A ball rolls down an inclined plane such that the distance (in centimeters) that it rolls in t seconds is given by s(t) = 2t^3 + 3t^2 + 4 for 0<=t<=3 At what time is the velocity 30cm/s?

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16-01-2017
Mathematics

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A ball rolls down an inclined plane such that the distance (in centimeters) that it rolls in t seconds is given by s(t) = 2t^3 + 3t^2 + 4 for 0<=t<=3 At what time is the velocity 30cm/s?

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LammettHash LammettHash · Answer 1 · 2017-01-16T18:55:36+01:00

The ball's velocity will be represented by the derivative of its distance function:

[tex]s'(t)=6t^2+6t[/tex]

Now find the times [tex]t[/tex] for which this is equal to 30, i.e. solve

[tex]6t^2+6t=30\implies t^2+t-5=0[/tex]

This has two solutions, [tex]t=-\dfrac{1\pm\sqrt{21}}2[/tex], but only one is positive and falls in the interval [tex][0,3][/tex]. So the velocity reaches 30 cm/s when [tex]t=-\dfrac{1-\sqrt{21}}2\approx1.79[/tex].