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--- repository:phineas_ferb [2021/02/16 23:32]
porcaro1 [Activity Information]
+++ repository:phineas_ferb [2021/02/17 19:08] (current)
porcaro1
@@ Line 17: / Line 17: @@
 ====Activity====
 ===Handout===
+{{ :repository:phinease_ferb_1.png?nolink&600|}}
 ** Phineas & Ferb **
 Phineas and Ferb just invented hovercraft skateboards, and Candace, being Candace, needs to take of picture of them to prove, once and for all to their mom, that Phineas and Ferb are still up to their crazy and dangerous shenanigans. Candace sees Phineas and Ferb riding their hoverboards 20 meters behind her, travelling at a constant velocity. She realizes that this is her opportunity; it takes her 20 seconds to grab her camera, bag, and hop on her electric scooter to chase them. She starts from rest and continues (miraculously) with constant acceleration.
   - Assuming Phineas and Ferb are travelling at 4 meters per second and Candace's scooter is capable of accelerating at 2 meters per second per second, where and when will Candace be able to take a photo of the boys?
@@ Line 28: / Line 30: @@
     - Final time
 ===Code===
-[[https://trinket.io/glowscript/b53b073702 | Link]]
+[[https://www.glowscript.org/#/user/porcaro1/folder/RepositoryPrograms/program/Phineas&Ferb-Incomplete | Link]]
 <code Python [enable_line_numbers="true"]>
 GlowScript 2.7 VPython
@@ Line 70: / Line 72: @@
 #Motion of P_F before Candace starts moving
 while t < t_CandaceDelay:
-    rate(500)
+   rate(500)
@@ Line 102: / 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.
+  - 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: $$x_{P\text{&}F}=60+4t$$ $$x_C=\dfrac{1}{2}*2*t^2$$ where $x_{P\text{&}F}$ describes Phineas and Ferb's position as a function of time and $x_C$ describes Candace's. If we set these equations equal to each other, we can solve for $t$, the time after Candace hopped on the electric scooter. We find $t=10$ seconds. Plugging this value into either of the two equations above will tell us when Candace reaches the boys. $60+4(10)=\dfrac{1}{2}*2*10^2=100$ meters. In summary, it will take Candace 30 seconds to reach the boys (20 seconds to grab her purse and 10 seconds to accelerate on her scooter) and she will reach Phineas and Ferb after travelling 100 meters.
+  - We know that acceleration is the rate at which velocity changes with respect to time; in other words $v=at$. Since we know Candace is accelerating at 2 meters per second per second and it takes her 10 seconds to reach Phineas and Ferb, we can solve for velocity and find that she was travelling 20 meters per second (almost 45 mph!) when she reaches the boys.
+  - See highlighted code below
+{{ :repository:phineas_ferb_2.png?nolink&600 |}}
 ===Code===
+[[https://www.glowscript.org/#/user/porcaro1/folder/RepositoryPrograms/program/Phineas&Ferb-Solution | Link]]
 <code Python [enable_line_numbers="true", highlight_lines_extra="42,45,52,55,56,61,62,63,64,65"]>
 GlowScript 2.7 VPython
@@ Line 178: / Line 184: @@
 ----
 ====See Also====
-  *
+  *[[colliding_crates | Colliding Crates]]
+  *[[head-on_collision | Head-On Collision]]
+  *[[rear-end_collision | Rear-End Collision]]