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A particle moves along a straight line with equation of motion s = f(t), where s is measured in meters and t in seconds. Find the velocity and speed when t = 3. f(t) = (t ^-1) - t
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i think the inverse of the t is kicking me
V(t)=S'(t) = -1*t^(-2) - 1. When t = 3 V(t) =[ - 1 / (3^2) ] - 1 V(t) = -1/9 - 1 V(t) = -10/9
Position = f(t) Velocity = f'(t) [ 1st derivative of position] Acceleration = f''(t) [2nd derivative of position, or 1st derivative of velocity] We can write the position function like this, if it helps. \[f(t) = t^{-1}-t = \frac{1}{t}-t\]
Speed is the same as velocity but it doesn't have direction, so you just need the absolute value of velocity Speed = | V(t) | = | - 10/9 | = 10/9
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