Physics Explained Through a Video Game/Motion in 2 Dimensions

Topic 1.5 - Motion in 2 Dimensions

There are situations where we may want to track the motion of an object that is moving in more than one dimension. Oftentimes, this is illustrated through projectile motion. Projectile motion is when an object travels through the air while only being accelerated by gravitational force (also known as the force due to gravity).

For now, all we need to know about forces in general is that they make objects accelerate in a certain direction. With gravitational force specifically, it results in all objects with mass to have an attractive force. This makes objects with mass to accelerate towards each other.

As a real life example, if one was to drop a banana peel on Earth, the banana peel will accelerate downwards until striking the ground. The Earth also would accelerate upwards towards the banana peel at the same time. However, the displacement of Earth because of the banana would be extremely small because the Earth has far, far more mass than the banana peel. As such, for almost all purposes, we can allow for the Earth's movement because of the banana's mass to be negligible.

For more of an introduction into gravitational attraction and other related information, check out this video by VSauce (from 2:40 to 7:27).

Consider the map Banana jungle by O_dot (below). We can see bananas accelerating towards the ground (in the downward direction) after they're released by the monkey. Until the bananas are either in contact with the ground or the jungle, the bananas are in free fall. This means that the bananas are only accelerating because of the force of gravity from the Earth. As such, the bananas have their vertical velocity become more negative while falling. However, their horizontal velocity stays the same.

A monkey throwing a banana that falls parabolically due to projectile motion.

This property can be seen by tracking the motion of the first banana thrown by the monkey (at about 2 seconds into the video). In this, the banana in free fall is accelerating at a constant rate downwards but has no acceleration horizontally. As such, note the following:

• The banana's horizontal velocity is constant after it is launched by the monkey until it hits the ground.
• However, the banana simultaneously has an increasing vertical velocity magnitude.
• Because of the above factors, the banana's path while falling becomes more vertical as time passes.
• This is because the ratio between the vertical velocity and the horizontal velocity is becoming greater in magnitude as time passes.

Air Resistance

In real life, air resistance affects all objects on Earth while they are in projectile motion in Earth's atmosphere. This is because objects collide directly with the air when moving differently than it. As such, the air can apply a drag force on the object. This force slows the object down and shortening its flight path. The drag force exerted on an object is dependent on a variety of factors, including the exact shape of the flying object and its velocity magnitude and direction.

In this content series, unless otherwise specified, we will assume that the drag force (air resistance) is negligible for all experiments to make calculations more manageable.

As explained above, air can apply a significant drag force on objects in some situations. For instance, in Emergency Landing by Raspy 667 (on the right), the visualized plane is falling at a constant vertical speed rather than accelerating towards the ocean. This specific example will be further explored in Unit 2: Forces and Newton's Laws. However, for now, it is important to discuss this from a kinematics perspective.

Gravitational force is making the plane accelerate downwards; however, an equally strong drag force is making the plane accelerate upwards. As such, because both of these forces are happening at the same time in the video, the plane is neither accelerating upwards or downwards. Instead, the plane has an acceleration of 0 m/s² until it lands. Because of this, the plane has a constant velocity (in the downwards direction) until it impacts the ocean.

Airborne in a Cavern

For this section, a physical situation will be broken down and fully explained. It is meant to provide an in-depth example for the following Practice Problems for this topic. For better understanding, it is recommended that you go through this section carefully beforehand.

With a stronger understanding of what makes objects in projectile motion accelerate, we'll now look into the trajectory of Clare (lime green color) in a modified version of Gemspark Cave by G3nius (down below).

Suppose Clare is launched upwards and towards the right in projectile motion. Assume that any vectors describing her motion have the components in (X,Y) notation in the Cartesian grid convention from Topic 1.1 - Displacement. It is given that Clare has an initial velocity vector of (10., 20.) (m/s). Also, assume that air resistance (drag force) is negligible and that the magnitude of Clare's acceleration due to gravity is 9.8 m/s².

Example 1: Calculating Angles

Calculate the angle (in degrees) at which Clare is initially launched at relative to the +X direction.

When Clare is initially launched, we're already given the initial velocity components. With this, we can calculate directly for the angle of her launch (with respect to the +X direction) by constructing a right angle triangle from the known initial vertical and horizontal velocity vectors.

More specifically, since the initial velocity is (10., 20.) m/s and is in (X,Y) notation, ${\vec {v}}_{initial,x}$has a magnitude of 10 m/s and is facing rightwards. Also, ${\vec {v}}_{initial,y}$would have a magnitude of 20. m/s and be pointed upwards. We can sketch these vector components onto Clare as shown on the right.

This illustration shows a vector component overlay representing Clare's initial velocity.

