Answer: ![p_{2}(t)=(-8t^{2}+3t-5,7t^{2}+10,8t^{2}+6t+12)](https://tex.z-dn.net/?f=p_%7B2%7D%28t%29%3D%28-8t%5E%7B2%7D%2B3t-5%2C7t%5E%7B2%7D%2B10%2C8t%5E%7B2%7D%2B6t%2B12%29)
Step-by-step explanation: Velocity is the first derivative of position function, so:
![p'_{1}(t)=(-16t+3,14t,16t+6)](https://tex.z-dn.net/?f=p%27_%7B1%7D%28t%29%3D%28-16t%2B3%2C14t%2C16t%2B6%29)
Since in the second scene acceleration is zero, velocity must be constant, which means velocity in the second scene is the same as in the first scene, i.e.,
![p'_{2}(t)=p'_{1}(t)](https://tex.z-dn.net/?f=p%27_%7B2%7D%28t%29%3Dp%27_%7B1%7D%28t%29)
![p'_{2}(t) = (-16t+3,14t,16t+6)](https://tex.z-dn.net/?f=p%27_%7B2%7D%28t%29%20%3D%20%28-16t%2B3%2C14t%2C16t%2B6%29)
To determine position function, integrate it:
![p_{2}(t) = \int\limits {(-16t+3,14t,16t+6)} \, dt](https://tex.z-dn.net/?f=p_%7B2%7D%28t%29%20%3D%20%5Cint%5Climits%20%7B%28-16t%2B3%2C14t%2C16t%2B6%29%7D%20%5C%2C%20dt)
![p_{2}(t) = (-8t^{2}+3t+c_{1},7t^{2}+c_{2},8t^{2}+6t+c_{3})](https://tex.z-dn.net/?f=p_%7B2%7D%28t%29%20%3D%20%28-8t%5E%7B2%7D%2B3t%2Bc_%7B1%7D%2C7t%5E%7B2%7D%2Bc_%7B2%7D%2C8t%5E%7B2%7D%2B6t%2Bc_%7B3%7D%29)
Now, the "initial condition" is that
![p_{2}(0)=p_{1}(1)](https://tex.z-dn.net/?f=p_%7B2%7D%280%29%3Dp_%7B1%7D%281%29)
![p_{2}(0) = (c_{1}c_{2},c_{3})](https://tex.z-dn.net/?f=p_%7B2%7D%280%29%20%3D%20%28c_%7B1%7Dc_%7B2%7D%2Cc_%7B3%7D%29)
![p_{1}(1)=(-8.1^{2}+3.1,7.1^{2}+3,8.1^{2}+6.1-2)](https://tex.z-dn.net/?f=p_%7B1%7D%281%29%3D%28-8.1%5E%7B2%7D%2B3.1%2C7.1%5E%7B2%7D%2B3%2C8.1%5E%7B2%7D%2B6.1-2%29)
![(c_{1},c_{2},c_{3})=(-5,10,12)](https://tex.z-dn.net/?f=%28c_%7B1%7D%2Cc_%7B2%7D%2Cc_%7B3%7D%29%3D%28-5%2C10%2C12%29)
![p_{2}(t)=(-8t^{2}+3t-5,7t^{2}+10,8t^{2}+6t+12)](https://tex.z-dn.net/?f=p_%7B2%7D%28t%29%3D%28-8t%5E%7B2%7D%2B3t-5%2C7t%5E%7B2%7D%2B10%2C8t%5E%7B2%7D%2B6t%2B12%29)
<u>Position function in the second scene is </u>
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