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    Topics || Problems

    Find the velocity of a point whose position is given by the equation \( s = 14t^3 + 5t -1\), where t = 2.

    Solution:

    \( \frac{ds}{dt} = velocity \)

    \( s = 14t^3 + 5t -1 \)

    \( \frac{ds}{dt} = 42t^2 +5 \)

    At t = 2

    \( \frac{ds}{dt} = 42(2)^2 +5 \)

    \( \frac{ds}{dt} = 173 \)