Hai người đi xe đạp từ A đến C. Người thứ nhất đi theo đường từ A đến B, rồi từ B đến C. Người thứ hai đi thẳng từ A đến C. Cả hai đều về đích một lúc.
Độ dịch chuyển của người thứ nhất là:
Hai người đi xe đạp từ A đến C. Người thứ nhất đi theo đường từ A đến B, rồi từ B đến C. Người thứ hai đi thẳng từ A đến C. Cả hai đều về đích một lúc.
Độ dịch chuyển của người thứ nhất là:
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Gọi $t$ là thời gian hai người đi từ A đến C.
Quãng đường người thứ nhất đi là $AB + BC = v_1t$
Quãng đường người thứ hai đi là $AC = v_2t$
Độ dịch chuyển của người thứ nhất là đoạn $AC$, bằng độ dịch chuyển của người thứ hai.
Ta có $AB = 2$ km, $BC = 2$ km.
Vì tam giác ABC vuông tại B nên $AC = \sqrt{AB^2 + BC^2} = \sqrt{2^2 + 2^2} = \sqrt{8} = 2\sqrt{2}$ km.
Tuy nhiên, đây là độ dài quãng đường, không phải độ dịch chuyển. Bài toán yêu cầu tính độ dịch chuyển của người thứ nhất. Vì cả hai người về đích cùng một lúc nên độ dịch chuyển của hai người bằng nhau. Người thứ hai đi thẳng từ A đến C, vậy độ dịch chuyển của người thứ nhất bằng độ dịch chuyển của người thứ hai, chính là đoạn AC.
Nhưng đề bài lại hỏi về độ dịch chuyển của người thứ nhất, tức là hỏi về đoạn đường AB + BC. Ta có thể hiểu là độ dài đường đi AB + BC = 2km + 2km = 4km
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