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Problem 22 Medium Difficulty

If a freely falling body starts from rest, then its displacement is given by $ s = \frac{1}{2} gt^2 $. Let the velocity after a time $ T $ be $ v_T $. Show that if we compute the average of the velocities with respect to $ t $ we get $ v_{ave} = \frac{1}{2} v_T $, but if we compute the average of the velocities with respect to s we get $ v_{ave} = \frac{2}{3} v_T $.

Answer

$\frac{2}{3} v_{T}$

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Video Transcript

recall the fact for this question that velocity is the derivative of position. In other words, ask prime of tea, as indicates position in this context. Therefore, the average lost you with respected teed one over tee times 1/2 This is 1/2 V of tea V s g t can t can be written this just simple algebraic manipulation as V over G. Therefore, V is the square root of two gs. Therefore we know that are average There's one over us Humanist es Juan the integral from us want us to the of STS is the same thing is one of her B minus A. And to grow from a bead on the function we're just using us Want us to in this context, therefore we end up with We're just putting in now what we know and then we have Diaz instead of GTX are variables s and this is a corpulent to 2/3 v of tea