MOTION IN A STRAIGHT LINE – Guided Solutions

Q1 In which of the following examples of motion, can the body be considered approximately a point object:
(a) a railway carriage moving without jerks between two stations.
(b) a monkey sitting on top of a man cycling smoothly on a circular track.
(c) a spinning cricket ball that turns sharply on hitting the ground.
(d) a tumbling beaker that has slipped off the edge of a table.

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Concept Theory

In mechanics, a body can be treated as a point object (particle) when its size is negligible compared with the distance travelled or the scale of motion.

In such cases, the internal structure, rotation, and shape of the body do not affect the analysis of motion.

Mathematically, if

\( \text{Size of object} \ll \text{Distance of motion} \)

then the body may be approximated as a point object.

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Solution Strategy (Thinking Map)

• Step 1 → Compare the size of the object with the distance travelled.
• Step 2 → If the size is negligible → treat as a point object.
• Step 3 → If the motion depends on rotation, tumbling, or size → cannot treat as a point object.

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Concept Illustration

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Solution

We examine each situation by comparing the object's size with the scale of motion.

(a) Railway carriage moving between two stations

The distance between stations is typically several kilometres, while the carriage length is only a few metres.

Since the carriage size is extremely small compared with the distance travelled, the carriage can be treated as a point object.

✔ Approximation valid

(b) Monkey sitting on a cyclist moving on a circular track

The radius of the circular track is much larger than the size of the monkey and cyclist.

Hence, their dimensions are negligible compared to the path of motion.

✔ Can be treated as a point object

(c) Spinning cricket ball turning sharply after hitting the ground

Here the motion depends on rotation, spin, and surface interaction.

The size and rotation of the ball significantly influence the motion.

✘ Cannot be treated as a point object

(d) Tumbling beaker falling from a table

The motion involves tumbling and rotation.

The orientation and size of the beaker affect how it falls and hits the ground.

✘ Cannot be approximated as a point object

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Final Answer

Bodies can be treated as point objects in cases:

(a) and (b)

Q3 A woman starts from her home at 9.00 am, walks with a speed of \(5\ km\ h^{-1}\) on a straight road up to her office \(2.5\ km\) away, stays at the office up to 5.00 pm, and returns home by an auto with a speed of \(25\ km\ h^{-1}\). Choose suitable scales and plot the x–t graph of her motion.

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Concept Theory

A position–time (x–t) graph represents how the position of an object changes with time.

• The slope of the graph represents velocity.
• A straight line with positive slope indicates uniform motion away from the origin.
• A horizontal line represents an object at rest.
• A straight line with negative slope represents motion towards the origin.

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Solution Strategy

• Step 1 → Calculate the time taken to reach the office.
• Step 2 → Mark the time interval when she remains at rest.
• Step 3 → Calculate the return journey time.
• Step 4 → Plot the corresponding segments on the x–t graph.

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Solution

1. Journey from home to office

Distance between home and office

\(x = 2.5\ km\)

Walking speed

\(v = 5\ km\ h^{-1}\)

Time taken

\[ t = \frac{x}{v} = \frac{2.5}{5} = 0.5\ h \]

\(0.5\ h = 30\ minutes\)

Therefore, she reaches the office at

9:30 am

2. Stay at the office

She remains at the office from

\(9:30\ am\) to \(5:00\ pm\)

During this period her position remains constant at

\(x = 2.5\ km\)

Hence the graph is a horizontal line.

3. Return journey

Distance from office to home

\(2.5\ km\)

Speed of auto

\(25\ km\ h^{-1}\)

Time taken

\[ t = \frac{2.5}{25} = 0.1\ h \]

\(0.1\ h = 6\ minutes\)

Thus she reaches home at approximately

5:06 pm

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Key Points for the Graph

• (9:00 am , 0 km)
• (9:30 am , 2.5 km)
• (5:00 pm , 2.5 km)
• (5:06 pm , 0 km)

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x–t Graph

Q4 A drunkard walking in a narrow lane takes 5 steps forward and 3 steps backward, followed again by 5 steps forward and 3 steps backward, and so on. Each step is 1 m long and requires 1 s. Plot the x–t graph of his motion. Determine graphically and otherwise how long the drunkard takes to fall in a pit 13 m away from the start.

