This physics problem tests your understanding of fundamental physical laws and their applications. The step-by-step solution below breaks down the problem using relevant equations and physical reasoning.
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Hapa kuna suluhisho la maswali yako: Kutokana na mlinganyo wa umbali: s = t^3 - 15t^2 + 63t - 10 a) Umbali uliosafirishwa wakati t=2. Step 1: Badilisha t=2 kwenye mlinganyo wa umbali. s(2) = (2)^3 - 15(2)^2 + 63(2) - 10 s(2) = 8 - 15(4) + 126 - 10 s(2) = 8 - 60 + 126 - 10 s(2) = 64 Umbali uliosafirishwa wakati t=2 ni mita 64. 64 m b) Umbali uliosafirishwa wakati wa sekunde ya 3. Umbali uliosafirishwa wakati wa sekunde ya 3 ni tofauti kati ya umbali kwa t=3 na umbali kwa t=2. Step 1: Tafuta umbali kwa t=3. s(3) = (3)^3 - 15(3)^2 + 63(3) - 10 s(3) = 27 - 15(9) + 189 - 10 s(3) = 27 - 135 + 189 - 10 s(3) = 71 Step 2: Kokotoa umbali uliosafirishwa wakati wa sekunde ya 3. Umbali = s(3) - s(2) = 71 - 64 = 7 mita. 7 m c) Muda wakati chembe imetulia papo hapo. Chembe imetulia wakati kasi yake V = 0. Kasi ni derivative ya umbali kwa heshima ya muda. Step 1: Tafuta mlinganyo wa kasi V(t). V = (ds)/(dt) = (d)/(dt)(t^3 - 15t^2 + 63t - 10) V = 3t^2 - 30t + 63 Step 2: Weka V=0 na utatue kwa t. 3t^2 - 30t + 63 = 0 Gawanya kwa 3: t^2 - 10t + 21 = 0 Step 3: Tafuta sababu za quadratic equation. (t-3)(t-7) = 0 Step 4: Tatua kwa t. t-3 = 0 au t-7 = 0 t = 3 au t = 7 Chembe imetulia papo hapo kwa t=3 sekunde na t=7 sekunde. t = 3 s au t = 7 s d) Kuongeza kasi wakati t=5. Kuongeza kasi a ni derivative ya kasi kwa heshima ya muda. Step 1: Tafuta mlinganyo wa kuongeza kasi a(t). a = (dV)/(dt) = (d)/(dt)(3t^2 - 30t + 63) a = 6t - 30 Step 2: Badilisha t=5 kwenye mlinganyo wa kuongeza kasi. a(5) = 6(5) - 30 a(5) = 30 - 30 a(5) = 0 Kuongeza kasi wakati t=5 ni 0 m/s^2. 0 m/s^2 3 done, 2 left today. You're making progress.