Motion in a Straight Line

A body can be treated as a point object when its size is much smaller than the distance it travels, i.e., the internal motion or rotation of the body is irrelevant to the problem.

(a) Railway carriage moving between two stations — YES, it can be treated as a point object.
The size of the carriage (~20 m) is negligible compared to the distance between two stations (several kilometres). The motion is smooth (no jerks), so internal details are unimportant.

(b) Monkey sitting on top of a man cycling on a circular track — YES, it can be treated as a point object.
The size of the monkey is negligible compared to the size of the circular track. The monkey is sitting still relative to the man, so no internal motion matters.

(c) Spinning cricket ball that turns sharply on hitting the ground — NO, it cannot be treated as a point object.
The spin (rotation) of the ball is crucial to understanding its sharp turn. The size and rotational motion of the ball are important here.

(d) Tumbling beaker that has slipped off the edge of a table — NO, it cannot be treated as a point object.
The beaker is tumbling (rotating), so its orientation and rotational motion are significant. It cannot be reduced to a point.

From the x–t graph (Fig. 2.9), we read the following information:
- The x-axis represents position (distance from school O) and the t-axis represents time.
- A's home P is at a smaller distance from school than B's home Q.
- B's graph starts from the origin (school) at an earlier time than A's graph.
- The slope of the x–t graph gives speed. B's line has a steeper slope than A's line.
- Both graphs end (reach home) at the same time.
- The two lines intersect once, meaning one child overtakes the other once.

(a) A lives closer to the school than B.
(Home P of A is at a smaller x-value than home Q of B on the graph.)

(b) B starts from the school earlier than A.
(B's graph begins at an earlier time on the t-axis.)

(c) B walks faster than A.
(The slope of B's x–t graph is steeper than A's, indicating greater speed.)

(d) A and B reach home at the same time.
(Both graphs terminate at the same value of t.)

(e) B overtakes A on the road once.
(The two lines cross once, meaning B, who started earlier but walks faster, overtakes A once.)

Given:
- Start time: 9:00 am, starting position: home (x = 0)
- Speed walking to office: $v_1 = 5\,\text{km h}^{-1}$
- Distance to office: $d = 2.5\,\text{km}$
- Stay at office: 9:30 am to 5:00 pm
- Return speed (auto): $v_2 = 25\,\text{km h}^{-1}$

Step 1: Time to walk to office
$$t_1 = \frac{d}{v_1} = \frac{2.5}{5} = 0.5\,\text{h} = 30\,\text{min}$$
She reaches office at 9:30 am.

Step 2: Stay at office
She stays from 9:30 am to 5:00 pm (i.e., for 7.5 hours). Position remains constant at $x = 2.5\,\text{km}$.

Step 3: Time to return home by auto
$$t_2 = \frac{d}{v_2} = \frac{2.5}{25} = 0.1\,\text{h} = 6\,\text{min}$$
She reaches home at 5:06 pm.

Key points for the x–t graph:

| Time | Position (km) |
|---|---|
| 9:00 am | 0 |
| 9:30 am | 2.5 |
| 5:00 pm | 2.5 |
| 5:06 pm | 0 |

Scale suggestion: x-axis: 1 cm = 1 hour (or 30 min); y-axis: 1 cm = 0.5 km.

Description of graph:
- From 9:00 am to 9:30 am: a straight line with positive slope (walking to office).
- From 9:30 am to 5:00 pm: a horizontal straight line at $x = 2.5\,\text{km}$ (staying at office).
- From 5:00 pm to 5:06 pm: a straight line with steep negative slope back to $x = 0$ (returning by auto).

Given:
- Each step = 1 m, takes 1 s
- Pattern: 5 steps forward (5 m in 5 s), then 3 steps backward (3 m in 3 s)
- Net displacement per cycle = 5 – 3 = 2 m in 8 s
- Pit is at 13 m from start.

