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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).$$r(t)=\left(3 t^{2}+t, 3 t-1\right)$$

## $$\mathbf{F}(t)=m \mathbf{a}(t)=m \mathbf{r}^{\prime \prime}(t)$$, $$\mathbf{F}(t)=\langle-60,0\rangle$$

Vectors

Vector Functions

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##### Top Calculus 3 Educators ##### Catherine R.

Missouri State University ##### Heather Z.

Oregon State University  ##### Kristen K.

University of Michigan - Ann Arbor

### Video Transcript

they give us a position function clients see, our of T is equal to three D squared Waas teak three T minus one and they want us to compute force. What? We're going to find the acceleration first. So let's start taking our derivatives well at 60 plus 103 on the acceleration is given by the second derivative. So we end up with six. And Europe. Well, we're going to be the force we need Mass on our mass is 10 kilograms. That's given. So our force, despite the time, is given by the vector 60 on zero constant. University of California, Berkeley

#### Topics

Vectors

Vector Functions

##### Top Calculus 3 Educators ##### Catherine R.

Missouri State University ##### Heather Z.

Oregon State University  ##### Kristen K.

University of Michigan - Ann Arbor