# The position of a particle moving along a line is given by $2t^3 -24t^2+90t + 7$ for $t >0$

For what values of $t$ is the speed of the particle increasing?

I tried to find the first derivative and I get $$6t^2-48t+90 = 0$$ $$t^2-8t+15 = 0$$

Which is giving me $t>5$ and $0 < t < 3$, but the book gives a different answer

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Calculus and algebra-precalculus? That seems contradictory :-) –  Aryabhata Mar 4 '12 at 19:11
Speed is increasing, is not the same as position is increasing. –  Aryabhata Mar 4 '12 at 19:12

Hint:

You're given position, $s$ as a function of time, $t$--That is $s(t)$. Differentiating this will give you velocity, $v(t)$.

To know when the velocity increases, you'll have to look at the derivative of velocity as a function of time, $t$. Note that this is the second derivative of $s(t)$.

So, \begin{align}speed(t)&=|6t^2-48t+90|\\(speed)'(t)&=?? \end{align}

And, the velocity $speed(t)$ increases when $(speed)'(t) \ge 0$. This means, the speed increases, when $t \in ??$.

To assist you, I'll add a graph while I leave it to you to do the actual differentiation to convince yourself.

$\hspace{1.5 in}$

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Also, I suppose we need to take the absolute value of the velocity to get the speed? –  Aryabhata Mar 4 '12 at 19:15
I think, yes, we'll have to take the absolute value for the velocity. : ( I'll edit the answer. Thanks for pointing out! –  user21436 Mar 4 '12 at 19:24

Speed = t^2 - 8t +15

Increasing speed means d(speed)/dt is positive.

Acceleration = d(speed)/dt = 2*t-8

So, speed is increasing for 2t-8 > 0, or, t > 4

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The velocity increases when its derivative is positive, that is the acceleration is positive, so derive the velocity and then you will get if I am not mistaken $t\geq 4$.

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You need to keep in mind that speed is the absolute value of velocity. If the velocity is positive and increasing, the speed will increasing. But if the velocity is negative and decreasing (becoming more and more negative), the speed is increasing. There are two other cases to consider, but I'll leave that to you.

For your problem, let's consider the motion of the point by looking at the velocity function (whose graph is a parabola opening upwards with zeros at $x=3$ and $x=5$). With the standard conventions:

From $t=0$ to $t=3$: The particle is moving to the right. Its speed is decreasing over $(0,3)$. (Note the velocity is positive and decreasing here.)

At $t=3$, the particle reverses direction. (The velocity is 0 here.)

From $t=3$ to $t=4$: The particle is moving to the left. Its speed is increasing over $(3,4)$. (The velocity is negative and decreasing here.)

From $t=4$ to $t=5$: The particle is moving to the left. Its speed is decreasing over $(4,5)$. (The velocity is negative and increasing here)

At $t=5$, the particle reverses direction. (The velocity is 0 here.)

From $t=5$ onwards: The particle is moving to the right. Its speed is increasing over $(5,\infty)$. (The velocity is positive and increasing here)

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Let's be careful. The velocity is $6(t^2-8t+15)$. This is $\ge 0$ when $t \ge 5$ and when $t\le 3$. So on $(5,\infty)$, and also on $(0,3)$, this is the speed. It is not the speed on $(3,5)$. There the speed is $-6(t^2-8t+15)$.

When $t > 5$ and also when $t< 3$, the derivative of speed is $6(2t-8)$, and is positive when $t>4$. So the speed is certainly increasing over the interval $(5,\infty)$.

In the interval $(3,5)$, the derivative of speed is $-6(2t-8)$. This is positive in the interval $(3,4)$.

So the speed is increasing on $(5,\infty)$ and on $(3,4)$.

Note that the derivative of speed does not exist at $3$ and at $5$.

Remark: Occasionally, I have asked questions of this nature, though not quite as sadistic. Here there is a double twist. Even if the student notices that the question doesn't ask where $s(t)$ is increasing, there is the velocity/speed trap as backup. Not a good idea, it only proves one can fool most of the people most of the time.

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