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- 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

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- anonymous

- schrodinger

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- Nayef

sorry wrong

- Nayef

I will try to solve

- anonymous

i got 4.5 to 8 also

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- Nayef

The time interval(s) when the particle is moving right: (0,2.5) & (6,8)

- 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

going faster: (3,4) & (6,8)

- Nayef

sorry
going faster: (3,4.5) & (6,8)

- Nayef

Slowing down: (0,3) & (4.5,6)

- 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

If V(t) positive , Then the particle going right , vise versa.

- Nayef

If V(t) and a(t) with different signs , then the particle decelerating, vise versa.
Hope you understand.

- anonymous

thank you it makes much sense

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