Motion in a Straight Line — NCERT Solutions

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

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

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

Summary Table:

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.

(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.

(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:

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

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

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

Summary Table:

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: $a_A > 0$, $a_B = 0$, $a_C > 0$, $a_D = 0$