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Concept Theory

In this motion the displacement does not increase uniformly because the drunkard alternates between forward and backward steps.

Therefore the x–t graph consists of zig-zag straight line segments:

• Positive slope → forward motion
• Negative slope → backward motion
• Each step corresponds to a change of 1 m in 1 s

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Solution Strategy

• Step 1 → Determine the net displacement per cycle.
• Step 2 → Find how many full cycles occur before the final forward run.
• Step 3 → Calculate the remaining forward steps to reach the pit.

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Q5 A car moving along a straight highway with speed of \(126 \,\text{km h}^{-1}\) is brought to a stop within a distance of \(200 \,\text{m}\). What is the retardation of the car (assumed uniform), and how long does it take for the car to stop?

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Concept Theory

When a body moving with an initial velocity slows down uniformly, the motion is described by the equations of uniformly accelerated motion.

• \(v = u + at\)
• \(v^2 - u^2 = 2as\)

Since the car is being stopped, the acceleration is opposite to the direction of motion, hence it represents retardation (negative acceleration).

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Solution Strategy

• Step 1 → Convert the speed from km h\(^{-1}\) to m s\(^{-1}\).
• Step 2 → Use \(v^2-u^2=2as\) to find acceleration.
• Step 3 → Use \(v=u+at\) to determine the stopping time.

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Given

Initial speed \[ u = 126\ \text{km h}^{-1} \] Stopping distance \[ s = 200\ \text{m} \] Final velocity \[ v = 0 \]

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1. Convert speed into SI units

\[ 126 \,\text{km h}^{-1} = \frac{126 \times 1000}{3600} \,\text{m s}^{-1} \] \[ = \frac{126 \times 5}{18} \] \[ = 35 \,\text{m s}^{-1} \] Thus \[ u = 35 \,\text{m s}^{-1} \]

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2. Retardation of the car

Using the equation \[ v^2 - u^2 = 2as \] Substituting values \[ 0^2 - (35)^2 = 2 \times a \times 200 \] \[ -1225 = 400a \] \[ a = \frac{-1225}{400} \] \[ a \approx -3.06 \,\text{m s}^{-2} \] The negative sign shows the acceleration is opposite to the motion.

Retardation = \(3.06 \,\text{m s}^{-2}\)

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3. Time taken to stop

Using \[ v = u + at \] Substituting values \[ 0 = 35 - 3.06t \] \[ 3.06t = 35 \] \[ t = \frac{35}{3.06} \] \[ t \approx 11.44\ \text{s} \]

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Final Answer

Retardation of the car:

\[ 3.06 \,\text{m s}^{-2} \]

Time taken to stop:

\[ 11.44 \,\text{s} \]

Q6

A player throws a ball upwards with an initial speed of \(29.4\,\text{m s}^{-1}\).
(a) What is the direction of acceleration during the upward motion of the ball?
(b) What are the velocity and acceleration of the ball at the highest point of its motion?
(c) Choose \(x=0\) and \(t=0\) to be the location and time of the ball at its highest point, vertically downward direction to be the positive direction of the x-axis, and give the signs of position, velocity and acceleration during the upward and downward motion.
(d) To what height does the ball rise and after how long does the ball return to the player's hands? (Take \(g=9.8\,\text{m s}^{-2}\) and neglect air resistance.)

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Concept Theory

When a body is projected vertically upward, the only force acting on it (neglecting air resistance) is gravity. Therefore the acceleration remains constant and directed downward throughout the motion.

• Acceleration due to gravity is constant.
• Velocity becomes zero momentarily at the highest point.
• Time of ascent equals time of descent in symmetric vertical motion.

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Solution Strategy

• Use concepts of gravitational acceleration.
• Apply kinematic equations of uniformly accelerated motion.
• Use symmetry of vertical motion to determine total time.

