V = X'(t) = -t^2 + 6 m/s Implies X(t) = -(1/3)t^3 + 6t The initial...

V = X'(t) = -t^2 + 6 m/s

Implies

X(t) = -(1/3)*t^3 + 6*t

The initial position is zero in this case.

This function will initially increase (this means the object is moving forward), then the negative force will be too strong, so the object will start moving backwards.

calculate the maxima and minima. Graph the function X(t). Describe the motion. Every detail.

Solutions

Expert Solution

Velocity of a particle as a function of time is given.Initially the particle is at x=0.. It is required to find points of maximum and minimum of x if they exist.It is also asked to draw the graph of x versus t.and to describe the motion of the particle in details.We determine the acceleration function.With the help of position , velocity and acceleration functions we answer the questions,

genius_generous answered 1 year ago

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V = X'(t) = -t^2 + 6 m/s Implies X(t) = -(1/3)*t^3 + 6*t The initial...

Solutions

Expert Solution

Related Solutions

V = X'(t) = -t^2 + 6 m/s Implies X(t) = -(1/3)t^3 + 6t The initial...