Question

A particle starts at time t=0 and moves along a number line so that its position, at time t greater than or equal to 0, is given by x(t)=(t-2)^3(t-6). The particle is moving to the right for
1) 0 < t < 5
2) 2 < t < 6
3) t >5
4) t greater than or equal to 0
5) Never

Answer 1

here we have to solve this inequality:
\[x\left( t \right) > 0 \Rightarrow {\left( {t - 2} \right)^3}\left( {t - 6} \right) > 0\]

Answer 2

Yes

Answer 3

i did that and i got (-infinity to 2) U (6, infinity)

Answer 4

yes! correct! More precisely, since \(t \geqslant 0 \), we have:
\[0 \leqslant t < 2 \cup t > 6\]

Answer 5

nevertheless it is not an option of yours

Answer 6

So the answer will be 5) because there is no option like that

Answer 7

I think taht maybe the right function \(x(t)\), can be this:
\[x\left( t \right) = {\left( {t - 2} \right)^3}\left( {6 - t} \right)\]

Answer 8

thank you this was more to just check my answer i wanted to make sure i understand what to do in a situation like this one

Answer 9

that*

Answer 10

I understand your concern however the function given was x(t)= (t-2)^3(t-6)

Answer 11

in that case I don't know how to think, since we have found the time interval, wherein the particle is moving to the right

Answer 12

try to ask to your teacher, then let me know his reply :)

Answer 13

I will try my best to do that Sir

Answer 14

ok! :)