dp/dt = 0 at changes
look for zeros of
-3t^2+2t +5 = 0
t = -1 and t = +5/3
from t = 0 to t = 5/3 velocity is +
p = 3 at 0
p = 6.51 + 3 = 9.51at t = 5/3
so moved +6.51
from t = 5/3 to t = 6 velocity is -
p = 9.51 at 5/3
p = -147 at t = 6
so moved 9.51 -147 = -137.5 from 5/3 to 6
total motion = 6.51 - 137.5 = -131
The position of a particle at any time t >= 0 is given by p(t) = -t^3 + t^2 + 5t + 3. How far does the particle travel during the first six seconds?
I calculated the distance traveled between the endpoints of the interval and the points at which the particle changes directions to get the answer 4481/27.
3 answers
Shouldn't 9.51 - 147 be 9.51 - (-147)?
Sorry, yes, add the absolute valuess