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
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.
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.
To find the interval in seconds where the particle is moving to the right, we need to determine when the velocity of the particle is positive. The velocity function is the derivative of the position function, s(t).
First, let's find the derivative of the position function, s(t):
s'(t) = d/dt [ (-t^3/3) + (13t^2/2) - 30t ]
= -(t^2) + 13t - 30
Now, let's find when the velocity function is positive:
-(t^2) + 13t - 30 > 0
To solve this inequality, we can factor the quadratic expression:
-(t - 3)(t - 10) > 0
Now, we have two critical points, t = 3 and t = 10. We divide the number line into three intervals: (-∞, 3), (3, 10), (10, ∞).
Testing each interval:
For t < 3, let's substitute t = 0:
-(0 - 3)(0 - 10) > 0
(3)(-10) > 0
-30 > 0
This is false, so the interval (-∞, 3) is not where the particle is moving to the right.
For t between 3 and 10, let's substitute t = 5:
-(5 - 3)(5 - 10) > 0
(2)(-5) > 0
-10 > 0
This is false, so the interval (3, 10) is not where the particle is moving to the right.
For t > 10, let's substitute t = 11:
-(11 - 3)(11 - 10) > 0
(8)(1) > 0
8 > 0
This is true, so the interval (10, ∞) is where the particle is moving to the right.
Therefore, the correct answer is:
d) the particle always moves to the right
To determine when the particle is moving to the right, we need to find the intervals of time where its velocity is positive.
The velocity is the derivative of the position function, so we need to find the derivative of s(t):
s'(t) = d/dt [(-t^3/3) + (13t^2/2) - 30t]
Using the power rule, we can differentiate each term:
s'(t) = -t^2 + 13t - 30
Now we need to find the intervals where s'(t) > 0. To do this, we solve the inequality -t^2 + 13t - 30 > 0.
Factorizing the quadratic equation, we get:
(-t + 3)(t - 10) > 0
Now we solve each factor separately:
For -t + 3 > 0, we have t > 3.
For t - 10 > 0, we have t > 10.
Since t represents time and cannot be negative, we only consider the solution t > 3.
Therefore, the particle is moving to the right for t > 3.
Option a) 3 < t < 9 is the correct answer.