A woman starts from her home at 9.00 am, walks with a speed of $5\ \text{km h}^{-1}$ on a straight road up to her office $2.5\ \text{km}$ away, stays at the office up to 5.00 pm, and returns home by an auto with a speed of $25\ \text{km h}^{-1}$. Choose suitable scales and plot the $x$-$t$ graph of her motion.

A woman starts from her home at 9.00 am, walks with a speed of 5km h⁻¹ on a straight road up to her office 2.5

Exercises · 2.3

5\ \text{km h}^{-1}

Meghalaya MboseTextbookSubjective· 3mImportance★★★★★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	$\lim_{h \to 0} \frac{s(t+h)-s(t)}{h}$

Watch out

Do not confuse instantaneous velocity with average velocity. A common exam trap: "A car travels 100 km in 2 hours. What is its velocity at the 1-hour mark?" The answer is not 50 km/h — that's the average. You need more information (or a position function) to find the instantaneous value.

Important

Instantaneous velocity is a vector — it has both magnitude (speed) and direction. If the object reverses direction, the instantaneous velocity changes sign. The speedometer only shows magnitude.

Why This Matters

Instantaneous velocity is the foundation of all of kinematics and dynamics. Newton's second law ( $F = ma$ ) uses acceleration, which is the instantaneous rate of change of velocity. Without this concept, you can't describe motion that changes — which is almost all real motion.

When you see a graph of position vs. time, the instantaneous velocity at any point is the slope of the tangent line at that point. That geometric interpretation will serve you well in both physics and calculus.

Instantaneous Velocity bridges the NCERT Class 11 Physics chapter on Motion in a Straight Line with the calculus taught in Class 11 Mathematics' Limits and Derivatives chapter, matching searches like "instantaneous velocity: definition and formula" or "kinematics important questions class 11 physics". This distinction from average velocity is a classic conceptual question in CBSE boards and a common numerical setup for JEE Main and NEET.

The key idea is to break the motion into three distinct segments (walk to office, stay at office, auto ride back) and plot position vs. time using the given speeds and distances. The final graph is a piecewise linear plot with slopes of $5\ \text{km/h}$ , $0$ , and $-25\ \text{km/h}$ .

Understanding the problem

We need to plot the position ( $x$ ) of the woman as a function of time ( $t$ ), measured from her home. The motion has three clear phases:

She walks from home to office at a steady speed.
She stays at the office for several hours.
She returns home by auto at a much higher speed.

The key physics here is instantaneous velocity — the slope of the $x$ - $t$ graph at any point gives the velocity at that instant. For uniform motion (constant speed), the graph is a straight line whose slope equals the velocity. When she is stationary, the slope is zero (horizontal line).

Let’s set up our coordinate system: let the home be at $x = 0$ and the office at $x = +2.5\ \text{km}$ . Time $t$ is measured in hours from 9:00 am.

Step-by-step solution

1. Find the time taken to walk to the office

Speed $v_1 = 5\ \text{km/h}$ , distance $d = 2.5\ \text{km}$ .

Time taken:

t_1 = \frac{d}{v_1} = \frac{2.5}{5} = 0.5\ \text{h}

So she reaches the office at $9:00\ \text{am} + 0.5\ \text{h} = 9:30\ \text{am}$ .

2. Plot the first segment (walking)

From $t = 0$ to $t = 0.5\ \text{h}$ , position goes from $x = 0$ to $x = 2.5\ \text{km}$ . The slope is $+5\ \text{km/h}$ (positive because she moves away from home).

3. The stay at the office

She stays from $9:30\ \text{am}$ to $5:00\ \text{pm}$ . That’s a duration of $7.5\ \text{hours}$ (from $t = 0.5\ \text{h}$ to $t = 8.0\ \text{h}$ , since 5:00 pm is 8 hours after 9:00 am).

During this time, position is constant at $x = 2.5\ \text{km}$ . The graph is a horizontal line — slope zero.

4. Find the time taken to return home

Speed of auto $v_2 = 25\ \text{km/h}$ , same distance $d = 2.5\ \text{km}$ .

Time taken:

t_2 = \frac{2.5}{25} = 0.1\ \text{h} = 6\ \text{minutes}

She leaves office at $t = 8.0\ \text{h}$ and reaches home at $t = 8.1\ \text{h}$ (i.e., 5:06 pm).

5. Plot the return segment

From $t = 8.0\ \text{h}$ to $t = 8.1\ \text{h}$ , position goes from $x = 2.5\ \text{km}$ back to $x = 0$ . The slope is $-25\ \text{km/h}$ (negative because she moves toward home).

