Ділення з остачею 4 клас с.67-68

Mathematics Lesson Structure and Warm-up
📌 The lesson focuses on performing division with a remainder, referencing page 67 of the textbook.
✍️ Students were instructed to record the date (April 7th), note "remote work," and write the topic: "Performing Division with Remainder" in their notebooks.
💡 The session began with a quote from Leonardo da Vinci emphasizing that "no human investigation can be called true science if it has not passed through mathematical proofs."

Division with Remainder Examples and Verification
➗ The first example reviewed was $560,234 \div 8$:
- The quotient was determined to be a five-digit number.
- Verification involved multiplication: $(70,29 \times 8) + 2 = 560,232 + 2 = 560,234$, confirming the calculation was correct.
- The second example worked through was $9,820 \div 24$, resulting in a quotient of 409 with a remainder of 4.

Problem Solving: Relative Motion (Doves Meeting)
🕊️ Original Problem: Two doves, $S = 2000 \text{ m}$ apart, flew towards each other. They met in $t = 40 \text{ s}$. If $V_1 = 20 \text{ m/s}$, find $V_2$.
- Combined Speed Calculation: The total distance covered per second is $S / t = 2000 \text{ m} / 40 \text{ s} = \mathbf{50 \text{ m/s}}$.
- Finding $V_2$: The second dove's speed is the combined speed minus the first dove's speed: $50 \text{ m/s} - 20 \text{ m/s} = \mathbf{30 \text{ m/s}}$.

Inverse Problem 1: Finding Initial Distance ($S$)
📐 Setup: Given $V_1 = 30 \text{ m/s}$, $V_2 = 20 \text{ m/s}$, and $t = 40 \text{ s}$. Find the initial distance $S$.
- Combined Speed: $V_1 + V_2 = 30 + 20 = \mathbf{50 \text{ m/s}}$.
- Total Distance: The initial distance is the combined speed multiplied by time: $50 \text{ m/s} \times 40 \text{ s} = \mathbf{2000 \text{ m}}$ (or $2 \text{ km}$).

Inverse Problem 2: Finding Time ($t$)
⏱️ Setup: Given $S = 2000 \text{ m} + 125 \text{ m} = 2125 \text{ m}$ (initial separation includes a remaining gap), $V_1 = 30 \text{ m/s}$, $V_2 = 20 \text{ m/s}$. Find the time $t$ when the gap is $125 \text{ m}$ remaining.
- Distance Covered: The distance covered by both doves is $S_{\text{total}} - S_{\text{remaining}} = 2125 \text{ m} - 125 \text{ m} = \mathbf{2000 \text{ m}}$.
- Time Calculation: Time is the distance covered divided by the combined speed: $2000 \text{ m} / 50 \text{ m/s} = \mathbf{40 \text{ seconds}}$.

Application Problem: Distance Remaining After Time $t$
📏 Setup: Initial distance $S = 2 \text{ km } 125 \text{ m}$ ($2125 \text{ m}$), $V_1 = 30 \text{ m/s}$, $V_2 = 20 \text{ m/s}$. Find the distance between them after $t = 15 \text{ s}$.
- Distance Covered in 15s: $(30 + 20) \times 15 = 50 \times 15 = \mathbf{750 \text{ m}}$.
- Remaining Distance: Subtract the covered distance from the initial separation: $2125 \text{ m} - 750 \text{ m} = \mathbf{1375 \text{ m}}$.

Key Points & Insights
➡️ When performing long division, remember to write a zero in the quotient if you must bring down two digits consecutively because the dividend is smaller than the divisor.
➡️ Verification of division with remainder is performed by multiplying the quotient by the divisor and adding the remainder to check if it equals the original dividend.
➡️ In relative motion problems, the time required to meet is found by dividing the initial distance by the sum of the speeds ($t = S / (V_1 + V_2)$).
➡️ For problems where a distance remains *after* time $t$, first calculate the distance covered ($S_{\text{covered}} = (V_1 + V_2) \times t$) and subtract this from the initial total distance.

📸 Video summarized with SummaryTube.com on Nov 05, 2025, 20:00 UTC