With this, we can consider the right angle triangle constructed by the ${\vec {v}}_{initial}$vector and its components. We can solve for the direction of a vector through considering its components and performing trigonometry.

With this, we can make a derivation for $\theta$ by noting that $tan(\theta )={\frac {{\textsf {opposite}}\;{\textsf {side}}}{{\textsf {adjacent}}\;{\textsf {side}}}}$.

From this, we can substitute in the known values for the sides of the drawn triangle. For reference, the side lengths represent the magnitudes of the components of ${\vec {v}}_{initial,x}$.

Therefore, $tan(\theta )={\frac {{\vec {v}}_{initial,y}}{{\vec {v}}_{initial,x}}}\implies tan(\theta )={\frac {20.}{10.}}{\textsf {(m/s)}}$.

Then, we can compose both sides of the equation by the inverse of $tan(x)$, $arctan(x)$. This allows for us to rewrite the equation as $\theta =arctan({\frac {20.}{10.}})\implies \theta =1.1071...\;{\textsf {radians}}\implies \theta =63\;{\textsf {degrees}}\;{\textsf {[2}}\;{\textsf {sigfigs]}}$

As such, we are able to calculate that at the instant when Clare is initially launched, she is traveling about 63 degrees above the +X direction.

Example 2: Considering Acceleration and Velocity Together

Consider the horizontal and vertical velocity of Clare while in the air. Because only the force of gravity is acting on Clare during then, in what direction is she accelerating towards? Also, how does this affect the horizontal and vertical velocities of Clare while in projectile motion?

With the understanding that there is only the force of gravity on Clare, the player is accelerating in the downwards direction while in projectile motion. Because of this, Clare only has a changing vertical velocity while her horizontal velocity stays constant.

To delve into this further, we know that the player's acceleration due to gravity is 9.8 m/s² in the downward direction. This is the only acceleration that is occurring on Clare. As such, the gravitation acceleration is acting fully in the vertical direction and not in the horizontal direction. Therefore, Clare has a vertical velocity that is decreasing by 9.8 m/s every second. This can be seen by the equation$v_{x,f}=v_{x,i}+a_{x}(t)$from Topic 1.4 - 1-D Uniform Accelerated Motion.

To get an intuition of Clare's changing velocity, check out Ohio State University's Bonk.io Physics Coding Project.

With this interactive demo (pictured on the right), we can find that if only gravitational force is acting on a player while airborne:

• The player's velocity vector will shift towards the -Y direction at a constant rate while in the air until the player bounces off the ground again.
• At all times in the video, the velocity vector's length in the horizontal direction remains the same.

What if the vertical acceleration was different?

In this example, we directly modify the overall vertical acceleration of the player while keeping the horizontal acceleration as 0.

In the clip's first half, the down arrow key is held (making the player have a more negative acceleration). As such, for when the player is airborne, the velocity vector will quickly move downwards.

This is in contrast to the clip's second half where the up arrow key is held (making the player have a less negative acceleration). In this test, the velocity vector noticeably moves downwards more slowly. Thus, we can see that by having a smaller acceleration magnitude in some direction, the velocity in that direction will change more slowly.

Example 3: Solving Multi-Step Kinematics Problems

Suppose Clare impacted the cave wall 3.5 seconds after launching with the previously specified conditions. What is Clare's net displacement magnitude during this time interval?

Although it appears that with only a time interval, the initial velocity vector, and the acceleration due to gravity, that calculating a precise value for the displacement of Clare is daunting, this problem becomes much simpler if we break it down.

Firstly, because the acceleration of Clare being vertical, the horizontal velocity stays the same for the entirety of the time interval. Because velocity is the rate at which displacement is occurring, if the velocity is constant in some direction x, ${\vec {d}}_{x}={\vec {v}}_{x}(t_{elapsed})$. Because ${\vec {v}}_{x}=10.\;{\textsf {m/s}}$and $t_{elapsed}=3.5\;{\textsf {s}}$, ${\vec {d}}_{x}=(10.)(3.5)\implies {\vec {d}}_{x}=35.\;{\textsf {m}}$.

Considering the vertical velocity, refer back to the general kinematic equation table given constant acceleration. To note, we already know ${\vec {v}}_{initial,y}$, ${\vec {a}}_{y,\;(gravity)}$, and $t_{elapsed}$. With this, we're looking for the vertical displacement ${\vec {d}}_{y}$, which is another way of saying $y_{final}-y_{initial}$. With the table, we can use the equation$x_{f}-x_{i}=v_{x,i}(t)+{\frac {1}{2}}a_{x}(t)^{2}$to solve this problem given the variables that we either know or need to solve for.

