Asked by Alice
Given the position function, s(t)= (-t^3/3)+(13t^2 /2)-30t, between t=0 and
t = 9, where s is given in feet and t is measured in seconds, find the interval in seconds where the particle is moving to the right.
a) 3 < t < 9
b) 5 < t < 9
c) the particle never moves to the right
d) the particle always moves to the right
t = 9, where s is given in feet and t is measured in seconds, find the interval in seconds where the particle is moving to the right.
a) 3 < t < 9
b) 5 < t < 9
c) the particle never moves to the right
d) the particle always moves to the right
Answers
Answered by
oobleck
if it is moving right, s is increasing
That means ds/dt > 0
so, where is
(-t^3/3)+(13t^2 /2)-30t
-t^2 + 13t - 30 > 0 ?
hint: between the roots, since the parabola opens downward.
Watch the domain given.
That means ds/dt > 0
so, where is
(-t^3/3)+(13t^2 /2)-30t
-t^2 + 13t - 30 > 0 ?
hint: between the roots, since the parabola opens downward.
Watch the domain given.
Answered by
bobpursley
ds/dt=positive going to right.
ds/dt= -t^2+13t-20
set ti to zero, solve for t to find boundries. Hint: between the zeroes it is positive.
ds/dt= -t^2+13t-20
set ti to zero, solve for t to find boundries. Hint: between the zeroes it is positive.
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