💬 👋 We’re always here. Join our Discord to connect with other students 24/7, any time, night or day.Join Here! # Show that for motion in a straight line with constant acceleration $a$, initial velocity $v_0$, and initial displacement $s_0$, the displacement after time $t$ is$$s = \dfrac{1}{2}at^2 + v_0t + s_0$$

## Distance with respect to time is equal to $s(t)=1 / 2 a t^{2}+v_{0} t+s_{0}$ Where $v_{0}$is initial speed and $s_{0}$ is the initial displacement.

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##### Top Calculus 2 / BC Educators  ##### Catherine R.

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### Video Transcript

So we want to show that for motion in a straight line with constant acceleration, initial velocity and initial displacement Vienna and S not, respectively, that we have our displacement after time. T. Is S. Equals one half a T squared because Vienna T plus asthma. So our constant acceleration A. Which means R a F. P function, it's just constant acceleration act. And that tells us that our velocity function is equal to 80 because aid constant plus Vienna, our velocity constant. Um in this case uh this would be plus C. But because we know that Vienna is the initial velocity will put that there. And we want to find our ask our position function that's going to be a T squared over two plus B, not T. Empty plus our initial position which is S not. So therefore we have proven what we wish to prove. California Baptist University

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Derivatives

Differentiation

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##### Top Calculus 2 / BC Educators  ##### Catherine R.

Missouri State University  Lectures

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