The position of a point on a line is given by the equation(t)= t3-6t2+9t-4, where s...

The position of a point on a line is given by the equation(t)= t3-6t2+9t-4, where s is measured in metres and t in seconds. What is the velocity of the point after 2 seconds? What is its acceleration after 4 seconds? Where is it when is first stops moving? How far has it travelled when its acceleration is 0?

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Solution:-

Note: Remaining part in (c)

From the graph drawn for velocity vs time, we can observe that the particle stops at t = 1sec for the first time. Now, we need to find where the particle is at this time,

So, the position of particle at t = 1sec:

S( t =1) = (1)^3 - 6*(1)^2 +9*(1) - 4 = 0, so particle is at the origin at t= 1sec. As S(t=0) = - 4, it means that the particle started to move from s= - 4 ( 4 unit left side of origin).

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milcah answered 1 year ago

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