A monkey climbs up a slippery pole for 3 seconds and subsequently slips for 3 seconds. Its velocity at time $t$ is given by $v(t) = 2t(3 - t)$; $0 < t < 3$ and $v(t) = -(t - 3)(6 - t)$ for $3 < t < 6$ s in m/s. It repeats this cycle till it reaches the height of 20 m. (a) At what time is its velocity maximum? (b) At what time is its average velocity maximum? (c) At what time is its acceleration maximum in magnitude? (d) How many cycles (counting fractions) are required to reach the top?

> [!TLDR] > The velocity function is piecewise quadratic over each 6 s cycle. The maximum velocity is $4.5\ \text{m/s}$ at $t=1.5\ \text{s}$, the maximum average velocity is $3.375\ \text{m/s}$ at $t=2.25\ \text{s}$, the acceleration reaches its largest magnitude of $6\ \text{m/s}^2$ at **both** $t=0$ and $t=3\ \text{s}$, and the net rise per 6 s cycle is $4.5\ \text{m}$, so $\dfrac{40}{9}$ cycles are needed to climb $20\ \text{m}$. ## Setting Up the Motion Over each 6 s cycle: - **Climbing phase** ($0<t<3$): $v_1(t) = 2t(3-t) = 6t - 2t^2$ (upward, positive). - **Slipping phase** ($3<t<6$): $v_2(t) = -(t-3)(6-t) = t^2 - 9t + 18$ (downward, so $v_2(t) \le 0$ throughout this interval — its roots are $t=3$ and $t=6$, and it opens upward, so it is negative strictly between them). ## (a) Time of maximum velocity $v_1(t) = 6t - 2t^2$ is a downward-opening parabola; its maximum is at its vertex: $$v_1'(t) = 6 - 4t = 0 \implies t = 1.5\ \text{s}, \qquad v_1(1.5) = 6(1.5) - 2(1.5)^2 = 9 - 4.5 = 4.5\ \text{m/s}$$ In the slipping phase $v_2(t) \le 0$ always, so it can never exceed $4.5\ \text{m/s}$. ## (b) Time of maximum average velocity Average velocity up to time $t$ is $V_{\text{avg}}(t) = \dfrac{S(t)}{t}$, where $S(t)$ is total displacement (integral of $v$). **For $0 3$:** since $v_2(t) \le 0$ throughout $(3,6)$, the total displacement $S(t)$ is *decreasing* while $t$ keeps increasing — both effects only reduce the ratio $S(t)/t$ further. So $S(t)/t < S(3)/3 = 9/3 = 3\ \text{m/s} < 3.375\ \text{m/s}$ for every $t$ in this range, and the average velocity can never exceed the value already found at $t=2.25\ \text{s}

A monkey climbs up a slippery pole for 3 seconds and subsequently slips for 3 seconds. Its velocity at time t

NCERT Exemplar · Q25

t

(a) At what time is its velocity maximum?

(b) At what time is its average velocity maximum?

(d) How many cycles (counting fractions) are required to reach the top?

Yanam CbseLong· 5mImportance★★★★★est

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Concept understanding — Instantaneous Velocity

Instantaneous Velocity: From "How Fast" to "How Fast Right Now"

You already know average velocity. If a car travels 120 km in 2 hours, its average velocity is 60 km/h. That tells you the overall rate, but it hides everything that happened in between — the traffic jams, the sudden bursts of speed, the moments the car was completely stopped.

Now imagine you want to know the car's velocity at exactly 10:15 AM, not averaged over an hour or a minute. That's instantaneous velocity — the velocity at a single instant of time.

The Intuition: Zooming In

Think of a speedometer needle. When you drive, the needle doesn't stay fixed at 60 km/h. It jumps up when you accelerate, drops when you brake. At any given moment, the needle points to a specific number. That number is your instantaneous speed (velocity, if direction matters).

But here's the puzzle: at a single instant, the car hasn't moved any distance. How can you have a speed if $\Delta t = 0$ ? You can't divide by zero.

The trick is to shrink the time interval smaller and smaller, and see what the average velocity approaches.

The Precise Definition

Let $s(t)$ be the position of an object at time $t$ . The average velocity over a time interval $[t, t+h]$ is:

v_{\text{avg}} = \frac{s(t+h) - s(t)}{h}

Now, let $h$ get closer and closer to 0 (but never equal to 0). If the average velocity settles down to a single number as $h \to 0$ , that number is the instantaneous velocity at time $t$ :

v(t) = \lim_{h \to 0} \frac{s(t+h) - s(t)}{h}

v(t) = \lim_{h \to 0} \frac{s(t+h) - s(t)}{h}

This limit is exactly the derivative of position with respect to time. In calculus notation: $v(t) = s'(t)$ .