To clarify, because the direction in the formula is merely a placeholder, we can use the equation with the direction being vertical. Thus, we are solving for vertical displacement with$y_{f}-y_{i}=v_{y,i}(t)+{\frac {1}{2}}a_{y}(t)^{2}$.

$y_{f}-y_{i}=v_{y,i}(t)+{\frac {1}{2}}a_{y}(t)^{2}$

$\implies {\vec {d}}_{y}=v_{y,i}(t)+{\frac {1}{2}}a_{y}(t)^{2}\quad {\text{[Definition of displacement.]}}$

${\vec {d}}_{y}=(20.)(3.5)+{\frac {1}{2}}(-9.8)(3.5)^{2}\quad {\text{[Substitution of known variables.]}}$

${\vec {d}}_{y}=9.975\;{\textsf {m}}\quad {\text{[Calculating; sigfigs left for later]}}$

With the two displacement components ${\vec {d}}_{x}$ and ${\vec {d}}_{y}$, we can solve for the net displacement through creating a right angle with the components and using the Pythagorean Theorem (as shown to the right) in order to solve for ${\vec {d}}_{net}$.

${\vec {d}}_{net}=\pm {\sqrt {({\vec {d}}_{y})^{2}+({\vec {d}}_{y})^{2}}}\;{\text{[Derived from Pythagorean Theorem**]}}$

${\vec {d}}_{net}={\sqrt {({\vec {d}}_{y})^{2}+({\vec {d}}_{y})^{2}}}\;{\text{[The displacement magnitude is needed.]}}$

${\vec {d}}_{net}={\sqrt {(35)^{2}+(9.975)^{2}}}\;{\text{[Substitution of variables]}}$

${\vec {d}}_{net}=36.\;{\textsf {m}}\;{\text{[Calculations; 2 sigfigs]}}$

Thus, Clare has a net displacement of 36. meters in the cave during the 3.5 second time interval between her launching and her collision with the cave wall.

**If needed, check out Topic 1.2 - Calculating Displacement for additional help with the derivation.

Practice Problems

Note: The (table of kinematic equations given constant acceleration) may be helpful for some question parts.

Question 1: Suppose a snake is sliding across boulders in the map Catalina by monkey butler. The snake falls off a large boulder with a horizontal velocity of -0.3 m/s and enters free fall motion until hitting the grassy ground 1.0 m below. While in free fall, the snake is accelerating downwards at a constant magnitude of 9.8 m/s². The snake experiences negligible air resistance.

Part a)

(i) For when the snake is free falling, which acceleration vector would model the overall acceleration of the snake? Pick one of the four choices from the image on the right.

(ii) Using the Cartesian Grid Convention, would the acceleration vector of the snake be in the positive or negative direction?

Part b) If we were to solve for the amount of time that would elapse during the snake's fall, which formula should we use with the given information?

Part c) Evaluate the amount of time that elapses during the snake's fall.

Part d) Using your response from part c), determine the horizontal displacement of the snake during its fall.

Question 1 Solutions:

By considering the initial prompt, we can collect the information that is already known about the snake's free fall.

• We're told that the snake has an initial horizontal velocity of -0.3 m/sec. Therefore, $v_{x,i}=-0.3\;{\text{m/s}}$.
• Likewise, we're told that the snake initially has no vertical velocity. Thus,$v_{y,i}=0.0\;{\text{m/s}}$.
• Because the snake fell a distance of 1.0 m downwards, it will have a vertical displacement of -1.0 m during the free fall. Thus, $\Delta y=-1.0\;{\text{m}}$
• Since the snake is told to be free fall motion, gravitational force is the only force acting on the snake. Thus, the gravitational acceleration on the snake (magnitude of 9.8 m/s²) is the overall acceleration of the snake. Because objects accelerate in the downwards direction due to gravity, by the Cartesian Grid Convention, we can allow for $a_{net}=-9.8\;{\text{m/s}}^{2}$.

Through considering the prompt carefully, we're now able to quickly answer Part a):

With Part a) (i), because acceleration due to gravity is the only type of acceleration acting on the snake while it is in free fall motion, the snake's acceleration vector can be represented by an arrow pointing downwards. Thus, Choice D in the diagram shown again on the right would be the correct answer.

To answer Part a) (ii), since the snake's acceleration is pointed downwards and we're using the Cartesian grid convention, the acceleration of the snake would be in the negative direction.

In order to answer Parts b) and c), we need to find a way to directly calculate for the time it takes for the snake to fall. By considering the (kinematic equation table), we can decide which equation would be the best to use in this situation.

One way to do this is by ruling out the equations that we can't use. We can accomplish this by asking ourselves one question: If we subsistence the known variables into some equation, would be be able to directly solve for a singular unknown value?