Tracking position after each cycle:

| Time (s) | Position (m) | Event |
|---|---|---|
| 0 | 0 | Start |
| 5 | 5 | After 5 forward steps |
| 8 | 2 | After 3 backward steps |
| 13 | 7 | After 5 forward steps |
| 16 | 4 | After 3 backward steps |
| 21 | 9 | After 5 forward steps |
| 24 | 6 | After 3 backward steps |
| 29 | 11 | After 5 forward steps |
| 32 | 8 | After 3 backward steps |
| 37 | 13 | After 5 forward steps — reaches pit! |

Step-by-step check for the last cycle (starting at t = 32 s, x = 8 m):
- At t = 33 s: x = 9 m
- At t = 34 s: x = 10 m
- At t = 35 s: x = 11 m
- At t = 36 s: x = 12 m
- At t = 37 s: x = 13 m ← Falls into pit

$$\boxed{\text{The drunkard falls into the pit after } 37\,\text{s}}$$

x–t graph description: The graph is a zigzag (piecewise linear) line. It rises steeply (slope = +1 m/s) for 5 s, then falls (slope = –1 m/s) for 3 s, repeating this pattern. The overall trend is a net upward drift of 2 m every 8 s, until the position reaches 13 m at t = 37 s.

Given:
- Initial velocity: $v_0 = 126\,\text{km h}^{-1} = 126 \times \frac{5}{18} = 35\,\text{m s}^{-1}$
- Final velocity: $v = 0$ (car stops)
- Distance: $x = 200\,\text{m}$
- Acceleration: uniform (retardation)

Step 1: Find retardation using $v^2 = v_0^2 + 2ax$
$$0 = (35)^2 + 2a(200)$$
$$0 = 1225 + 400a$$
$$a = -\frac{1225}{400} = -3.0625\,\text{m s}^{-2}$$

$$\boxed{\text{Retardation} = 3.06\,\text{m s}^{-2} \approx 3.1\,\text{m s}^{-2}}$$

Step 2: Find time using $v = v_0 + at$
$$0 = 35 + (-3.0625)\,t$$
$$t = \frac{35}{3.0625} \approx 11.43\,\text{s}$$

$$\boxed{t \approx 11.4\,\text{s}}$$

Result: The retardation of the car is approximately $3.06\,\text{m s}^{-2}$ and it takes approximately $11.4\,\text{s}$ to stop.

Given:
- Initial speed (upward): $v_0 = 29.4\,\text{m s}^{-1}$
- $g = 9.8\,\text{m s}^{-2}$

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(a) Direction of acceleration during upward motion:

During the entire flight (upward and downward), the only acceleration acting on the ball is due to gravity, which always acts vertically downward (towards the Earth), regardless of the direction of motion.

$$\boxed{\text{Acceleration is directed vertically downward during upward motion.}}$$

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(b) Velocity and acceleration at the highest point:

At the highest point, the ball momentarily comes to rest.
- Velocity = 0 (the ball stops momentarily before reversing direction)
- Acceleration = $g = 9.8\,\text{m s}^{-2}$, directed vertically downward (gravity never stops acting)

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(c) Signs of position, velocity, and acceleration:

Coordinate system: Origin at highest point, positive direction = vertically downward, $t = 0$ at highest point.

During upward motion (ball moving from player's hand to highest point):
- The ball is below the highest point (origin), so position x < 0 → negative
- The ball moves upward, which is opposite to positive direction → velocity v < 0 → negative
- Acceleration due to gravity is downward = positive direction → a > 0 → positive

During downward motion (ball falling from highest point back to player's hand):
- The ball is below the highest point (origin), so position x > 0 → positive
- The ball moves downward = positive direction → velocity v > 0 → positive
- Acceleration due to gravity is downward = positive direction → a > 0 → positive

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(d) Maximum height and time to return:

Maximum height: Using $v^2 = v_0^2 - 2gH$ (taking upward as positive, $v = 0$ at top):
$$0 = (29.4)^2 - 2(9.8)H$$
$$H = \frac{(29.4)^2}{2 \times 9.8} = \frac{864.36}{19.6} = 44.1\,\text{m}$$

$$\boxed{H = 44.1\,\text{m}}$$

Time to reach highest point: Using $v = v_0 - gt$:
$$0 = 29.4 - 9.8\,t_1$$
$$t_1 = \frac{29.4}{9.8} = 3\,\text{s}$$

By symmetry of projectile motion (no air resistance), the time to come back down equals the time to go up.

Total time = $2t_1 = 2 \times 3 = 6\,\text{s}$

$$\boxed{\text{The ball rises to a height of }44.1\,\text{m and returns to the player's hands after }6\,\text{s.}}$$

(a) A particle with zero speed at an instant may have non-zero acceleration at that instant.

