Posted by Anonymous on Friday, December 9, 2011 at 4:31am.
displacement = integral v dt
= -t^3/3 + 2 t^2 - 3 t
from -2 to +6
do the arithmetic
distance traveled
that means always use absolute value of any part of the integral
when is velocity negative and when is it positive?
0 = t^2 - 4 t + 3
0 = (t-3)(t-1)
velocity 0 at t = 1 and t = 3
so:
take absolute value of integral from
-2 to + 1
add to absolute value of integral from +1 to +3
add to absolute value of integral from +3 to + 6
thank you
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