antiderivatives: acelleration --> velocity --> position?

Question

I need to calculate velocity [v(t)] and position [s(t)] from acceleration [a(t)] starting at 0 feet/second. I have to do this with ***antiderivitives*** (even if there is an easier way).

a(t) = 60t

am I correct in calculating?:
v(t) = (60t^2)/2
and
s(t) = (60/2)(t^3/3) = 10t^3
?????

s(0)=0

ignore:
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Answer 1

Looks right to me. To check this, simply take the derivative of 10t^3 and compare with your equation of velocity, and take the derivative of that, comparing with the equation of acceleration.

BUT: Always remember to add the constant term, C! Remember, you do not know the starting position.

Answer 2

It´s OK but as you don´t know the position at t=0 you must add a constant which is s(0)
v(t) has no problems as you know v(0)=0

Answer 3

v = int(a(t))dt = int(60t)dt = 60(t^2)/2 = 30t^2. Since v = 0 when t = 0, this is correct.

s = int(v(t))dt = int(30(t^2))dt = 30int(t^2)dt = 30(t^3)/3 = 10t^3. Since s = 0 when t = 0, you are correct again.

Answer 4

yes ..

but you should add an integration constant to s(t) becuase you dont know that you started at s=0 [although you do know the starting velocity was 0 ft/sec]

Answer 5

yes, and in each case there could be a constant of integration which defines the value of velocity and position at time zero, if they are not zero.

Answer 6

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