To cut to the chase, we can't easily introduce more unknown values and be able to solve for the unknown variable that we're asked to look for. By using this process of elimination, we can decide against using Equations 1, 2, and 4. This is because all of them would introduce the unknown variable $v_{y,f}$. Because we don't know what the final vertical velocity of the snake is, choosing one of these equations would either be introducing additional work or be leading us to a dead-end.

As such, let's consider Equation 3 because of a lack of an alternative: $y_{f}-y_{i}=v_{y,i}(t)+{\frac {1}{2}}a_{y}(t)^{2}$. We know the vertical displacement of the snake, which would be equal to $y_{f}-y_{i}$ . Thus, by definition, we can rewrite the equation as: $\Delta y=v_{y,i}(t)+{\frac {1}{2}}a_{y}(t)^{2}$. Additionally we know that the initial vertical velocity, $v_{y,i}$, is equal to 0. If we substitute this into our equation, then the term $v_{y,i}(t)$would also equal 0. Thus, we can remove this term from the equation.

To clarify, even if we were considering a different situation where $v_{y,i}(t)\neq 0$, we could still eventually arrive with directly solving for the time elapsed. However, this would involve dealing with simplifying a quadratic equation, such as through using the quadratic formula. After this, we would also likely need to eliminate an extraneous solution. For this particular problem with the snake, the solving process will be simplified because of the term $v_{y,i}(t)$ being removable.

As such, we can write Equation 3 in the form $\Delta y={\frac {1}{2}}a_{y}(t)^{2}$ for the situation with the snake. Because the vertical acceleration (which is$a_{net}$) and the vertical displacement ($\Delta y$) are both known, we can solve directly for the time elapsed. Therefore, Equation 3 can be used to solve for the time elapsed, answering Part b).

Because of the derivations made above, we can use the rewritten form of Equation 3, $\Delta y={\frac {1}{2}}a_{y}(t)^{2}$, and use algebra to directly solve for $t_{elapsed}$as shown below:

$\Delta y={\frac {1}{2}}a_{y}(t)^{2}$

$\implies (t)^{2}={\frac {2\Delta y}{a_{y}}}\;{\text{[Isolating for the time variable.]}}$

$\implies t=\pm {\sqrt {\frac {2\Delta y}{a_{y}}}}\;{\text{[Square rooting both sides of the equation.]}}$

$\implies t=\mathbf {+} {\sqrt {\frac {2\Delta y}{a_{y}}}}\;{\text{[Seeking a positive elapsed time.]}}$

$\implies t={\sqrt {\frac {2(-1.0)}{-9.8}}}\;{\text{[Substitution of variables.]}}$

$\implies \mathbf {t=0.45} \;{\textbf {sec}}\;{\text{[Computation; 2 sigfigs]}}$

Thus, the snake falls for 0.45 sec between the time at which it falls off the boulder and when it impacts the grassy ground, answering Part c).

Finally, in order to approach Part d), we need to be aware that the horizontal velocity of the snake is constant. This is because there are no other forces acting on the snake apart from gravitational force. As such, we can consider our initial horizontal velocity, $v_{x,i}=-0.3\;{\text{m/s}}$. If our horizontal velocity is constant, then it will remain the same at all times while the snake is falling. Therefore, $v_{\mathbf {x} }=-0.3\;{\text{m/s}}$.

As introduced in Topic 1.3 - Speed and Average Velocity, we know that $v_{x,\;average}={\frac {\Delta x}{\Delta t}}$. Because $v_{x}$ would be equal to $v_{x,\;average}$during the snake's fall, we can substitute $v_{x}$into the expression and simplify, as shown below:

$v_{x,\;average}={\frac {\Delta x}{\Delta t}}$

$\implies v_{x}={\frac {\Delta x}{\Delta t}}\;{\text{[Substitution of variables.]}}$

$\implies \Delta {x}=v_{x}\Delta t\;{\text{[Algebra.]}}$

$\implies \Delta {x}=(-0.3)(0.4517...)\;{\text{[Substitution.]}}$

$\implies \Delta {x}={\text{0.14}}\;{\text{m}}\;{\text{[Computation; 2 sigfigs]}}$

Thus, the snake has a horizontal displacement of -0.14 m during the fall, answering Part d).

Question 2: Suppose a great white shark swallows a krill in the map SHARK!! by O_dot. The krill accelerates rapidly past the shark's jaws before disappearing in the shark's stomach 0.500 seconds later. The krill's acceleration vector relative to the shark has a constant magnitude of 19.5 m/s² and is pointed below the -X direction by 5.00 degrees. To note, the krill is assumed to initially be at rest when entering the shark's jaws.