TRUE.

*Reason:* Speed and acceleration are independent quantities. A particle can be momentarily at rest while still having a net force (and hence acceleration) acting on it.

*Example:* A ball thrown vertically upward has zero speed at the highest point, but the acceleration due to gravity ($g = 9.8\,\text{m s}^{-2}$ downward) is non-zero at that instant.

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(b) A particle with zero speed may have non-zero velocity.

FALSE.

*Reason:* Speed is the magnitude of velocity. If speed = 0, then $|v| = 0$, which means velocity = 0. It is impossible to have zero speed with non-zero velocity.

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(c) A particle with constant speed must have zero acceleration.

TRUE (for one-dimensional motion only).

*Reason:* In one-dimensional motion, the direction of motion is fixed (either forward or backward along a line). If speed (magnitude of velocity) is constant and direction is fixed, then velocity is constant, and hence acceleration $a = \frac{dv}{dt} = 0$.

*Note:* This would be FALSE in two or three dimensions (e.g., uniform circular motion has constant speed but non-zero centripetal acceleration). But for one-dimensional motion, it is TRUE.

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(d) A particle with positive value of acceleration must be speeding up.

FALSE.

*Reason:* The sign of acceleration depends on the chosen positive direction of the axis, not on whether the particle is speeding up or slowing down. If the particle moves in the negative direction (negative velocity) and has positive acceleration, it is actually slowing down.

*Example:* A ball thrown upward: if upward is taken as negative and downward as positive, then a = +g > 0. During the upward journey, the ball has negative velocity and positive acceleration — it is decelerating (slowing down), not speeding up.

Given:
- Initial height: $H = 90\,\text{m}$, dropped from rest ($u = 0$)
- At each bounce, speed after collision = $\frac{9}{10}$ × speed before collision
- $g = 9.8\,\text{m s}^{-2} \approx 10\,\text{m s}^{-2}$ (for easier calculation; use $g = 10\,\text{m s}^{-2}$)

Step 1: Speed just before first collision (1st fall)
$$v_1 = \sqrt{2gH} = \sqrt{2 \times 10 \times 90} = \sqrt{1800} \approx 42.4\,\text{m s}^{-1}$$

Using $g = 9.8\,\text{m s}^{-2}$:
$$v_1 = \sqrt{2 \times 9.8 \times 90} = \sqrt{1764} = 42\,\text{m s}^{-1}$$

Time to fall to floor (1st fall):
$$t_1 = \frac{v_1}{g} = \frac{42}{9.8} \approx 4.29\,\text{s}$$

Step 2: Speed just after 1st bounce:
$$v_1' = \frac{9}{10} \times 42 = 37.8\,\text{m s}^{-1}$$

Time for 1st bounce (up and down):
$$t_{b1} = \frac{2v_1'}{g} = \frac{2 \times 37.8}{9.8} \approx 7.71\,\text{s}$$

This is too long; the ball would not complete a full bounce within 12 s after the first fall at ~4.29 s. Let us recalculate:

Total time after 1st bounce ends = $4.29 + 7.71 = 12\,\text{s}$.

So within 0 to 12 s, the ball:
- Falls freely from $t = 0$ to $t \approx 4.29\,\text{s}$: speed increases linearly from 0 to 42 m/s.
- Bounces back (1st bounce): speed instantaneously becomes 37.8 m/s at $t \approx 4.29\,\text{s}$, then decreases linearly to 0 at the top, then increases linearly back to 37.8 m/s at $t \approx 12\,\text{s}$.

Description of speed-time graph:

| Segment | Time interval | Speed behaviour |
|---|---|---|
| Free fall (1st) | $0$ to $4.29\,\text{s}$ | Increases linearly: $0 \to 42\,\text{m s}^{-1}$ |
| 1st bounce (up) | $4.29\,\text{s}$ to $8.14\,\text{s}$ | Decreases linearly: $37.8 \to 0\,\text{m s}^{-1}$ |
| 1st bounce (down) | $8.14\,\text{s}$ to $12\,\text{s}$ | Increases linearly: $0 \to 37.8\,\text{m s}^{-1}$ |

Key features of the graph:
- The graph consists of straight line segments (since acceleration = $g$ = constant).
- At each collision, there is a sudden drop in speed (from 42 to 37.8 m/s).
- The slope of each segment has magnitude $g = 9.8\,\text{m s}^{-2}$.
- The graph is V-shaped between bounces (speed decreases to zero at maximum height, then increases again).

