applications of indefinite integrals

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

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?