Part a)

(i) Consider if the krill's position was recorded by a dot every 0.100 seconds between when it entered the shark's jaws ${\text{(}}\;t_{initial}=0.000\;{\text{sec)}}$to its arrival at the shark's stomach ${\text{(}}\;t_{final}=0.500\;{\text{sec)}}$. Which of the three choices below would best represent the krill's changing position during this time interval?

• 
• 
• 

(ii) Referencing the diagram to the right, calculate the magnitude of the krill's acceleration in the horizontal direction.



Part b) Calculate the horizontal displacement of the krill between 0.000 and 0.500 sec.

Part c) Suppose a researcher is solving for the krill's overall speed when it arrives in the shark's stomach ${\text{(at t = 0.500 sec)}}$. What should the researcher use as the acceleration value when solving for the krill's overall speed? Also, write a justification for your choice.

The krill's acceleration vector, as given before Part a.

The horizontal component of the acceleration vector, as calculated in Part a (ii).

Neither.

Part d)* In a new situation, suppose the researcher later spots the shark swallowing a second krill. Both krills travel into the shark with the same acceleration and travel the exact same path. However, the second krill travels into the shark with an initial velocity magnitude of 2.0 m/sec. The initial velocity is directed 5.0 degrees below the -X axis. How much time elapses for the second krill to reach the shark's stomach from the shark's jaws?

*Note: For this question part, you may end up with a quadratic equation. In this case, move all of your terms onto the same side of the equation. Then, enter your variables into this customized Desmos graphing calculator.

Question 2 Solutions:

Just as we did in Question 1, considering the initial prompt allows for us to quickly get an idea of what exactly is going on. By reading the prompt:

• We're told that the krill travels from the shark's mouth to stomach in 0.500 sec. Therefore, $t_{elapsed}=0.500\;{\text{s}}$.
• Also, we're told that the krill is accelerating at 19.5 m/s² directed 5 degrees below the -X axis. Thus,${\vec {a}}_{net}=19.5\;{\text{m/s}}^{2}$and $\theta =5.00^{\circ }$.
• Additionally, because we know that the krill is initially at rest, ${\vec {v}}_{net,i}={\vec {v}}_{x,i}={\vec {v}}_{y,i}=0.000\;{\text{m/s}}$.

With this, we know that the krill is starting from rest and is accelerating such that it is quickly moving leftwards and downwards. Thus, with Part a) (i), we need to find the choice where the average velocity of the krill is increasing as time passes.

Because the dots displayed on Choices A, B, and C are meant to be position records of the krill at 0.000, 0.100, 0.200, ... , and 0.500 seconds. Therefore, 0.100 seconds passes between each recorded dot. Since ${\vec {v}}_{net,avg}={\frac {{\vec {d}}_{net}}{\Delta t}}$such that ${\vec {d}}_{net}={\vec {v}}_{net,avg}\Delta t$, if we know that the average velocity between each time interval is increasing, then the displacement between each subsequent dot must be greater.

In other words, the krill is becoming faster because it is accelerating from rest. As such, the dots should be spaced further apart as the change in time between the dots are the same but the krill's velocity is increasing.

When considering our choices, only Choice B would satisfy the situation as the dots are becoming further. To rule out the other choices, Choice A incorrectly has the dots all evenly spaced, implying that the krill has a zero acceleration and that it doesn't start from rest. Also, Choice C incorrectly shows the dots becoming closer to one another as time passes. This would falsely imply that the krill does not start from rest and that the krill's acceleration would not be in the left-down direction.

Thus, Choice B (displayed on the right) is the answer for Part a) (i).

For the following section, Part a) (ii), we're presented with a computational problem with the support of a diagram (reproduced on the right) and need to calculate for the magnitude of the krill's horizontal acceleration. To note, we're already given the krill's net acceleration, $a_{net}=19.5\;{\text{m/s}}^{2}$and its angle relative to the -X axis, $\theta =5.00^{\circ }$. However, we need to figure out the horizontal component of already known net acceleration.

Using the provided diagram, we can see that the components construct a right angle triangle in which $a_{x}$extends in the horizontal direction. Using trigonometry, because $cos(\theta )={\frac {{\textsf {adjacent}}\;{\textsf {side}}}{{\textsf {hypotenuse}}\;{\textsf {side}}}}$in which $a_{x}\iff {\textsf {adjacent}}\;{\textsf {side}}$and $a_{net}\iff {\textsf {hypotenuse}}\;{\textsf {side}}$, we can rewrite this equation as $cos(\theta )={\frac {a_{x}}{a_{net}}}$and then solve for $a_{x}$.

