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

Question

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

nayef · Answer

sorry wrong

nayef · Answer

I will try to solve

anonymous · Answer

i got 4.5 to 8 also

nayef · Answer

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

irishboy123 · Answer

$v(t) = t^2 - 9t + 18$

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 · Answer

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

nayef · Answer

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

nayef · Answer

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

nayef · Answer

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 · Answer

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

nayef · Answer

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

anonymous · Answer

thank you it makes much sense