SOLUTION: A particle is moving according to an equation of motion. Its position after t seconds is given by s(t) = 2t^3 - 11t^2 + 13t - 1. After 1 second its position is at 3. Where is the o

Question 1105949: A particle is moving according to an equation of motion. Its position after t seconds is given by s(t) = 2t^3 - 11t^2 + 13t - 1. After 1 second its position is at 3. Where is the original position of the particle? Explain.
How far from the starting point is the particle at the end of 5 seconds?
Answer by ikleyn(52781) (Show Source): You can put this solution on YOUR website!
.
A particle is moving according to an equation of motion. Its position after t seconds is given by s(t) = 2t^3 - 11t^2 + 13t - 1.
After 1 second its position is at 3.
Where is the original position of the particle? Explain. How far from the starting point is the particle at the end of 5 seconds?
~~~~~~~~~~~~~~~~~~~~

0.  The info  "After 1 second its position is at 3." is EXCESSIVE, UNNECESSARY and is not used in the solution.

    It is overlayed by this statement: 

        "A particle is moving according to an equation of motion. 
         Its position after t seconds is given by s(t) = 2t^3 - 11t^2 + 13t - 1."



1.  "Where is the original position of the particle?"

    Answer:  Substitute t= 0 into the formula and calculate s(0).



2.  "How far from the starting point is the particle at the end of 5 seconds?".

    Answer:  Substitute t= 5 into the formula and calculate s(5).

             Then calculate  delta_s = s(5) - s(0).


             It will be your final answer.

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