$cos(\theta )={\frac {a_{x}}{a_{net}}}$

$a_{x}=cos(\theta )\;a_{net}\;{\text{[Algebra.]}}$

$a_{x}=cos(5.00^{\circ })*19.5\;{\text{[Substitution of variables]}}$

$a_{x,magnitude}=19.4\;{\text{m/s}}^{2}\;{\text{[Calculations; 3 sigfigs]}}$

Thus, the krill has a horizontal acceleration magnitude of 19.4 m/s² when it is being swallowed by the shark, answering Part a) (ii). Note that if we weren't solving for the magnitude of the krill's horizontal acceleration, then this value would have been negative as the krill by the Cartesian grid convention. This is because the krill is accelerating to the left.

When beginning to approach Part b), in order to approach solving for the horizontal displacement, we need to consider what we currently know about the krill's horizontal movement.

For one, we're already given that the krill starts from rest. Thus, we know that ${\vec {v}}_{x,i}=0.000\;{\text{m/s}}$. Also, we had just solved for the krill's horizontal acceleration in Part a) (ii). Therefore, we also know that because of the Cartesian grid convention, $a_{x}=-19.4\;{\text{m/s}}^{2}$. Finally, we already know the time elapsed for the krill, $t_{elapsed}=0.500\;{\text{s}}$.

As seen in the solution for Question 1: Part b), we can use the kinematic equations table and eliminate Equations 1, 2, and 4 because of the final velocity being unknown. This is with the understanding that by introducing more unknown values, we can't solve directly for any individual unknown equation, unless we were simultaneously using multiple equations.

After this, we can simplify for the change in displacement in a very similar fashion to Question 1: Part b) by using Equation 3. Note that with the current shark and krill problem, we're instead evaluating for the horizontal displacement, not the change in time.

$x_{f}-x_{i}=v_{x,i}(t_{elapsed})+{\frac {1}{2}}a{x}(t)^{2}\;{\text{[Equation 3.]}}$

$\Delta x=v_{x,i}(t_{elapsed})+{\frac {1}{2}}a{x}(t)^{2}\;{\text{[Definition of displacement.]}}$

$\Delta x={\cancelto {0}{(0.000)(0.500)}}+{\frac {1}{2}}(19.5)(0.500)^{2}\;{\text{[Substitution of variables.]}}$

$\Delta x=2.4375\;{\text{[Calculations.]}}$

$\Delta x=2.44\;{\text{m}}\;{\text{[3 sigfigs; units]}}$

Thus, the krill traveled 2.44 m horizontally on its way from the shark's jaws to the shark's stomach, answering Part b).

The next section of the question, Part (c) is conceptual in nature and requires a justification. Justifying a conceptual answer doesn't mean that we need to quantitatively calculate the situation. Instead, we need to make an argument of which acceleration value should we use when solving for the krill's overall speed at $t=0.500\;{\text{s}}$.

When considering the overall speed of the krill, this is just another way of saying the magnitude of the overall velocity. We're aware that the krill is moving for the full time interval between 0.000 and 0.500 seconds 5.00 degrees below the -X direction. This is because the krill is starting from rest, motionless, and then only accelerates in this direction. As such, there's both a horizontal and a vertical component for the velocity vector.

Another way of looking at this problem is by recalling Choice B from Part a) (i). In this, the krill's path is marked by a straight line of dots.

Because of the krill's path, we understand that if we were to calculate for the krill's final velocity, while already knowing the initial velocity and some acceleration value, we can use Equation 1 from the kinematic equations table in which $v_{x,f}=v_{x,i}+a_{x}t$ where x is some direction. If we were to use this equation to solve for the overall final velocity, then we would need to substitute in the overall acceleration, $a_{net}$, into this equation. This is because $a_{net}$shares the same direction angle as the velocity.

Therefore, one possible response to Part c) would be:

PRIMARY SAMPLE RESPONSE

Choice: The krill's acceleration vector, as given before Part a.


Reason: Suppose we choose to use the equation $v_{f}=v_{i}+at$to solve for the overall final velocity. This would be solved in the direction angle of 5 degrees below the -X direction. Because the vectors of velocity during the time interval and the overall acceleration share the same direction, we could use the overall acceleration to solve for the final overall velocity.

To note, selecting "The horizontal component of the acceleration vector, as calculated in Part a (ii)" or "Neither" as your choice for Part c) does not necessarily mean that you're incorrect. There are alternate solutions to this problem, two of which are provided below. The previous derivation and the "primary sample response" above, however, may showcase a simpler approach.

ALTERNATE SAMPLE RESPONSE #1

Choice: The horizontal component of the acceleration vector, as calculated in Part a (ii).


