Motion and rate of change derivatives. Attached is the question and where I am stuck, I would really appreciate the help :)
1 answer:
Answer:
A. 3 < t < 9
Step-by-step explanation:
When the particle is moving to the right, the velocity is positive.
To find the velocity, take the derivative of s(t) with respect to t.
s(t) = -t³/3 + 13t²/2 − 30t
s'(t) = -t² + 13t − 30
Find when the velocity is 0.
0 = -t² + 13t − 30
0 = t² − 13t + 30
0 = (t − 3) (t − 10)
t = 3, 10
Check the sign of s'(t) in each interval.
0 < t < 3, s'(t) < 0
3 < t < 9, s'(t) > 0
The particle moves to the right when 3 < t < 9.
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Answer:
88
Step-by-step explanation:
The first part is 8 x 40=320
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Question 1:
y=-x+7
y=2x-2
(-x+7)=2x-2
-x+7+x=2x-2+x
7=2x-2+x
7+2=2x-2+x+2
9=3x
9/3=3x/3
3=x
y=-x+7
y=-(3)+7
y=4
Question 2:
y-1=2x
y+2x=5
y-1=2x
y-1+1=2x+1
y=2x
(2x+1)+2x=5
2x+1+2x=5
4x+1=5
4x+1-1=5-1
4x=4
4x/4=4/4
x=1
y-1=2(1)
y-1+1=2+1
y=3
Answer: x=1 y=3