Watch out

A common mistake is to think the return slope should be steeper than the outward slope — which is correct — but students sometimes forget that the sign is negative. The auto moves faster, so the line is steeper, but downward.

6. Choose suitable scales for the graph

Time axis ( $t$ ): from $0$ to $8.5\ \text{h}$ . A scale of $1\ \text{cm} = 1\ \text{h}$ works well.
Position axis ( $x$ ): from $0$ to $3\ \text{km}$ . A scale of $1\ \text{cm} = 0.5\ \text{km}$ is convenient.

7. Draw the graph

The $x$ - $t$ graph consists of three straight line segments:

Time interval (h)	Position (km)	Slope (km/h)	Description
$0$ to $0.5$	$0$ to $2.5$	$+5$	Walking to office
$0.5$ to $8.0$	$2.5$ (constant)	$0$	At office
$8.0$ to $8.1$	$2.5$ to $0$	$-25$	Auto ride back

The graph is a rising line, then a flat line, then a steeply falling line.

Tip

You don’t need to plot every point — just the three key points: $(0,0)$ , $(0.5, 2.5)$ , $(8.0, 2.5)$ , and $(8.1, 0)$ . Connect them with straight lines.

✓Final answer

x

Concept: Building an x-t Graph from a Clock-Time Timeline

Method: Clock-Time Bookkeeping (reason in real time-of-day, convert to $t$ only for the axis)

Rather than immediately defining $t$ as "hours since 9:00 am" and working entirely in that variable, this method tracks the woman's day using the actual clock times she experiences — exactly as you would plan a real schedule — and only converts to a numeric time axis at the very last step, when the graph itself needs to be drawn.

Steps

List the known clock-time events directly:
- Leaves home: 9:00 am
- Walking speed: $5\ \text{km h}^{-1}$ , office distance $2.5\ \text{km}$
- Stays at office until: 5:00 pm
- Return trip: by auto at $25\ \text{km h}^{-1}$ , same $2.5\ \text{km}$
Find the clock time she reaches the office, using $\text{time} = \text{distance}/\text{speed}$ :

\frac{2.5\ \text{km}}{5\ \text{km h}^{-1}} = 0.5\ \text{h} = 30\ \text{min}

$9{:}00\ \text{am} + 30\ \text{min} = \mathbf{9{:}30\ \text{am}}$ — she is at the office ( $x=2.5\ \text{km}$ ) by this clock time.

She remains at the office ( $x=2.5\ \text{km}$ , unchanged) from 9:30 am to 5:00 pm — no calculation needed, this interval is given directly as clock times.
Find the clock time she gets home, using the same distance-over-speed relation for the auto:

\frac{2.5\ \text{km}}{25\ \text{km h}^{-1}} = 0.1\ \text{h} = 6\ \text{min}

$5{:}00\ \text{pm} + 6\ \text{min} = \mathbf{5{:}06\ \text{pm}}$ — back at $x=0$ .

Only now convert every clock-time event to the numeric $t$ -axis (hours after 9:00 am), purely as a labelling step for the graph:

Clock time $t$ (h after 9 am) Position $x$ (km)
9:00 am $0$ $0$
9:30 am $0.5$ $2.5$
5:00 pm $8.0$ $2.5$
5:06 pm $8.1$ $0$
Plot the three straight segments through these four points: rising (slope $+5$ ) from $(0,0)$ to $(0.5,2.5)$ ; flat from $(0.5,2.5)$ to $(8.0,2.5)$ ; steeply falling (slope $-25$ ) from $(8.0,2.5)$ to $(8.1,0)$ .

Clock time	$t$ (h after 9 am)	Position $x$ (km)
9:00 am	$0$	$0$
9:30 am	$0.5$	$2.5$
5:00 pm	$8.0$	$2.5$
5:06 pm	$8.1$	$0$

Why reason in clock time first

The problem statement itself is given entirely in clock times (9:00 am, 5:00 pm) — reasoning directly in that language avoids an unnecessary translation step at the start and matches how you'd actually plan the schedule; the $t$ -axis conversion is deferred to exactly where it's needed, for the graph, and nowhere else.

Final Answer

The $x$ - $t$ graph is the three-segment line through $(0,0) \to (0.5\ \text{h},\,2.5\ \text{km}) \to (8.0\ \text{h},\,2.5\ \text{km}) \to (8.1\ \text{h},\,0)$ , with slopes $+5$ , $0$ , $-25\ \text{km h}^{-1}$ respectively — reached here entirely from clock-time bookkeeping (9:00 am $\to$ 9:30 am $\to$ 5:00 pm $\to$ 5:06 pm).

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