(a) Magnitude of displacement vs. total path length:

Displacement is the shortest distance between the initial and final positions of a particle. Its magnitude depends only on the start and end points.

Total path length (distance) is the actual length of the path traversed by the particle, regardless of direction.

*Example:* A person walks 4 m east, then 3 m west.
- Displacement = $4 - 3 = 1\,\text{m}$ (east); magnitude = 1 m
- Total path length = $4 + 3 = 7\,\text{m}$

Why path length ≥ |displacement|:

In one-dimensional motion, if a particle moves from $x_1$ to $x_2$ without reversing direction, path length = $|x_2 - x_1|$ = |displacement|.

If the particle reverses direction at some point, the path length includes the extra distance travelled back and forth, while displacement only accounts for the net change. Hence:
$$\text{Total path length} \geq |\text{Displacement}|$$

Equality holds when the particle moves in one direction only (no reversal of motion).

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(b) Magnitude of average velocity vs. average speed:

$$|\bar{v}| = \frac{|\text{Displacement}|}{\text{Time interval}}, \quad \bar{s} = \frac{\text{Total path length}}{\text{Time interval}}$$

*Example (same as above):* Total time = 7 s.
- $|\bar{v}| = \frac{1}{7}\,\text{m s}^{-1}$
- $\bar{s} = \frac{7}{7} = 1\,\text{m s}^{-1}$

Clearly \bar{s} > |\bar{v}|.

Why average speed ≥ |average velocity|:

Since total path length $\geq$ |displacement| (from part a), dividing both sides by the same positive time interval:
$$\frac{\text{Total path length}}{\Delta t} \geq \frac{|\text{Displacement}|}{\Delta t}$$
$$\therefore\quad \bar{s} \geq |\bar{v}|$$

Equality holds when the particle moves in one direction only (no reversal), so that path length = |displacement|.

Setting up the problem:

- Distance to market: $d = 2.5\,\text{km}$
- Speed to market: $v_1 = 5\,\text{km h}^{-1}$
- Speed returning: $v_2 = 7.5\,\text{km h}^{-1}$

Time to reach market:
$$t_1 = \frac{2.5}{5} = 0.5\,\text{h} = 30\,\text{min}$$

Time to return home from market:
$$t_2 = \frac{2.5}{7.5} = \frac{1}{3}\,\text{h} = 20\,\text{min}$$

Total time for round trip: $30 + 20 = 50\,\text{min}$

---

(i) 0 to 30 min (= 0 to 0.5 h):

The man just reaches the market. He has walked 2.5 km in the forward direction.

- Displacement = 2.5 km (towards market)
- Total path length = 2.5 km

(a) Magnitude of average velocity:
$$|\bar{v}| = \frac{\text{Displacement}}{\text{Time}} = \frac{2.5}{0.5} = 5\,\text{km h}^{-1}$$

(b) Average speed:
$$\bar{s} = \frac{\text{Path length}}{\text{Time}} = \frac{2.5}{0.5} = 5\,\text{km h}^{-1}$$

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(ii) 0 to 50 min (= 0 to 5/6 h):

The man completes the round trip (reaches home).

- Displacement = 0 (starts and ends at home)
- Total path length = $2.5 + 2.5 = 5\,\text{km}$

(a) Magnitude of average velocity:
$$|\bar{v}| = \frac{0}{5/6} = 0\,\text{km h}^{-1}$$

(b) Average speed:
$$\bar{s} = \frac{5}{5/6} = 5 \times \frac{6}{5} = 6\,\text{km h}^{-1}$$

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(iii) 0 to 40 min (= 0 to 2/3 h):

The man reaches the market at 30 min, then walks back for 10 min at 7.5 km h⁻¹.

Distance covered on return in 10 min $= \frac{10}{60} \times 7.5 = 1.25\,\text{km}$

So at $t = 40\,\text{min}$, he is $2.5 - 1.25 = 1.25\,\text{km}$ from home (towards market).

