2017-02-14 13:56:37dvhtlbz
applications of indefinite integrals
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applications of indefinite integrals
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v=∫adt ∴v=18t-(3/2)t+C when 0<=t<=4 When t=4, v=50 50=72-(3/2)(16)+C C=2 ∴v=18t-(3/2)t+2 u can be otbtained when t=0 ∴u=2 When t=5, a=t+2 ∴v=(1/2)t+2t+C When t=4, v=50 50=8+8+C C=32 ∴v=(1/2)t+2t+32 When t=5, v=25/2+10+32=109/2 m/s When 0=<=t<=4 v=18t-(3/2)t+2 ∴s=9t-(1/2)t+2t+C Total distance travelled in the first 4 seconds=144-32+8=120m When t>=4 v=(1/2)t+2t+32 s=(1/6)t+t+32t+C ∴Distance travelled in the 5th second=(1/6)(125-64)+9+32=307/6m ∴Total distance=120+307/6=1027/6 m
其他解答:
when 0 <= t <= 4 du/dt = 18 - 3t when 4 <= t du/dt = t + 2 when 0 <= t <= 4 u = -3/2t^2 +18t +c 50 =(-3/2)4+18*4 +c c = 2 when 0 <= t <= 4 u = (-3/2)t^2 +18t +2 -----------------------(1) when 4 <= t u = 1/2t^2 + 2t +c 50 = 1/2*16 + 8 + c c = 34 when 4 <= t u = 1/2t^2 + 2t + 34 ------------------------(2) put t = 0 into (1) u = 2 (m/s) put t=5 into (2) u = 56.5 when 0 <= t <= 4 f(t) = [-3/6t^3 +9t^2 +2t] f(4) - f(0) = 120 when 4 <= t f(t)= 1/6t^3 + t^2 + 34t f(5) - f(4) = 53.167 distance = 120 + 53.167 = 173.167 2008-06-25 18:41:07 補充： E個係中幾題目?7638E7481407D16B
applications of indefinite integrals
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最佳解答:v=∫adt ∴v=18t-(3/2)t+C when 0<=t<=4 When t=4, v=50 50=72-(3/2)(16)+C C=2 ∴v=18t-(3/2)t+2 u can be otbtained when t=0 ∴u=2 When t=5, a=t+2 ∴v=(1/2)t+2t+C When t=4, v=50 50=8+8+C C=32 ∴v=(1/2)t+2t+32 When t=5, v=25/2+10+32=109/2 m/s When 0=<=t<=4 v=18t-(3/2)t+2 ∴s=9t-(1/2)t+2t+C Total distance travelled in the first 4 seconds=144-32+8=120m When t>=4 v=(1/2)t+2t+32 s=(1/6)t+t+32t+C ∴Distance travelled in the 5th second=(1/6)(125-64)+9+32=307/6m ∴Total distance=120+307/6=1027/6 m
其他解答:
when 0 <= t <= 4 du/dt = 18 - 3t when 4 <= t du/dt = t + 2 when 0 <= t <= 4 u = -3/2t^2 +18t +c 50 =(-3/2)4+18*4 +c c = 2 when 0 <= t <= 4 u = (-3/2)t^2 +18t +2 -----------------------(1) when 4 <= t u = 1/2t^2 + 2t +c 50 = 1/2*16 + 8 + c c = 34 when 4 <= t u = 1/2t^2 + 2t + 34 ------------------------(2) put t = 0 into (1) u = 2 (m/s) put t=5 into (2) u = 56.5 when 0 <= t <= 4 f(t) = [-3/6t^3 +9t^2 +2t] f(4) - f(0) = 120 when 4 <= t f(t)= 1/6t^3 + t^2 + 34t f(5) - f(4) = 53.167 distance = 120 + 53.167 = 173.167 2008-06-25 18:41:07 補充： E個係中幾題目?7638E7481407D16B