Reason: We can use the equation$v_{x,f}=v_{x,i}+a_{x}t$to solve for the overall final velocity in the horizontal (X) direction. Then, because we know the direction angle of the overall velocity vector at the final time, we can use trigonometry to solve for the krill's final overall speed.

ALTERNATE SAMPLE RESPONSE #2

Choice: Neither.


Reason: We can calculate first for the acceleration of the krill in the vertical (Y) direction with trigonometry. This is because we are given the direction angle at which the krill's acceleration is occurring. After this, we can use the equation$v_{y,f}=v_{y,i}+a_{y}t$to solve for the overall final velocity in the Y direction. Then, because we know the direction angle of the overall velocity vector at the final time, we can use trigonometry to solve for the krill's final overall speed.

The final section, Part d), was particularly challenging. As such, the derivations have been attempted to be explained as thoroughly as possible. Two methods are presented below. Note that there may be versions of these approaches that arrive at the same solution.

ANALYTICAL METHOD for Part d):

First, we need to interpret what has changed in the modified situation with the second krill. From the prompt for Part d), we understand that although the path and acceleration of the second krill is the same, it will have a different initial velocity of 2.00 m/s. This new velocity is directed 5.00 degrees below the -X direction.

Because of this, we should expect for the second krill to travel faster to the shark's stomach. The second krill, with a greater initial velocity and the same acceleration, will be traveling faster than the first krill at all times between its entrance into the shark's jaws until reaching the shark's stomach. Therefore, our answer for Part d) should be less than 0.500 seconds, the given time elapsed for the first krill.

Since we have a known initial velocity,$v_{i}=2.0{\text{m/s}}$, a net acceleration ${\vec {a}}_{net}=19.5\;{\text{m/s}}^{2}$, and a horizontal displacement (found in Part b) $\Delta x=2.44\;{\text{m}}$; we can start by trying to see if there's any equations can we can use through the kinematic equations table. With only these variables, we would be forced to introduce another unknown in order to solve an equation for the time elapsed. To avoid this, we can consider if we can solve for another new variable beforehand. This would make it possible for us to then solve directly for the time elapsed.

One possibility would be to consider Equation 1: $v_{net,f}=v_{net,i}+a_{net}*(t)$. Although we don't yet know the final velocity, we could attempt solve for it with Equation 4: $(v_{net,f})^{2}=(v_{net,i})^{2}+2a_{net}(d_{net})$. Once again, we run into the situation where we're introducing a new, unknown variable. This time, however, we have already solved for the horizontal displacement in part b). Also, because we are told that the paths of the two krills are the same, we can use trigonometry to solve for the net displacement of the second krill.

Since the velocity and acceleration vectors are pointed 5.00 degrees under the -X direction for the entirety of the elapsed time for both krills, we can write the equation $cos(\theta )={\frac {\Delta x}{d_{net}}}$. Then, we can solve directly for $d_{net}$such that:

$cos(\theta )={\frac {\Delta x}{d_{net}}}$

$\implies d_{net}={\frac {\Delta x}{cos(\theta )}}\;{\text{[Algebra.]}}$

$\implies d_{net}={\frac {2.4375}{cos(5.00)}}\;{\text{[Substituting variables.]}}$

$\implies d_{net}=2.4468...\;{\text{m}}\;{\text{[Calculating; keeping sigfigs for now]}}$

Then, we can again look back at Equation 4 where $(v_{net,f})^{2}=(v_{net,i})^{2}+2a_{net}(d_{net})$and solve directly for the final velocity of the second krill.

$(v_{net,f})^{2}=(v_{net,i})^{2}+2a_{net}(d_{net})$

$\implies (v_{net,f})={\sqrt {(v_{net,i})^{2}+2a_{net}(d_{net})}}\;{\text{[Squaring both sides.]}}$

$\implies (v_{net,f})={\sqrt {(2.00)^{2}+2(19.5)(2.4468...)}}\;{\text{[Substituting variables.]}}$

$\implies v_{net,f}=9.9712...\;{\text{m/s}}\;{\text{[Calculations; keeping sigfigs for now.]}}$

Finally, we now have the variables necessary to solve directly for the time elapsed with Equation 1: $v_{net,f}=v_{net,i}+a_{net}*(t)$as shown below.

$v_{net,f}=v_{net,i}+a_{net}(t)$

$\implies t={\frac {(v_{net,f}-v_{net,i})}{a_{net}}}\;{\text{[Solving for time.]}}$

$\implies t={\frac {(9.9712...-2.00)}{19.5}}\;{\text{[Substitution.]}}$

$\implies t=0.4088...\;{\text{[Calculations.]}}$

$\implies t=0.408\;{\text{sec}}\;{\text{[Units; 3 sigfigs]}}$

Thus, 0.408 seconds elapsed for the second krill between when it entered the shark's jaws to it arriving at the shark's stomach. This answers Part d) through an analytical method.

