The distance covered by moving particle through point O is given by the equation, s=t3-15t2+63t−10. Find (a)…

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The distance covered by moving particle through point O is given by the equation, s=t3-15t2+63t−10. Find (a) distance covered when t=2 (b) the distance covered during the 3rd second (c) the time when the partiec<0xC2><0xAD>ul is momentarily at rest. (d) The acceleration when t=5.

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Hapa kuna suluhisho la maswali yako: Kutokana na mlinganyo wa umbali: s = t³ − 15t² + 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(t³ − 15t² + 63t − 10); V = 3t² − 30t + 63; Step 2: Weka V=0 na utatue kwa t. 3t² − 30t + 63 = 0 Gawanya kwa 3: t² − 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(3t² − 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/s²

The distance covered by moving particle through point O is given by the equation, s=t3−15t2+63t−10. Find (a) distance covered when t=2 (b) the distance covered during the 3rd second (c) the time when the partiec<0xC2><0xAD>ul is momentarily at rest. (d) The acceleration when t=5.

Answers

Step-by-step solution

Part a· 1 step

Badilisha $t=2$ kwenye mlinganyo wa umbali.

Part b· 2 steps
$\text{7 m}$

Tafuta umbali kwa $t=3$ .

Kokotoa umbali uliosafirishwa wakati wa sekunde ya 3.

Part c· 4 steps
$\text{t = 3 s au t = 7 s}$

Tafuta mlinganyo wa kasi $V(t)$ .

Weka $V=0$ na utatue kwa $t$ .

Tafuta sababu za quadratic equation.

Tatua kwa $t$ .

Part d· 2 steps
$\text{0 m/s}^2$

Tafuta mlinganyo wa kuongeza kasi $a(t)$ .

Badilisha $t=5$ kwenye mlinganyo wa kuongeza kasi.

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The distance covered by moving particle through point O is given by the equation, s=t3−15t2+63t−10. Find (a) distance covered when t=2 (b) the distance covered during the 3rd second (c) the time when the partiec<0xC2><0xAD>ul is momentarily at rest. (d) The acceleration when t=5.

Answers

Step-by-step solution

Part a· 1 step

Badilisha t=2t=2t=2 kwenye mlinganyo wa umbali.

Part b· 2 steps7 m\text{7 m}7 m

Tafuta umbali kwa t=3t=3t=3.

Kokotoa umbali uliosafirishwa wakati wa sekunde ya 3.

Part c· 4 stepst = 3 s au t = 7 s\text{t = 3 s au t = 7 s}t = 3 s au t = 7 s

Tafuta mlinganyo wa kasi V(t)V(t)V(t).

Weka V=0V=0V=0 na utatue kwa ttt.