I am working on a problem and I am stuck. Here is theproblem: A particle moves a

Question

I am working on a problem and I am stuck. Here is theproblem:

A particle moves along the x-axis according to the equationX=2.00+3.00t-1.00t2 where x is in meters and t isin seconds. At t=3.00s find:

A: Position of the particle
B: It's Velocity
C: It's Acceleration

Cramster has the solutions to this problem, and I was able tofigure out part A and part B. Part A is simple substitution. Part B you just take the derivative of (2+3t-1t2)and then plug in t=3. According to the solution here onCramster for part C shows dv/dt. Well we found v in part Bwhich is 3.00, but I am unclear where the -2.00t comes from. Any clarification would be great! Thanks.

Explanation / Answer

given    X=2.00+3.00t-1.00t2 dX /dt = 0+3 - 2t                                            here from the law of derivation we have d (xn) /dt = n* xn-1 again we have         d2 X /dt2 = -2 we know that d2 X /dt2 = a    , acceleration so         a =-2   m/s2 which is constant , do notdepends up on the time again we have         d2 X /dt2 = -2 we know that d2 X /dt2 = a    , acceleration so         a =-2   m/s2 which is constant , do notdepends up on the time

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