- Displacement = 1.25 km (from home towards market)
- Total path length = $2.5 + 1.25 = 3.75\,\text{km}$
- Time = $\frac{2}{3}\,\text{h}$

(a) Magnitude of average velocity:
$$|\bar{v}| = \frac{1.25}{2/3} = 1.25 \times \frac{3}{2} = 1.875\,\text{km h}^{-1}$$

(b) Average speed:
$$\bar{s} = \frac{3.75}{2/3} = 3.75 \times \frac{3}{2} = 5.625\,\text{km h}^{-1}$$

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Summary Table:

| Interval | $|\bar{v}|$ (km h⁻¹) | $\bar{s}$ (km h⁻¹) |
|---|---|---|
| 0–30 min | 5 | 5 |
| 0–50 min | 0 | 6 |
| 0–40 min | 1.875 | 5.625 |

Explanation:

Instantaneous velocity is defined as the limit of average velocity as the time interval $\Delta t \to 0$:
$$v = \lim_{\Delta t \to 0} \frac{\Delta x}{\Delta t}$$

Instantaneous speed is defined as the limit of average speed as $\Delta t \to 0$:
$$\text{speed} = \lim_{\Delta t \to 0} \frac{\text{path length}}{\Delta t}$$

As $\Delta t \to 0$, the path length covered by the particle in the infinitesimally small interval $\Delta t$ approaches the magnitude of the displacement $|\Delta x|$. This is because in an infinitesimally small time interval, the particle cannot reverse its direction of motion — it moves in essentially one direction. Therefore, the path length and the magnitude of displacement become equal in the limit.

Hence:
$$\lim_{\Delta t \to 0} \frac{\text{path length}}{\Delta t} = \lim_{\Delta t \to 0} \frac{|\Delta x|}{\Delta t} = \left|\lim_{\Delta t \to 0} \frac{\Delta x}{\Delta t}\right| = |v|$$

$$\boxed{\therefore\quad \text{Instantaneous speed} = |\text{Instantaneous velocity}|}$$

The distinction between average speed and magnitude of average velocity arises because over a finite time interval the particle may reverse direction, making path length > |displacement|. This possibility vanishes as $\Delta t \to 0$.

Note: The graphs (a) to (d) in Fig. 2.10 are described based on standard NCERT content. The graphs are:
- (a) Position–time graph showing a curve that doubles back (multi-valued x for a given t)
- (b) Speed–time graph showing negative speed
- (c) Position–time graph showing a vertical portion (infinite velocity)
- (d) Total path length–time graph showing a decreasing portion

(a) Cannot represent one-dimensional motion.

*Reason:* The graph shows that for a single value of time $t$, the particle has two different positions. This is physically impossible — a particle cannot be at two places at the same time. (The x–t graph is multi-valued in x for a given t.)

(b) Cannot represent one-dimensional motion.

*Reason:* The graph shows negative values of speed. Speed is the magnitude of velocity and is always non-negative ($\text{speed} \geq 0$). A negative speed has no physical meaning.

(c) Cannot represent one-dimensional motion.

*Reason:* The graph shows a vertical portion, meaning the position changes instantaneously (in zero time). This would imply infinite velocity, which is physically impossible.

(d) Cannot represent one-dimensional motion.

*Reason:* The graph shows total path length decreasing with time. Total path length is a cumulative quantity — it can only increase or remain constant with time. It can never decrease, as a particle cannot 'un-travel' a path.

Answer: No, it is NOT correct to say so.

Reason:

The x–t graph describes the motion of a particle along a straight line (one-dimensional motion). The x-axis represents the position of the particle along a straight line, and the t-axis represents time. The shape of the curve on the x–t graph does NOT represent the actual path (trajectory) of the particle in space.

- A straight line on the x–t graph means uniform velocity (constant speed), not that the particle moves in a straight line in space (it always moves in a straight line in 1D motion).
- A parabolic curve on the x–t graph means uniformly accelerated motion, not that the particle moves on a parabolic path in space.

Suitable physical context for this graph:

The graph shows:
- For t < 0: a straight line through the origin, indicating the particle was at $x = 0$ for all t < 0 (at rest at the origin), OR it could represent uniform motion.
- For t > 0: a parabolic curve, indicating uniformly accelerated motion starting from rest at $t = 0$.

