12:I[70017,["/_next/static/immutable/chunks/2gtpnylaz50zo.js","/_next/static/immutable/chunks/22kg_k_j3c_ao.js","/_next/static/immutable/chunks/3s-qhi50ildb6.js"],""] 13:I[16632,["/_next/static/immutable/chunks/2gtpnylaz50zo.js","/_next/static/immutable/chunks/22kg_k_j3c_ao.js","/_next/static/immutable/chunks/3s-qhi50ildb6.js"],"default"] 14:I[3261,["/_next/static/immutable/chunks/2gtpnylaz50zo.js","/_next/static/immutable/chunks/22kg_k_j3c_ao.js","/_next/static/immutable/chunks/3s-qhi50ildb6.js"],"default"] 15:I[44578,["/_next/static/immutable/chunks/2gtpnylaz50zo.js","/_next/static/immutable/chunks/22kg_k_j3c_ao.js","/_next/static/immutable/chunks/3s-qhi50ildb6.js"],"default"] 16:T18d7,{ "consigne": "Pour chaque situation, choisis d'abord la relation, puis calcule. ⚠️ Avec un écho, le son fait un aller-retour, donc il parcourt deux fois la distance.", "donnees": [ { "cle": "air", "tex": "v_{\\text{air}} = 300\\ \\mathrm{m.s^{-1}}" }, { "cle": "note", "tex": "Dans ce défi, on prend 300 m.s⁻¹ au lieu de 340 m.s⁻¹ pour calculer de tête." } ], "questions": [ { "enonce": "Un son parcourt une distance d pendant une durée Δt. Quelle relation donne sa vitesse ?", "choix": [ "v = \\dfrac{d}{\\Delta t}", "v = d \\times \\Delta t", "v = \\dfrac{\\Delta t}{d}" ], "bonne": 0, "explication": "Une vitesse est la distance parcourue divisée par la durée du parcours.", "donnees": [] }, { "enonce": "Un son parcourt 900 m en 3 s. Quelle est sa vitesse ?", "reponse": 300, "tolerancePct": 2, "unite": "m.s⁻¹|m/s", "explication": "v = 900 / 3 = 300 m.s⁻¹.", "donnees": [] }, { "enonce": "Tu connais la vitesse v et la durée Δt. Quelle relation donne la distance parcourue ?", "choix": [ "d = \\dfrac{\\Delta t}{v}", "d = \\dfrac{v}{\\Delta t}", "d = v \\times \\Delta t" ], "bonne": 2, "explication": "À partir de v = d/Δt, on trouve que la distance vaut vitesse × durée.", "donnees": [] }, { "enonce": "À 300 m.s⁻¹, quelle distance le son parcourt-il en 5 s ?", "reponse": 1500, "tolerancePct": 2, "unite": "m", "explication": "d = 300 × 5 = 1 500 m.", "donnees": [ "air", "note" ] }, { "enonce": "Tu connais la distance d et la vitesse v. Quelle relation donne la durée du trajet ?", "choix": [ "\\Delta t = d \\times v", "\\Delta t = \\dfrac{d}{v}", "\\Delta t = \\dfrac{v}{d}" ], "bonne": 1, "explication": "À partir de v = d/Δt, on trouve que la durée vaut distance / vitesse.", "donnees": [] }, { "enonce": "Combien de temps le son met-il pour parcourir 1 200 m ?", "reponse": 4, "tolerancePct": 2, "unite": "s", "aide": "Tu cherches une durée, donc tu divises la distance par la vitesse.", "explication": "Δt = 1 200 / 300 = 4 s.", "donnees": [ "air", "note" ] }, { "enonce": "Tu vois l'éclair, puis tu entends le tonnerre avec un retard Δt. Quelle relation donne la distance d de l'orage ?", "image": "/content/seconde/le-son/orage-eclair-tonnerre.svg", "choix": [ "d = v \\times \\Delta t", "d = \\dfrac{v \\times \\Delta t}{2}", "d = \\dfrac{v}{\\Delta t}" ], "bonne": 0, "explication": "Le son fait un aller simple de l'orage jusqu'à toi, donc d = v × Δt.", "donnees": [] }, { "enonce": "Pendant un orage, tu vois l'éclair puis tu entends le tonnerre 2 s plus tard. À quelle distance est l'orage ?", "image": "/content/seconde/le-son/orage-eclair-tonnerre.svg", "reponse": 600, "tolerancePct": 2, "unite": "m", "aide": "La