"deriviting" ???
v(t) = 3t^2 + 2t + 1
the t of the vertex of this parabola is -2/6 = -1/3
and it opens up
but t ≥ 0 to be a real situation
so at t = 0, v = 1 unit of distance/unit of time
According to your equation, this is a very interesting hockey puck.
for values of t, the velocity will be forever increasing, since
3t^2 + 2t + 1 will be always positive
I suspect a typo in your equation , there should be some negative t terms.
An ice hockey puck is sliding along a straight line so that its position is s(t) =
t3
3 + t2 + t −
7/3 . At which times t is the velocity of the puck increasing? I tried deriviting the equation then solving for t and plugging in numbers to see if it was positive or negative and couldn't get it.
5 answers
sorry the equation is 1/3(t^2) +t^2 +t -7/3
so, follow Reiny's reasoning with the new function.
the answer is >-1 but when i derived the equation there were still no negative numbers because -7/3 is a constant so it went to 0. after deriving the equation i got t to be -1 so i pluggedin -2 and 0 and got both to be positive(or increasing) all solutions is not an option for this question though...
assuming a typo, the position
s = 1/3(t^3) +t^2 +t -7/3
v = ds/dt = t^2 + 2t + 1
But, you want to know when the speed is increasing. That is, dv/dt > 0.
dv/dt = 2t+2
That is, dv/dt > 0 when t > -1
Not sure what negative t means in this context, but the puck is always speeding up for t>0.
s = 1/3(t^3) +t^2 +t -7/3
v = ds/dt = t^2 + 2t + 1
But, you want to know when the speed is increasing. That is, dv/dt > 0.
dv/dt = 2t+2
That is, dv/dt > 0 when t > -1
Not sure what negative t means in this context, but the puck is always speeding up for t>0.
2 answers