A Concrete Example

Suppose a ball is dropped from rest, and its height (in meters) after $t$ seconds is $s(t) = 4.9t^2$ (ignoring air resistance).

Average velocity from $t=2$ to $t=2.1$ seconds:

v_{\text{avg}} = \frac{4.9(2.1)^2 - 4.9(2)^2}{0.1} = \frac{4.9(4.41 - 4)}{0.1} = \frac{4.9 \times 0.41}{0.1} = 20.09 \text{ m/s}

Average velocity from $t=2$ to $t=2.01$ :

v_{\text{avg}} = \frac{4.9(2.01)^2 - 4.9(2)^2}{0.01} = \frac{4.9(4.0401 - 4)}{0.01} = 19.649 \text{ m/s}

Average velocity from $t=2$ to $t=2.001$ :

v_{\text{avg}} = \frac{4.9(2.001)^2 - 4.9(2)^2}{0.001} = 19.6049 \text{ m/s}

The numbers are converging to 19.6 m/s. That's the instantaneous velocity at $t=2$ seconds.

Using the derivative: $v(t) = 9.8t$ , so $v(2) = 19.6$ m/s. Matches perfectly.

Key Takeaways for Exams

Concept	Meaning	Formula
Average velocity	Total displacement ÷ total time	$\frac{\Delta s}{\Delta t}$
Instantaneous velocity	Velocity at a single moment	$\lim_{h \to 0} \frac{s(t+h)-s(t)}{h}$

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The velocity function is piecewise quadratic over each 6 s cycle. The maximum velocity is $4.5\ \text{m/s}$ at $t=1.5\ \text{s}$ , the maximum average velocity is $3.375\ \text{m/s}$ at $t=2.25\ \text{s}$ , the acceleration reaches its largest magnitude of $6\ \text{m/s}^2$ at both $t=0$ and $t=3\ \text{s}$ , and the net rise per 6 s cycle is $4.5\ \text{m}$ , so $\dfrac{40}{9}$ cycles are needed to climb $20\ \text{m}$ .

Setting Up the Motion

Over each 6 s cycle:

Climbing phase ( $0<t<3$ ): $v_1(t) = 2t(3-t) = 6t - 2t^2$ (upward, positive).
Slipping phase ( $3<t<6$ ): $v_2(t) = -(t-3)(6-t) = t^2 - 9t + 18$ (downward, so $v_2(t) \le 0$ throughout this interval — its roots are $t=3$ and $t=6$ , and it opens upward, so it is negative strictly between them).

(a) Time of maximum velocity

$v_1(t) = 6t - 2t^2$ is a downward-opening parabola; its maximum is at its vertex:

v_1'(t) = 6 - 4t = 0 \implies t = 1.5\ \text{s}, \qquad v_1(1.5) = 6(1.5) - 2(1.5)^2 = 9 - 4.5 = 4.5\ \text{m/s}

In the slipping phase $v_2(t) \le 0$ always, so it can never exceed $4.5\ \text{m/s}$ .

(b) Time of maximum average velocity

Average velocity up to time $t$ is $V_{\text{avg}}(t) = \dfrac{S(t)}{t}$ , where $S(t)$ is total displacement (integral of $v$ ).

For $0 < t \le 3$ : $S_1(t) = \displaystyle\int_0^t (6u - 2u^2)\,du = 3t^2 - \dfrac{2}{3}t^3$ , so

V_{\text{avg}}(t) = 3t - \frac{2}{3}t^2

Maximizing: $\dfrac{dV_{\text{avg}}}{dt} = 3 - \dfrac{4}{3}t = 0 \implies t = \dfrac{9}{4} = 2.25\ \text{s}$ , giving

V_{\text{avg}}(2.25) = 3(2.25) - \frac{2}{3}(2.25)^2 = 6.75 - 3.375 = 3.375\ \text{m/s}

For $t > 3$ : since $v_2(t) \le 0$ throughout $(3,6)$ , the total displacement $S(t)$ is decreasing while $t$ keeps increasing — both effects only reduce the ratio $S(t)/t$ further. So $S(t)/t < S(3)/3 = 9/3 = 3\ \text{m/s} < 3.375\ \text{m/s}$ for every $t$ in this range, and the average velocity can never exceed the value already found at $t=2.25\ \text{s}$ .

So the average velocity is maximum at $t = 2.25\ \text{s}$ .

(c) Time of maximum acceleration magnitude

Climbing phase: $a_1(t) = v_1'(t) = 6 - 4t$ . This is a straight line, so $|a_1(t)|$ is largest at the two ends of the interval:

a_1(0) = 6\ \text{m/s}^2, \qquad a_1(3) = 6 - 12 = -6\ \text{m/s}^2 \ (\text{magnitude } 6)

and it passes through $0$ at $t=1.5\ \text{s}$ in between.

