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--- repository:phineas_ferb [2021/02/17 18:55]
porcaro1 [Answer Key]
+++ repository:phineas_ferb [2021/02/17 18:55]
porcaro1 [Answer Key]
@@ Line 104: / Line 104: @@
 ====Answer Key====
 ===Handout===
-  - The process for determining where and when Candace will be able to take a photo of the boys (i.e. when Candace's position intersects with Phineas and Ferb's) starts by breaking the problem down into two parts: before and after Candace begins moving. If we define Candace as being at the origin ($x=0$), then we must first determine where Phineas and Ferb are once Candace gets on her scooter. To do this, we can use the following kinematic equation (which will helpful for the entire problem): $x=x_0 + vt + at^2$, where $x$ is current position, $x_0$ is initial position, $v$ is velocity, $a$ is acceleration, and $t$ is time. We know Phineas and Ferb start 20 m to the left of Candace, they are travelling at a constant velocity of 4 m/s, and it takes Candace 20 seconds to get on her scooter, therefore we can find where Phineas and Ferb are in relation to Candace: $x=-20 + 4(20) + 0(20)^2 = 60$. Phineas and Ferb are 60 m to the right of Candace when she starts her scooter. Next (similarly to the [[https://www.msuperl.org/wikis/icsam/doku.php?id=repository:head-on_collision | Head-On Collision]] and [[https://www.msuperl.org/wikis/icsam/doku.php?id=repository:rear-end_collision | Rear-End Collision]] activities) we will create a system of equations describing the motion of Phineas and Ferb and Candace and solve for the unknown variables. $$x_{P2}=60+4t$$ $$x_C=\dfrac{1}{2}*2*t^2$$
+  - The process for determining where and when Candace will be able to take a photo of the boys (i.e. when Candace's position intersects with Phineas and Ferb's) starts by breaking the problem down into two parts: before and after Candace begins moving. If we define Candace as being at the origin ($x=0$), then we must first determine where Phineas and Ferb are once Candace gets on her scooter. To do this, we can use the following kinematic equation (which will helpful for the entire problem): $x=x_0 + vt + at^2$, where $x$ is current position, $x_0$ is initial position, $v$ is velocity, $a$ is acceleration, and $t$ is time. We know Phineas and Ferb start 20 m to the left of Candace, they are travelling at a constant velocity of 4 m/s, and it takes Candace 20 seconds to get on her scooter, therefore we can find where Phineas and Ferb are in relation to Candace: $x=-20 + 4(20) + 0(20)^2 = 60$. Phineas and Ferb are 60 m to the right of Candace when she starts her scooter. Next (similarly to the [[https://www.msuperl.org/wikis/icsam/doku.php?id=repository:head-on_collision | Head-On Collision]] and [[https://www.msuperl.org/wikis/icsam/doku.php?id=repository:rear-end_collision | Rear-End Collision]] activities) we will create a system of equations describing the motion of Phineas and Ferb and Candace and solve for the unknown variables. $$x_{P\text{&}F}=60+4t$$ $$x_C=\dfrac{1}{2}*2*t^2$$
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