation a=-3 v² m/s² where v in m/s. If it moves along a straight line and has velocity 10 m/s and position s = 8m when t=0, determine its velocity and position when t= 3 s. Where the velocity become zero. Discuss briefly.​

Explanation:

Given: a = -3v^2

By definition, the acceleration is the time derivative of velocity v:

Re-arranging the expression above, we get

Integrating this expression, we get

Since v = 10 when t = 0, that gives us k = -1/10. The expression for v can then be written as

or

We also know that

or

We can integrate this to get s:

Let u = 30t +1

du = 30dt

so

So we can now write s as

We know that when t = 0, s = 8 m, therefore k = 8 m.

Next, we need to find the position and velocity at t = 3 s. At t = 3 s,

Note: velocity approaches zero as t -->

fgravity=10x10=100n case 1

fgravity=30x20=600n case 2

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