Avenger6 said:
Hi, i am seriously struggling with the following questions:
I'm stuck on how I can integrate a second time in question 9. In question 12 I have integrated to find the displacement formula however I'm stuck on how I can substitute pi into the formula. An I'm lost all together in question 13. Any help is greatly appreciated
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9)
a = (3t + 1)^2
= 9t^2 + 6t + 1
v = 3t^3 + 3t^2 + t + c1
When t=0, v=0
0 = 3.0^3 + 3.0^2 + c1
c1 = 0
Therefore, v = 3t^3 + 3t^2 + t
Now, x = 3/4 . t^4 + t^3 + (t^2)/2 + c2
When t=0, x = -2 (2m left to the origin)
-2 = 3/4 . t^4 + t^3 + (t^2)/2 + c2
So c2 = -2
Therefore, x=3/4 . t^4 + t^3 + (t^2)/2 -2
When t=4, x = 3/4 . 4^4 + 4^3 + (4^2)/2 - 2
Therefore, x = I don't have the calculator here. Work this out for me please. Thanks
12)
a = -9sin3t
v = 3cos3t + c
When t=0, v = 5
i.e. 5 = 3cos3.0 + c1
c1 = 2
Therefore, v = 3cos3t + 2
Now, x = sin3t + 2t + c2
When t=0, x = -3 (since it is left)
i.e. -3 = sin3.0 + 2.0 + c2
Therefore, c2 = -3
So, x = sin3t + 2t -3
When t = pi,
x = sin3pi + 2pi - 3
= 2pi - 3
13) v = t/(t^2 + 3)
v = 1/2 2t/(t^2+3)
Therefore, x = 1/2 ln(t^2+3) + c (by integration)
When t=0, x=0
i.e. 0 = 1/2 ln(0^2 + 3) + c
So, c= -1/2 ln(3)
Now, x = 1/2 ln(t^2 + 3) -1/2 ln(3)
When t=10, x = 1/2 ln(10^2 + 3) - 1/2 ln (3)
So x = 1/2 ln(103) - 1/2 ln(3)
= don't have the calculator. Sorry!
