I'm stuck on this problem:
The vertical position of a particle, in feet, is given by the function
$s(t) = t^3 – 3t^2 – 4$, where t is measured in seconds. How far does
the particle travel between t = 0 and t = 5 seconds?
So I took the derivative/velocity of the function, getting $v(t) = 3t^2 – 6t = 3t(t-2)$.
I set it equal to 0, and the critical points were t = 2 and t = 0. I put them on a number line, solving for velocity left of 0, between 0 and 2, and right of 2 and got v(-1) = negative, v(1) = positive, and v(3) = positive. I just needed to know the signs of velocity, and clearly the particle changed direction.
Next, I took the position s(t) of the particle at each of those points t = 0, t = 2, and t = 5, obtaining s(0) = -2, s(2) = -6, and s(5) = 48, if I'm not wrong. I believe this question's asking for total distance, so adding up 2 + 6 + 48 gives me 56 feet. But this is apparently the wrong answer, and I'm not sure why. It is asking for distance, not displacement, right? Or am I doing something wrong?
Can someone please help? Thanks.
Best Answer
$d = |s(0) - s(2)| + |s(5) - s(2)|$
$s(0) = -4$
$s(2) = -8$
$s(5) = 46$
All of your results are off by 2, but it shouldn't make a difference in the final answer.
$d = 4+54 = 58$