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Solution

(a) Direction of acceleration

During the entire motion, the only acceleration acting on the ball is the acceleration due to gravity.

This acceleration always acts vertically downward towards the centre of the Earth, irrespective of whether the ball is moving upward or downward.

Direction of acceleration: downward.

(b) Velocity and acceleration at the highest point

At the highest point, the ball momentarily comes to rest.

\[ v = 0 \]

However, gravity continues to act on the ball, so acceleration remains

\[ a = 9.8\,\text{m s}^{-2} \]

directed vertically downward.

(c) Sign convention for position, velocity and acceleration

The highest point is chosen as the origin (\(x=0\)), and the downward direction is taken as positive.

Motion	Position \(x\)	Velocity \(v\)	Acceleration \(a\)
Upward motion	Negative	Negative	Positive
Downward motion	Positive	Positive	Positive

Acceleration remains positive because it is always directed downward.

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(d) Maximum height reached

Initial velocity

\[ u = 29.4\,\text{m s}^{-1} \]

Acceleration

\[ a = -9.8\,\text{m s}^{-2} \]

At the highest point

\[ v = 0 \] Using \[ v^2 - u^2 = 2ah \] \[ 0 - (29.4)^2 = 2(-9.8)h \] \[ h = \frac{(29.4)^2}{2\times9.8} \] \[ h = 44.1\,\text{m} \]

Maximum height = \(44.1\,\text{m}\)

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Time to reach the highest point Using \[ v = u - gt \] \[ 0 = 29.4 - 9.8t \] \[ t = \frac{29.4}{9.8} \] \[ t = 3\,\text{s} \]

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Total time of flight

For symmetric vertical motion

\[ T = 2t \] \[ T = 2\times3 \] \[ T = 6\,\text{s} \]

The ball returns to the player's hands after \(6\,\text{s}\).

Q7 Read each statement below carefully and state with reasons and examples if it is true or false. A particle in one-dimensional motion:
(a) with zero speed at an instant may have non-zero acceleration at that instant
(b) with zero speed may have non-zero velocity
(c) with constant speed must have zero acceleration
(d) with positive value of acceleration must be speeding up

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Concept Theory

In one-dimensional motion:

• Speed is the magnitude of velocity.
• Velocity describes both magnitude and direction of motion.
• Acceleration represents the rate of change of velocity.

The sign of velocity and acceleration determines whether the object is speeding up or slowing down.

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Solution Strategy

• Recall definitions of speed, velocity and acceleration.
• Analyse whether each statement is consistent with physical examples.

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Solution

(a) Zero speed but non-zero acceleration

A particle may momentarily have zero speed but still experience acceleration.

Example: A ball thrown vertically upward has zero speed at the highest point, yet gravity continues to act on it.

Acceleration remains \[ 9.8\,\text{m s}^{-2} \] directed downward.

Statement (a) — True

(b) Zero speed but non-zero velocity

Speed is the magnitude of velocity.

If speed is zero, the magnitude of velocity is zero, so the velocity itself must also be zero.

Statement (b) — False

(c) Constant speed implies zero acceleration

In one-dimensional motion, constant speed means velocity remains constant (because the direction does not change).

Since acceleration is the rate of change of velocity, the acceleration must be zero.

Example: A car moving along a straight road at a steady speed.

Statement (c) — True

(d) Positive acceleration implies speeding up

Positive acceleration means velocity is increasing in the positive direction.

However, if the velocity is negative while acceleration is positive, the particle actually slows down.

Example: A ball moving upward while gravity acts downward.

Statement (d) — False

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Final Summary

Statement	Answer
(a)	True
(b)	False
(c)	True
(d)	False

Q8 A ball is dropped from a height of 90 m on a floor. At each collision with the floor, the ball loses one tenth of its speed. Plot the speed–time graph of its motion between \(t = 0\) to \(12\,\text{s}\).

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Concept Theory

When a body falls freely under gravity, its speed increases uniformly with time. During a collision with the ground, the direction of velocity reverses and the speed decreases due to loss of energy.

