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Find the force acting on an object of mass 10 kg with the given position function (in units of meters and seconds).

$$r(t)=\left\langle 20 t-3,-16 t^{2}+2 t+30\right\rangle$$

$\langle 0,-320\rangle$

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Oregon State University

University of Michigan - Ann Arbor

Idaho State University

Boston College

Yeah, This time they give us a position function of 20 t mine in ST a negative 16 t squared Waas to T plus 30. And, well, let's see, they want us to compute the force. So hurry and take some derivatives to get the acceleration. We'll have 20 in the first component negative 30 30 plus student and the second component, ANA. The second derivative, which is the acceleration, will be zero on Get a 32 well to get the first for granted. The mess they've given that it's on a kilogram well that are forced back here with respect to time is given by Let's See, the constant vector is zero on negative 320.

University of California, Berkeley