Question

A cyclist pedals along a straight road with velocity v(t)=12t^2-48t+36 (in mph) for 0
13 years ago

Answer 1

this is a strange way of asking solve \(12t^2-48t+36>0\)

Answer 2

because if the velocity is positive, that is the same as saying she moves in the positive direction
start by dividing by 12 and solve
\[t^2-4t+3>0\]

Answer 3

+ve 3<t<1
-ve 1<t<3

Answer 4

ok so on the interval \(0\le t\le 3\) it is negative on \((1,3)\) and positive on \((0,1)\)
btw there is no such interval as \(3<t<1\)
it is two intervals
\((-\infty, 1)\cup (3,\infty)\)

Answer 5

@satellite73 can you show me how you got the answers

Answer 6

\(t^2-4t+3=(t-1)(t-3)\) is zero at \(t=1,t=3\)
since this is a parabola that opens up, it is negative between the zeros and positive outside of them

Answer 7

if you want to solve it algebraically
(t-1)(t-3) > 0 when both terms are positive:
(t-1)> 0 and (at the same time) t-3>0
this requires t>1 and t>3, which simplifies to t>3

or both terms are negative
t-1<0 and t-3<0
t< 1 and t<3. this becomes just t<1

so positive v for t<1 or t>3