In a speed–time graph:

• Straight line segments represent uniform acceleration due to gravity.
• Sudden vertical drops represent instantaneous changes in speed during collisions.

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Solution Strategy

• Determine the speed just before the first collision.
• Calculate the time taken to reach the ground.
• Find the reduced speed after the bounce.
• Determine the time for the upward and downward motion after the bounce.
• Plot the corresponding points on the speed–time graph.

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Q9 Explain clearly, with examples, the distinction between:
(a) magnitude of displacement (sometimes called distance) over an interval of time, and the total length of path covered by a particle over the same interval;
(b) magnitude of average velocity over an interval of time, and the average speed over the same interval.
[Average speed of a particle over an interval of time is defined as the total path length divided by the time interval]. Show in both (a) and (b) that the second quantity is either greater than or equal to the first. When is the equality sign true?
(Consider one-dimensional motion only.)

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Concept Theory

In one-dimensional motion, two types of quantities are often compared:

• Quantities based on displacement (straight-line change in position).
• Quantities based on actual path length travelled.

Since the path taken can include forward and backward motion, the total path length is always greater than or equal to the magnitude of displacement.

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Solution

(a) Magnitude of displacement vs total path length

The magnitude of displacement is the straight-line distance between the initial and final positions of a particle.

\[ |\Delta x| = |x_{\text{final}} - x_{\text{initial}}| \]

The total path length is the complete distance travelled along the path, including any back-and-forth motion.

Example: If a particle moves from \(x=0\) m to \(x=10\) m and then returns to \(x=4\) m:

Magnitude of displacement:

\[ |4 - 0| = 4\ \text{m} \]

Total path length:

\[ 10 + 6 = 16\ \text{m} \]

Thus,

\[ \text{path length} \geq \text{magnitude of displacement} \]

Equality occurs only when the motion is entirely in one direction without reversing.

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(b) Magnitude of average velocity vs average speed

The magnitude of average velocity is defined as

\[ |\bar{v}| = \frac{|\Delta x|}{\Delta t} \]

where \( \Delta x \) is the displacement.

The average speed is defined as

\[ \text{average speed} = \frac{\text{total path length}}{\Delta t} \]

Using the previous example, suppose the motion occurs in \(7\,\text{s}\).

Magnitude of average velocity:

\[ |\bar{v}| = \frac{4}{7}\ \text{m s}^{-1} \]

Average speed:

\[ \text{average speed} = \frac{16}{7}\ \text{m s}^{-1} \]

Thus,

\[ \text{average speed} \geq |\bar{v}| \]

Equality occurs only when the particle moves in a single direction without reversing, so that

\[ \text{path length} = |\text{displacement}| \]

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Final Conclusion

• \(\text{Path length} \geq |\text{displacement}|\)
• \(\text{Average speed} \geq |\text{average velocity}|\)

Equality holds only when the particle moves in one direction without changing its direction.

Q10 A man walks on a straight road from his home to a market 2.5 km away with a speed of 5 km h\(^{-1}\). Finding the market closed, he instantly turns and walks back home with a speed of 7.5 km h\(^{-1}\). What is the
(a) magnitude of average velocity, and
(b) average speed of the man over the interval of time (i) 0 to 30 min, (ii) 0 to 50 min, (iii) 0 to 40 min ?

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Concept Theory

• Average velocity = displacement / time
• Average speed = total distance / time

Average velocity depends on the net displacement, whereas average speed depends on the total distance travelled.

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Given

Distance to market \[ 2.5\ \text{km} \] Speed towards market \[ 5\ \text{km h}^{-1} \] Speed while returning \[ 7.5\ \text{km h}^{-1} \]

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1. Time to reach the market

\[ t=\frac{\text{distance}}{\text{speed}} \] \[ t=\frac{2.5}{5}=0.5\ \text{h}=30\ \text{min} \] Thus he reaches the market at **30 min**.