A suitable context: A ball is thrown horizontally from a cliff at $t = 0$. Before $t = 0$, the ball was at rest (or moving uniformly). After $t = 0$, it falls under gravity with uniform acceleration, so its vertical position varies as $x = \frac{1}{2}gt^2$ — a parabola on the x–t graph. The particle always moves along a straight line (vertical direction) in 1D.

Given:
- Speed of police van: $v_p = 30\,\text{km h}^{-1} = 30 \times \frac{5}{18} = \frac{25}{3}\,\text{m s}^{-1} \approx 8.33\,\text{m s}^{-1}$
- Speed of thief's car: $v_t = 192\,\text{km h}^{-1} = 192 \times \frac{5}{18} = \frac{160}{3}\,\text{m s}^{-1} \approx 53.33\,\text{m s}^{-1}$
- Muzzle speed of bullet (relative to the gun/van): $v_b = 150\,\text{m s}^{-1}$

Step 1: Speed of bullet relative to ground

The bullet is fired in the direction of motion of the police van. The speed of the bullet relative to the ground is:
$$v_{\text{bullet, ground}} = v_p + v_b = 8.33 + 150 = 158.33\,\text{m s}^{-1}$$

Step 2: Speed of bullet relative to thief's car

The speed relevant for damaging the thief's car is the speed of the bullet relative to the thief's car:
$$v_{\text{rel}} = v_{\text{bullet, ground}} - v_t = 158.33 - 53.33 = 105\,\text{m s}^{-1}$$

Let us verify with exact fractions:
$$v_{\text{rel}} = \left(150 + \frac{25}{3}\right) - \frac{160}{3} = 150 + \frac{25 - 160}{3} = 150 - \frac{135}{3} = 150 - 45 = 105\,\text{m s}^{-1}$$

$$\boxed{\text{The bullet hits the thief's car with a speed of }105\,\text{m s}^{-1}.}$$

Note: The graphs in Fig. 2.12 are described based on standard NCERT content for this question.

(a) x–t graph (shows position increasing, then decreasing, then increasing again — like a ball bouncing or a person going and returning):

Suitable situation: A ball thrown upward from the ground, reaching a maximum height, and then falling back. The position first increases (ball going up), reaches a maximum, then decreases (ball coming down). Alternatively, a person walking away from home, stopping, and returning.

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(b) v–t graph (shows velocity starting at some positive value, decreasing to zero, becoming negative, then returning to zero — like a bouncing ball's velocity):

Suitable situation: A ball thrown upward with some initial velocity. The velocity decreases uniformly (due to gravity) from a positive value to zero at the highest point, then becomes negative (downward motion) and increases in magnitude until it hits the ground. If the ball bounces elastically, the velocity reverses sign at each bounce.

Alternatively: A car moving forward, braking to a stop, then reversing and braking to a stop again.

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(c) a–t graph (shows acceleration as a constant positive value, then suddenly zero, then constant negative — or a step function):

Suitable situation: A ball in free fall (constant acceleration $g$ downward) hits the floor and bounces. During free fall, $a = -g$ (constant). At the moment of collision with the floor, the acceleration is large and positive (floor exerts a large upward force). After bouncing, $a = -g$ again. This gives a spike in the a–t graph at the moment of collision.

Alternatively: A car accelerating uniformly, then moving at constant speed (zero acceleration), then decelerating uniformly.

Concept: In simple harmonic motion (SHM), the acceleration is always directed towards the mean position (origin), i.e., $a \propto -x$. The velocity is the slope of the x–t graph.

From the x–t graph of SHM (sinusoidal curve), we read the approximate positions:

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At $t = 0.3\,\text{s}$:
- From the graph: the particle is at a negative position (x < 0), i.e., on the negative side of the origin.
- The curve is moving towards more negative values (slope is negative), so velocity v < 0 (negative).
- Since $a = -\omega^2 x$ and x < 0, we get a > 0 → acceleration is positive.

Signs: x < 0, v < 0, a > 0

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At $t = 1.2\,\text{s}$:
- From the graph: the particle is at a positive position (x > 0), near or at maximum positive displacement.
- The curve has a small positive slope (moving back towards zero from positive maximum), so velocity v < 0 (the particle is returning towards origin, slope is negative near maximum).

Actually at $t = 1.2\,\text{s}$, the particle is at positive $x$ and moving back (slope negative): v < 0.
- Since x > 0, a = -\omega^2 x < 0 → acceleration is negative.

