anonymous
  • anonymous
A particle is moving with velocity v(t) = t2 – 9t + 18 with distance, s measured in meters, left or right of zero, and t measured in seconds, with t between 0 and 8 seconds inclusive. The position at time t = 0 sec is 1 meter right of zero, that is, s(0) = 1. The time interval(s) when the particle is moving right The time interval(s) when the particle is going faster slowing down
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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Nayef
  • Nayef
sorry wrong
Nayef
  • Nayef
I will try to solve
anonymous
  • anonymous
i got 4.5 to 8 also

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More answers

Nayef
  • Nayef
The time interval(s) when the particle is moving right: (0,2.5) & (6,8)
IrishBoy123
  • IrishBoy123
\(v(t) = t^2 - 9t + 18\) the first thing you should do is plot this: ie \(v(t) vs \ t\). then you have a visual for what is to follow. eg it is moving right if \(v(t) > 0\)
Nayef
  • Nayef
going faster: (3,4) & (6,8)
Nayef
  • Nayef
sorry going faster: (3,4.5) & (6,8)
Nayef
  • Nayef
Slowing down: (0,3) & (4.5,6)
Nayef
  • Nayef
All that can be understood from drawing V(t) and a(t). where a(t) is the acceleration , we can get it from diffrentating V(t)
Nayef
  • Nayef
If V(t) positive , Then the particle going right , vise versa.
Nayef
  • Nayef
If V(t) and a(t) with different signs , then the particle decelerating, vise versa. Hope you understand.
anonymous
  • anonymous
thank you it makes much sense

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