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.
"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.
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.
To determine when the velocity of the ice hockey puck is increasing, we need to find the values of t for which the derivative of the position function, s(t), is positive.
Let's start by finding the derivative of the position function to obtain the velocity function, v(t). Differentiate each term of the function separately:
s(t) = (1/3)t^3 + t^2 + t - 7/3
Differentiating, we get:
v(t) = (d/dt)((1/3)t^3 + t^2 + t - 7/3)
= (1/3)(d/dt)(t^3) + (d/dt)(t^2) + (d/dt)(t) - (d/dt)(7/3)
= (1/3)(3t^2) + (2t) + 1
= t^2 + 2t + 1
Now that we have the velocity function, v(t), we need to find the values of t for which the velocity is increasing. In other words, we want to find the values of t where the derivative of the velocity function, v'(t), is positive.
Taking the derivative of the velocity function, we get:
v'(t) = (d/dt)(t^2 + 2t + 1)
= 2t + 2
To find the values of t when the velocity is increasing, we need to solve the inequality:
v'(t) > 0
2t + 2 > 0
Subtracting 2 from both sides, we get:
2t > -2
Dividing by 2, we have:
t > -1
Therefore, the velocity of the puck is increasing for values of t greater than -1.
To summarize, the velocity of the ice hockey puck is increasing when t is greater than -1.