Spec Investigation by Kathan Desai Assessment

Assignment Overview

Part A

Vector equations:

a⃗=(−51)+t(2030)=(−5+20t1+30t)\vec{a} = \left( -5 \atop 1 \right) + t \left( 20 \atop 30 \right) = \left( -5 + 20t \atop 1 + 30t \right)a=(1−5)+t(3020)=(1+30t−5+20t) b⃗=(32)+t(2119)=(3+21t2+19t)\vec{b} = \left( 3 \atop 2 \right) + t \left( 21 \atop 19 \right) = \left( 3 + 21t \atop 2 + 19t \right)b=(23)+t(1921)=(2+19t3+21t)

Table 1: Vector a⃗\vec{a}a

Mathematically, show that they will collide.

When t = 2, d = 0

DC→=(−5+40t22+7t)−(5+35t−2+19t)=(−10+5t24−12t)\overrightarrow{DC} = \left( -5 + 40t \atop 22 + 7t \right) - \left( 5 + 35t \atop -2 + 19t \right) = \left( -10 + 5t \atop 24 - 12t \right)DC=(22+7t−5+40t)−(−2+19t5+35t)=(24−12t−10+5t) D=(5−2)direction=(35t19t)at t=2 (75,36)D = \left( 5 \atop -2 \right) \quad direction = \left( 35t \atop 19t \right) \quad \text{at } t=2 \, (75,36)D=(−25)direction=(19t35t)at t=2(75,36) C=(−522)direction=(40t7t)at t=2 (75,36)C = \left( -5 \atop 22 \right) \quad direction = \left( 40t \atop 7t \right) \quad \text{at } t=2 \, (75,36)C=(22−5)direction=(7t40t)at t=2(75,36)

Distance

d=(−10+5t)2−(24−12t)2d = \sqrt{(-10 + 5t)^2 - (24 - 12t)^2}d=(−10+5t)2−(24−12t)2

When t = 2

d=(−10+5(2))2−(24−12(2))2=0d = \sqrt{(-10 + 5(2))^2 - (24 - 12(2))^2} = 0d=(−10+5(2))2−(24−12(2))2=0

Part B

• Investigate another situation (Yacht C and Yacht D) with different initial positions and velocity vectors, where the yachts collide.

When t = 2, d = 0

Yacht D=(5−2),direction=(35t19t)at t=2 (75,36)\text{Yacht D} = \left( 5 \atop -2 \right), \quad \text{direction} = \left( 35t \atop 19t \right) \quad \text{at } t=2 \, (75,36)Yacht D=(−25),direction=(19t35t)at t=2(75,36) Yacht C=(−522),direction=(40t7t)at t=2 (75,36)\text{Yacht C} = \left( -5 \atop 22 \right), \quad \text{direction} = \left( 40t \atop 7t \right) \quad \text{at } t=2 \, (75,36)Yacht C=(22−5),direction=(7t40t)at t=2(75,36) DC→=(−5+40t22+9t)−(5+35t−2+19t)=(10−10t30−20t)\overrightarrow{DC} = \left( -5 + 40t \atop 22 + 9t \right) - \left( 5 + 35t \atop -2 + 19t \right) = \left( 10 - 10t \atop 30 - 20t \right)DC=(22+9t−5+40t)−(−2+19t5+35t)=(30−20t10−10t)

Assessment Requirements – Summary

The assessment required students to:

1. Work with vector equations to represent the paths of moving objects (yachts).

2. Part A – Show mathematically that two given yachts (with defined vector equations) collide by proving their displacement vectors and positions coincide at a specific time ttt.

   • Derive the vector equations of motion.

   • Calculate the relative displacement vector.

   • Show that the distance between yachts becomes zero at t=2t=2t=2.

3. Part B – Investigate another situation with different initial positions and velocity vectors for the yachts.

   • Formulate the new vector equations.

   • Compute relative displacement.

   • Show mathematically that collision occurs at t=2t=2t=2.

4. Conclude by confirming collisions occur through vector analysis and distance calculation.

Step-by-Step Approach Guided by Academic Mentor

Step 1: Understanding Vector Equations (Part A)

• The mentor began by explaining how to express the position of moving objects (yachts) using vector equations of motion .

• For Yacht A (a⃗\vec{a}a) and Yacht B (b⃗\vec{b}b), the student learned to write:

  a⃗=(−5,1)+t(20,30),b⃗=(3,2)+t(21,19)\vec{a} = (-5, 1) + t(20, 30), \quad \vec{b} = (3, 2) + t(21, 19)a=(−5,1)+t(20,30),b=(3,2)+t(21,19)

• This step reinforced the concept that initial position and velocity form the foundation of motion equations.

Step 2: Establishing the Condition for Collision

• The mentor highlighted that for a collision, the yachts must occupy the same position at the same time .

• The relative displacement vector DC→\overrightarrow{DC}DC was introduced, representing the difference between the positions of the two yachts at time ttt.

• The student calculated:

  DC→=(−10+5t, 24−12t)\overrightarrow{DC} = (-10 + 5t, \, 24 - 12t)DC=(−10+5t,24−12t)

• The condition for collision: the distance between yachts = 0 .

Step 3: Distance Calculation (Part A)

• The mentor guided the student to apply the distance formula:

  d=(−10+5t)2+(24−12t)2d = \sqrt{(-10 + 5t)^2 + (24 - 12t)^2}d=(−10+5t)2+(24−12t)2

• Substituting t=2t = 2t=2, the student showed that d=0d=0d=0, confirming collision at coordinates (75, 36).

Step 4: Investigating Another Situation (Part B)

• The mentor introduced new initial positions and velocity vectors for Yachts C and D.

• The new equations were written:

  D=(5,−2)+t(35,19),C=(−5,22)+t(40,7)D = (5, -2) + t(35, 19), \quad C = (-5, 22) + t(40, 7)D=(5,−2)+t(35,19),C=(−5,22)+t(40,7)

• Relative displacement was derived:

  DC→=(10−10t, 30−20t)\overrightarrow{DC} = (10 - 10t, \, 30 - 20t)DC=(10−10t,30−20t)

• Again, applying t=2t=2t=2, both yachts reached the same point (75, 36), proving collision.

Step 5: Reflection and Learning Outcomes

Through the mentor’s step-by-step guidance, the student achieved the following:

• Learned how to formulate vector equations for motion.

• Understood the mathematical condition for collision (positions coincide, distance = 0).

• Applied distance formula and displacement vectors effectively.

• Developed problem-solving skills in both direct given situations (Part A) and investigated cases with altered conditions (Part B).

• Strengthened conceptual knowledge of vectors, relative motion, and mathematical reasoning.

Final Outcome:
The student successfully proved, in both Part A and Part B, that the yachts collided at t=2t=2t=2. The mentor’s structured approach ensured clarity, built step-by-step understanding, and helped the student meet all assessment requirements while covering essential learning objectives in vectors and motion.

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