Signs: x > 0, v < 0, a < 0

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At $t = -1.2\,\text{s}$:
- From the graph (extending to negative time): the particle is at a negative position (x < 0).
- The slope of the curve at $t = -1.2\,\text{s}$ is positive (particle moving towards origin from negative side), so velocity v > 0 (positive).
- Since x < 0, a = -\omega^2 x > 0 → acceleration is positive.

Signs: x < 0, v > 0, a > 0

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Summary Table:

| Time | Position $x$ | Velocity $v$ | Acceleration $a$ |
|---|---|---|---|
| $t = 0.3\,\text{s}$ | Negative | Negative | Positive |
| $t = 1.2\,\text{s}$ | Positive | Negative | Negative |
| $t = -1.2\,\text{s}$ | Negative | Positive | Positive |

Concept:
- Average speed in an interval = total path length / time interval. Since the time intervals are equal, the interval with the greatest path length (greatest change in $|x|$) has the greatest average speed.
- The slope of the x–t graph gives the velocity. The sign of the slope gives the sign of average velocity.
- Average velocity = $\Delta x / \Delta t$ = (change in position) / (time interval).

From Fig. 2.14 (standard NCERT graph), the three intervals are labelled 1, 2, and 3:

- Interval 1: The x–t curve has a moderate positive slope. The displacement is positive but moderate.
- Interval 2: The x–t curve is nearly flat (small slope). The displacement is small.
- Interval 3: The x–t curve has a steep slope (large change in x). The displacement is large.

Greatest average speed: Interval 3
(The magnitude of the slope, and hence the path length covered, is greatest in interval 3.)

Least average speed: Interval 2
(The curve is nearly flat, meaning very little displacement — least path length.)

Signs of average velocity:
- Interval 1: Slope is positive → average velocity is positive.
- Interval 2: Slope is negative (x decreases) → average velocity is negative.
- Interval 3: Slope is positive (x increases steeply) → average velocity is positive.

Summary:
- Greatest average speed: Interval 3
- Least average speed: Interval 2
- Sign of average velocity: Interval 1 → positive; Interval 2 → negative; Interval 3 → positive.

Concept:
- The particle moves along a constant direction, so we take that direction as positive.
- Speed = magnitude of velocity; since direction is constant, speed = velocity (all positive).
- Average acceleration = change in speed / time interval (since direction is constant).
- The magnitude of average acceleration in an interval = |slope of speed-time graph| in that interval.
- Average speed in an interval = area under speed-time graph / time interval.

From Fig. 2.15 (standard NCERT graph), the speed-time graph shows:
- Interval 1: Speed increases (positive slope — acceleration)
- Interval 2: Speed is roughly constant or changes slowly
- Interval 3: Speed decreases steeply (large negative slope — deceleration)

Greatest average acceleration (magnitude): Interval 3
(The speed-time graph has the steepest slope in interval 3, meaning the greatest rate of change of speed.)

Greatest average speed: Interval 2
(The speed values are highest during interval 2 — the area under the graph is greatest.)

Signs of $v$ and $a$ in each interval:

Since the particle always moves in the positive direction:
- $v$ is positive in all three intervals (speed is always positive and direction is positive).

For acceleration:
- Interval 1: Speed is increasing → a > 0 (positive)
- Interval 2: Speed is approximately constant → $a \approx 0$ (or slightly positive/negative depending on graph)
- Interval 3: Speed is decreasing → a < 0 (negative)

Accelerations at points A, B, C, D:

The acceleration at any point = slope of the speed-time graph at that point.

- Point A: The graph has a finite positive slope → a > 0 (positive, finite)
- Point B: The graph appears to have zero slope (local maximum of speed, curve is flat) → $a = 0$
- Point C: The graph has a finite positive slope (speed increasing) → a > 0
- Point D: The graph has zero slope (local maximum or the curve is flat at D) → $a = 0$

*Note: At points B and D, the speed-time curve has zero slope (tangent is horizontal), so the instantaneous acceleration is zero at these points.*

Summary:
- Greatest magnitude of average acceleration: Interval 3
- Greatest average speed: Interval 2
- v > 0 in all intervals; a > 0 in interval 1, $a \approx 0$ in interval 2, a < 0 in interval 3
- Accelerations: a_A > 0, $a_B = 0$, a_C > 0, $a_D = 0$