Question

An animation has two scenes, each of them being one second long. In the first scene, an object moves along the path p⃗ 1(t)=(−8t2+3t,7t2+3,8t2+6t−2) for 0≤t≤1. At the end of the scene, the object’s acceleration immediately becomes zero, and remains zero throughout the second scene. The vector-valued function p⃗ 2(t) for the object’s path in the second scene also uses 0≤t≤1, so that t=0 in the second scene is the same moment as t=1 in the first scene. What is p⃗ 2(t)?

Answer 1

Answer: $p_{2}(t)=(-8t^{2}+3t-5,7t^{2}+10,8t^{2}+6t+12)$

Step-by-step explanation: Velocity is the first derivative of position function, so:

$p'_{1}(t)=(-16t+3,14t,16t+6)$

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.,

$p'_{2}(t)=p'_{1}(t)$

$p'_{2}(t) = (-16t+3,14t,16t+6)$

To determine position function, integrate it:

$p_{2}(t) = \int\limits {(-16t+3,14t,16t+6)} \, dt$

$p_{2}(t) = (-8t^{2}+3t+c_{1},7t^{2}+c_{2},8t^{2}+6t+c_{3})$

Now, the "initial condition" is that

$p_{2}(0)=p_{1}(1)$

$p_{2}(0) = (c_{1}c_{2},c_{3})$

$p_{1}(1)=(-8.1^{2}+3.1,7.1^{2}+3,8.1^{2}+6.1-2)$

$(c_{1},c_{2},c_{3})=(-5,10,12)$

$p_{2}(t)=(-8t^{2}+3t-5,7t^{2}+10,8t^{2}+6t+12)$

Position function in the second scene is $p_{2}(t)=(-8t^{2}+3t-5,7t^{2}+10,8t^{2}+6t+12)$