Given the position function, s of t equals negative t cubed divided by 3 plus 13 times t squared divided by 2 minus 30 times t, 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.
3 < t < 9
5 < t < 9
The particle never moves to the right.
The particle always moves to the right. <- This is my answer
what's with all the words?
s(t) = -t^3/3 + 13t^2/2 - 30t
when the particle is mofing to the right, s(t) is increasing, so ds/dt > 0
-t^2 + 13t - 30 > 0
3 < t < 10
Since our domain is 0<t<9, our interval is thus
3<t<9
Ah, the particle that always moves to the right. I love a consistent traveler! Unfortunately, that's not the case here.
To find the interval where the particle is moving to the right, we need to determine when its velocity is positive. Velocity, v(t), is the derivative of the position function, s(t).
Let's calculate the velocity function first by taking the derivative of the position function:
v(t) = -3t^2/3 + 26t - 30.
Now, let's find when the velocity is positive. Solving the inequality v(t) > 0, we get t > 5.
So, the particle is moving to the right between t = 5 and t = 9. The correct answer is 5 < t < 9. Keep on rolling, little particle!
To determine whether the particle is moving to the right, we need to analyze the velocity function. The velocity function, v(t), is the derivative of the position function, s(t).
First, let's find the velocity function by taking the derivative of the position function.
s(t) = -t^3/3 + 13t^2/2 - 30t
Taking the derivative:
v(t) = d/dt[-t^3/3 + 13t^2/2 - 30t]
= -t^2 + 13t - 30
Now, we need to find the intervals where v(t) is positive since positive velocity indicates motion to the right.
To find the intervals where v(t) is positive, we need to solve the inequality v(t) > 0.
-v^2 + 13t - 30 >0
To solve this quadratic inequality, we can factorize it:
-(t - 3)(t - 10) > 0
Now, we can analyze the sign of this inequality by looking at the values of t that make it true. We break the number line into three intervals based on the critical points t = 3 and t = 10.
For t < 3:
When t < 3, both factors are negative, so the inequality -(t - 3)(t - 10) > 0 is false.
For 3 < t < 10:
When 3 < t < 10, the first factor (t - 3) becomes positive, and the second factor (t - 10) remains negative. Therefore, -(t - 3)(t - 10) > 0 is true in this interval.
For t > 10:
When t > 10, both factors are positive again, making the inequality false.
From this analysis, we can conclude that the particle is moving to the right during the interval 3 < t < 10.
Therefore, the correct answer is: 3 < t < 10
To determine the interval in which the particle is moving to the right, we need to analyze the behavior of the velocity function, v(t), which is the derivative of the position function, s(t).
The velocity function can be found by differentiating the position function with respect to t. So, let's differentiate s(t) to find v(t):
s(t) = -t^3/3 + 13t^2/2 - 30t
Differentiating both sides with respect to t:
v(t) = d/dt (-t^3/3) + d/dt (13t^2/2) - d/dt (30t)
v(t) = -t^2 + 13t - 30
Now, to find the intervals where the particle is moving to the right, we need to identify the values of t for which v(t) is positive (greater than zero).
To solve the inequality v(t) > 0, we can factorize the quadratic equation:
v(t) = -(t - 3)(t - 10)
We need to determine the values of t that make v(t) positive, which means we need to find the intervals where t - 3 > 0 and t - 10 > 0.
For t - 3 > 0, we have t > 3, which means the particle is moving to the right after t = 3.
For t - 10 > 0, we have t > 10, which is not within the given interval of t = 0 to t = 9.
Hence, the correct answer is: The particle is moving to the right for t > 3. Therefore, the interval is t > 3.