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2. Time to return home

\[ t=\frac{2.5}{7.5} \] \[ t=\frac{1}{3}\ \text{h}=20\ \text{min} \] Thus he reaches home again at \[ 30 + 20 = 50\ \text{min} \]

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Required calculations

(i) Interval: 0–30 min

At 30 min he reaches the market. Displacement \[ 2.5\ \text{km} \] Distance travelled \[ 2.5\ \text{km} \] Time \[ 0.5\ \text{h} \] Average velocity \[ \frac{2.5}{0.5}=5\ \text{km h}^{-1} \] Average speed \[ \frac{2.5}{0.5}=5\ \text{km h}^{-1} \]

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(ii) Interval: 0–50 min

At 50 min he returns home. Displacement \[ 0 \] Total distance travelled \[ 2.5+2.5=5\ \text{km} \] Time \[ \frac{5}{6}\ \text{h} \] Average velocity \[ 0 \] Average speed \[ \frac{5}{5/6}=6\ \text{km h}^{-1} \]

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(iii) Interval: 0–40 min

40 min \(=\frac{2}{3}\) h After reaching the market at 30 min, he walks back for **10 min**. \[ 10\ \text{min}=\frac{1}{6}\ \text{h} \] Distance covered during return \[ 7.5\times\frac{1}{6}=1.25\ \text{km} \] Position relative to home \[ 2.5-1.25=1.25\ \text{km} \] Displacement \[ 1.25\ \text{km} \] Total distance travelled \[ 2.5+1.25=3.75\ \text{km} \] Average velocity \[ \frac{1.25}{2/3}=1.875\ \text{km h}^{-1} \] Average speed \[ \frac{3.75}{2/3}=5.625\ \text{km h}^{-1} \]

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Final Results

Average velocity

• 0–30 min → \(5\ \text{km h}^{-1}\)
• 0–50 min → \(0\)
• 0–40 min → \(1.875\ \text{km h}^{-1}\)

Average speed

• 0–30 min → \(5\ \text{km h}^{-1}\)
• 0–50 min → \(6\ \text{km h}^{-1}\)
• 0–40 min → \(5.625\ \text{km h}^{-1}\)

Q11 In Exercises 2.9 and 2.10, we have carefully distinguished between average speed and magnitude of average velocity. No such distinction is necessary when we consider instantaneous speed and magnitude of velocity. The instantaneous speed is always equal to the magnitude of instantaneous velocity. Why?

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Concept Theory

Average quantities are calculated over a finite time interval, while instantaneous quantities are defined over an infinitesimally small time interval.

Over finite intervals, a particle may change direction or move along a curved path, so the distance travelled can be greater than the magnitude of displacement. Hence,

\[ \text{average speed} \ge |\text{average velocity}| \]

However, instantaneous quantities are defined in the limit as the time interval approaches zero.

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Solution

Instantaneous speed is defined as

\[ v_{\text{speed}} = \lim_{\Delta t \to 0} \frac{\Delta s}{\Delta t} \]

where \( \Delta s \) is the small distance travelled in time \( \Delta t \).

Instantaneous velocity is the vector

\[ \vec{v} = \lim_{\Delta t \to 0} \frac{\Delta \vec{x}}{\Delta t} \]

and its magnitude is

\[ |\vec{v}| = \lim_{\Delta t \to 0} \frac{|\Delta \vec{x}|}{\Delta t} \]

When \( \Delta t \to 0 \), the motion during that extremely small interval is effectively along a straight line. Therefore the distance travelled becomes equal to the magnitude of displacement:

\[ \Delta s = |\Delta \vec{x}| \]

Hence,

\[ \text{instantaneous speed} = |\text{instantaneous velocity}| \]

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Example

Consider a car moving along a circular track. Over a full lap the average velocity may be zero because the displacement is zero, while the average speed is non-zero.

But at any particular instant, the car has a velocity directed along the tangent to the circular path, and the speedometer reading gives the magnitude of this velocity. Thus,

instantaneous speed = magnitude of instantaneous velocity.