GRAPHICAL METHOD for Part d):

Note: This solution method involves using this custom Desmos graph. Have the graph open alongside the solution presented below.

Another way to solve Part d) involves a partial graphical approach. However, this method also begins by analytically considering the variables that we currently have. These include a net initial velocity,$v_{i}=2.0{\text{m/s}}$, a net acceleration ${\vec {a}}_{net}=19.5\;{\text{m/s}}^{2}$, and a horizontal displacement (found in Part b) $\Delta x=2.44\;{\text{m}}$.

After collecting these terms, we refer to the kinematic equations table and consider Equation 3: $x_{net,f}-x_{net,i}=v_{net,i}(t)+{\frac {1}{2}}a_{net}(t)^{2}$. Although we are introducing an additional unknown variable (net displacement) $x_{net,f}-x_{net,i}\iff d_{net}$into our problem by considering Equation 3, we can directly solve for this value beforehand with trigonometry. This is because we already know the horizontal displacement and the krill's angle of displacement.

Using the same approach as the Analytical Solution, we can find that the net displacement of the second krill with the horizontal displacement. As such, with holding onto significant figures, $d_{net}=2.4468...\;{\text{m}}$. For the derivation of this, please refer to the Analytical solution.

Now, we can directly solve for the time elapsed by using Equation 3 as shown below:

$x_{net,f}-x_{net,i}=v_{net,i}(t)+{\frac {1}{2}}a_{net}(t)^{2}$

$\implies d_{net}=v_{net,i}(t)+{\frac {1}{2}}a_{net}(t)^{2}$[Definition of displacement.]

$\implies 0=-d_{net}+v_{net,i}(t)+{\frac {1}{2}}a_{net}(t)^{2}$[Setting the quadratic equation equal to 0.]

By allowing for the quadratic equation above to equal 0, we can figure out for which values of $t_{elapsed}$the equation is true. To explain, $t_{elapsed}$ is the only remaining unknown variable in the equation above. Thus, we can evaluate if a certain value of $t_{elapsed}$is the answer by checking if it satisfies the equation.

We could do their either by using the quadratic formula or by using a graphing calculator to look for the equation's intersection points with the X-Axis (where the equation would equal 0). However, because a graphing calculator was provided for this question, the rest of the solution will focus on graphically solving the problem.

To solve this with the graphing calculator, we need to first decide which function should be used to model the equation that we have derived. On the custom graph, we have four possible graph selections. However, only Function Choice 1: $f\left(t\right)=\ -\left(x_{NetDisplacement}\right)+\ v_{NetInitial}\left(t\right)+{\frac {1}{2}}a_{net}\left(t\right)^{2}$contains the same variables from the derived equation. To clarify, with the other functions, they:

• Have the horizontal displacement in the equation instead of the net displacement and/or
• Neglect that the net initial velocity is non-zero.

Following this, we scroll down and enter in our values for $x_{NetDisplacement}\iff d_{net}$,$v_{NetInitial}\iff v_{net,i}$, and $a_{net}$. Note that $x_{NetHorizontal}\iff d_{x}$can be skipped as it is not in our selected function.

Using the information gathered so far, the entered values should be:

$x_{NetDisplacement}\iff d_{net}$:$2.4468...\;{\text{m}}$
$v_{NetInitial}\iff v_{net,i}$: $2.00\;{\text{m}}$
$\quad a_{net}$: $19.5\;{\text{m/s}}^{2}$

Note: To prevent confusion, all of these values must share being either all positive or all negative. This is because the values above all are vectors in the same direction. Thus, it's our choice depending on whether we want to follow the Cartesian grid convention. 

For this solution, we all for the values to all be positive. However, both ways result in the same answer. 

With this, we are able to create a parabolic curve that intersects the X-axis at $t\approx -0.614\;{\text{sec}}$ and $t\approx 0.409\;{\text{sec}}$. As explained on the Desmos graph, only the positive solution is correct. We can confirm this logically as the krill arrived at the shark's stomach after passing through the shark's jaws. Thus, a positive change in time must of occurred.

Therefore, with three significant figures, the second krill traveled from the shark's jaws to its stomach in $t=0.409\;{\text{sec}}$ , also answering Part d).*

*Note: Answers here may vary a small amount with the Analytical Solution depending on the amount of precision used by the graph. In this case, we were only off in the last digit from the Analytical Solution, which found that $t=0.408\;{\text{sec}}$.