Slipping phase: $a_2(t) = v_2'(t) = 2t - 9$ :

a_2(3) = 6 - 9 = -3\ \text{m/s}^2, \qquad a_2(6) = 12 - 9 = 3\ \text{m/s}^2

so $|a_2(t)| \le 3\ \text{m/s}^2$ throughout — smaller than the climbing phase's peak. …

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Concept: Locating Extrema by Completing the Square; Computing Areas via the Parabola-Segment Rule — No Derivatives At All

Method: Vertex Form + Archimedes' Parabola-Segment Area (a purely algebraic/geometric route, avoiding calculus throughout)

The stored answer differentiates each piecewise velocity function to find maxima and checks the sign of successive derivatives. This method instead relies on two purely algebraic/geometric facts — a quadratic's vertex is visible directly from its completed-square form, and the area between a parabola and a chord connecting its two roots equals $\tfrac23\times\text{base}\times\text{height}$ (the classical parabola-segment rule) — to answer every part without ever writing $v'(t)$ or $a'(t)$ .

Setup

Climbing phase, $0<t<3$ : $v_1(t) = 6t-2t^2$ . Slipping phase, $3<t<6$ : $v_2(t) = t^2-9t+18$ .

(a) Time of maximum velocity — by completing the square

Complete the square on $v_1(t)$ :

v_1(t) = -2t^2+6t = -2\left(t^2-3t\right) = -2\left[(t-1.5)^2 - 2.25\right] = -2(t-1.5)^2 + 4.5

Since $-2(t-1.5)^2 \le 0$ always, this is manifestly maximised — without differentiating — exactly when the squared term is zero, i.e. at $t=1.5\ \text{s}$ , giving $v_{1,\max}=4.5\ \text{m/s}$ .

Rule out the slipping phase directly from its sign, no calculus needed: $v_2(t) = (t-3)(t-6)$ has roots at $t=3,6$ and opens upward (positive leading coefficient), so it is $\le0$ strictly between its roots — the slipping-phase velocity is never positive, so it can never beat $4.5\ \text{m/s}$ .

(b) Time of maximum average velocity — same vertex trick, on a different quadratic

Get the displacement via the parabola-segment area rule, not integration bookkeeping. For $0<t\le3$ , displacement up to time $t$ is the area under $v_1$ from $0$ to $t$ ; rather than integrating, directly complete the square on $S_1(t)/t$ : using $S_1(t) = 3t^2-\tfrac23t^3$ (obtainable either by direct antiderivative or the segment rule below — kept here since $V_{\text{avg}}$ itself needs to be a function of $t$ , not the fixed segment area):

V_{\text{avg}}(t) = \frac{S_1(t)}{t} = 3t - \frac{2}{3}t^2 = -\frac23\left(t^2-\frac{9}{2}t\right) = -\frac23\left[\left(t-\frac94\right)^2 - \frac{81}{16}\right] = -\frac23\left(t-\frac94\right)^2 + \frac{27}{8}

Maximised, again by inspection of the completed square, at $t=\dfrac94=2.25\ \text{s}$ , giving $V_{\text{avg}} = \dfrac{27}{8}=3.375\ \text{m/s}$ .

Confirm nothing later beats this, using the sign fact from Step 2 alone. Since $v_2(t)\le0$ throughout the slip, $S(t)$ can only decrease for $t>3$ while $t$ itself keeps growing — both effects only shrink the ratio $S(t)/t$ further, so the running average can never recover past its value at $t=3$ ( $=9/3=3<3.375$ ), let alone exceed the peak already found. No new calculus needed — just the established sign of $v_2$ .

(c) Time of maximum $|a|$ — using the "linear function is extremal at its endpoints" rule

State the rule. A linear function's magnitude, over a closed interval, is always maximised at one of the interval's two endpoints — never in the interior (a straight line has no interior turning point at all). This needs no derivative: it follows because a linear function is monotonic (or constant), so its extreme values over an interval are exactly its two boundary values.
Apply it to $a_1(t) = 6-4t$ (linear, since $v_1$ is quadratic) over $[0,3]$ : endpoints give $a_1(0)=6$ and $a_1(3)=6-12=-6$ (magnitude $6$ ) — the maximum magnitude, $6\ \text{m/s}^2$ , occurs at both ends.
Apply the same rule to $a_2(t)=2t-9$ over $[3,6]$ : endpoints give $a_2(3)=-3$ and $a_2(6)=3$ — maximum magnitude only $3\ \text{m/s}^2$ , smaller than the climbing phase's peak. …

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Setting Up the Motion

(a) Time of maximum velocity

(b) Time of maximum average velocity

(c) Time of maximum acceleration magnitude

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