Q12 Look at the graphs (a) to (d) (Fig. 2.10) carefully and state, with reasons, which of these cannot possibly represent one-dimensional motion of a particle.

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Concept Used

• At any instant of time, a particle can have only one position.
• Similarly, it can have only one velocity at a given time.
• Speed (magnitude of velocity) cannot be negative.
• Total path length can only increase or remain constant with time.

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Q13 Figure 2.11 shows the \(x\)-\(t\) plot of one-dimensional motion of a particle. Is it correct to say from the graph that the particle moves in a straight line for \(t<0\) and on a parabolic path for \(t>0\)? If not, suggest a suitable physical context for this graph.

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Q14 A police van moving on a highway with a speed of 30 km h\(^{-1}\) fires a bullet at a thief’s car speeding away in the same direction with a speed of 192 km h\(^{-1}\). If the muzzle speed of the bullet is 150 m s\(^{-1}\), with what speed does the bullet hit the thief’s car? (Note: Obtain the speed relevant for damaging the thief’s car.)

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Concept Used

The relevant speed for impact is the relative speed of the bullet with respect to the thief’s car.

Relative velocity is given by

\[ v_{A/B}=v_A-v_B \]

when both bodies move in the same direction.

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Given

Police van speed \[ v_p = 30\ \text{km h}^{-1} \] Thief’s car speed \[ v_t = 192\ \text{km h}^{-1} \] Muzzle speed of bullet \[ v_m = 150\ \text{m s}^{-1} \]

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Step 1: Convert vehicle speeds into SI units

\[ v_p=\frac{30\times1000}{3600} \] \[ v_p \approx 8.33\ \text{m s}^{-1} \] \[ v_t=\frac{192\times1000}{3600} \] \[ v_t \approx 53.33\ \text{m s}^{-1} \]

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Step 2: Bullet speed relative to the ground

The muzzle speed is measured relative to the police van. Therefore the bullet’s speed relative to the ground is

\[ v_b = v_m + v_p \] \[ v_b = 150 + 8.33 \] \[ v_b = 158.33\ \text{m s}^{-1} \]

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Step 3: Bullet speed relative to the thief’s car

Since both the bullet and the thief’s car move in the same direction, the relative speed is

\[ v_{\text{bullet w.r.t car}} = v_b - v_t \] \[ = 158.33 - 53.33 \] \[ = 105\ \text{m s}^{-1} \]

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Final Answer

The bullet hits the thief’s car with a speed of \(105\ \text{m s}^{-1}\).

Q15 Suggest a suitable physical situation for each of the following graphs (Fig. 2.12).

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Solution

A possible physical situation corresponding to each graph is described below.

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(a) x–t graph

In this graph the slope of the \(x\)-\(t\) curve represents velocity.

• Initially the graph has a constant positive slope → motion with constant velocity in the positive \(x\)-direction.
• The slope gradually becomes zero → the particle momentarily comes to rest.
• The slope then becomes negative → the particle moves in the opposite direction with constant velocity.

Physical example: A ball rolling on a smooth floor towards a wall, stopping momentarily on collision, and then rebounding back with a smaller speed.

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(b) v–t graph

The velocity changes sign at regular intervals while its magnitude gradually decreases.

This indicates repeated reversals of direction with decreasing speed.

Physical example: A ball dropped on a hard floor. After each bounce it rebounds upward with smaller speed, falls again, and the process repeats until it eventually comes to rest.

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(c) a–t graph

The acceleration is zero for most of the time but shows a sharp peak for a very short duration.

This represents a brief impulsive force acting on the particle.

Physical example: A cricket ball moving horizontally that is suddenly struck by a bat. During the very short contact time the ball experiences a large acceleration, while before and after the impact the acceleration is nearly zero.

Q16 Figure 2.13 gives the \(x\)-\(t\) plot of a particle executing one-dimensional simple harmonic motion. (You will learn about this motion in more detail in Chapter 13). Give the signs of position, velocity and acceleration of the particle at \(t = 0.3\,\text{s},\; 1.2\,\text{s},\; -1.2\,\text{s}\).

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