lumière arrive presque instantanément, mais le son met 2 s pour venir jusqu'à toi. Le son fait un aller simple, sans retour.", "explication": "d = 300 × 2 = 600 m.", "donnees": [ "air", "note" ] }, { "enonce": "Tu cries face à un obstacle, et l'écho te revient au bout d'une durée Δt. Quelle relation donne la distance d jusqu'à l'obstacle ?", "choix": [ "d = \\dfrac{2 \\times v}{\\Delta t}", "d = v \\times \\Delta t", "d = \\dfrac{v \\times \\Delta t}{2}" ], "bonne": 2, "explication": "Pendant Δt, le son parcourt l'aller et le retour, soit 2 × d. La distance à l'obstacle vaut donc v × Δt / 2.", "donnees": [] }, { "enonce": "Tu cries face à une falaise, et l'écho te revient 2 s après. À quelle distance est la falaise ?", "image": "/content/seconde/le-son/echo-aller-retour.svg", "reponse": 300, "tolerancePct": 2, "unite": "m", "aide": "En 2 s, le son fait l'aller et le retour, soit deux fois la distance jusqu'à la falaise. Tu divises donc par 2.", "explication": "Distance parcourue par le son : 300 × 2 = 600 m (aller-retour). La falaise est à 600 / 2 = 300 m.", "donnees": [ "air", "note" ] }, { "enonce": "Tu cries face à un autre obstacle, et l'écho te revient au bout d'une durée Δt. Quelle relation donne la distance d jusqu'à l'obstacle ?", "choix": [ "d = \\dfrac{v}{2 \\times \\Delta t}", "d = \\dfrac{v \\times \\Delta t}{2}", "d = 2 \\times v \\times \\Delta t" ], "bonne": 1, "explication": "Comme pour tout écho, le son fait l'aller et le retour, donc d = v × Δt / 2.", "donnees": [] }, { "enonce": "Tu cries, et l'écho te revient 4 s plus tard. À quelle distance est l'obstacle ?", "reponse": 600, "tolerancePct": 2, "unite": "m", "aide": "Le son fait l'aller et le retour, donc pense à diviser par 2.", "explication": "Distance parcourue par le son : 300 × 4 = 1 200 m (aller-retour). L'obstacle est à 1 200 / 2 = 600 m.", "donnees": [ "air", "note" ] }, { "enonce": "Tu cries face à un obstacle situé à la distance d. Quelle relation donne la durée entre ton cri et le retour de l'écho ?", "choix": [ "\\Delta t = \\dfrac{2 \\times d}{v}", "\\Delta t = \\dfrac{d}{2 \\times v}", "\\Delta t = 2 \\times d \\times v" ], "bonne": 0, "explication": "Le son parcourt l'aller et le retour, soit 2 × d. La durée du trajet est donc 2 × d / v.", "donnees": [] }, { "enonce": "Tu cries face à un mur situé à 900 m. Au bout de combien de temps entends-tu l'écho ?", "reponse": 6, "tolerancePct": 2, "unite": "s", "aide": "Le son doit faire l'aller et le retour : 2 × 900 m.", "explication": "Trajet total : 2 × 900 = 1 800 m, donc Δt = 1 800 / 300 = 6 s.", "donnees": [ "air", "note" ] } ] } 17:T82f,

v_{\text{air}} = 300\ \mathrm{m.s^{-1}}

18:T68f,

v = \dfrac{d}{\Delta t}

19:T43d,

v = d \times \Delta t

1a:T68f,

v = \dfrac{\Delta t}{d}

1b:T68f,

d = \dfrac{\Delta t}{v}

1c:T68f,

d = \dfrac{v}{\Delta t}

1d:T43d,

d = v \times \Delta t

1e:T43d,

\Delta t = d \times v

1f:T682,

\Delta t = \dfrac{d}{v}

20:T682,

\Delta t = \dfrac{v}{d}

21:T75a,

d = \dfrac{v \times \Delta t}{2}

22:T767,

d = \dfrac{2 \times v}{\Delta t}

23:T75c,

d = \dfrac{v}{2 \times \Delta t}

24:T56f,

d = 2 \times v \times \Delta t

25:T75a,

\Delta t = \dfrac{2 \times d}{v}

26:T75c,

\Delta t = \dfrac{d}{2 \times v}

27:T56f,

\Delta t = 2 \times d \times v

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