NCERT Solutions for Class 11 Physics Chapter 3 Motion in a Straight Line

The Motion in a Straight Line or Rectilinear Motion refers to the motion of an object along a straight line. In this type of motion, the position of the object changes linearly with time along a specific path.

Examples:

• • A car moving along a straight road.
  • A stone dropped from a certain height (neglecting air resistance).
  • A child running on a straight track.

Position (x)

The position of an object is its location in space relative to a chosen reference point or origin. It is a vector quantity, meaning it has both magnitude and direction.

Position Time Graph of Stationary Object

Here the displacement does not change with the change in time.

Position Time Graph of an Object in Uniform Motion

The change in displacement is uniform, equal displacement in equal span of time.

Displacement (\(\Delta x\))

The change in position of the object.

\(\Delta x = x_f – x_i\)

• \(x_f\): Final position
• \(x_i\): Initial position

Distance

Distance is the total length of the path traveled by an object, regardless of its direction. Unlike displacement, distance is a scalar quantity and only has magnitude.

Velocity (v)

The rate of change of displacement with time.

Average velocity: \( \bar{v} = \frac{\Delta x}{\Delta t} \)

• \(\Delta t\): Time interval

Instantaneous velocity: The velocity at a specific instant of time, \( v = \lim_{\Delta t \to 0} \frac{\Delta x}{\Delta t} = \frac{dx}{dt} \)

Acceleration (a)

The rate of change of velocity with time.

Average acceleration: \( \bar{a} = \frac{\Delta v}{\Delta t} \)

• \(\Delta v\): Change in velocity

Instantaneous acceleration: The acceleration at a specific instant of time, \( a = \lim_{\Delta t \to 0} \frac{\Delta v}{\Delta t} = \frac{dv}{dt} \)

Equations of Motion for Uniform Acceleration

For rectilinear motion with uniform (constant) acceleration, the following kinematic equations are used:

First Equation: \( v = u + at \)

• \(u\): Initial velocity
• \(v\): Final velocity
• \(a\): Acceleration
• \(t\): Time

Derivation: Acceleration \((a) = (v-u)/ (t)\)

or, \(v-u = at\)

or, \(v = u + at \)

Second Equation:  \( s = ut + \frac{1}{2}at^2 \)

• \( u \): Initial velocity
• \( v \): Final velocity
• \( a \): Constant acceleration
• \( t \): Time
• \( s \): Displacement

1. Definition of Velocity

Velocity (\( v \)) is defined as the rate of change of displacement (\( s \)) with respect to time (\( t \)).

\[
v = \frac{ds}{dt}
\]

2. Using the First Equation of Motion

From the first equation of motion, we know that:

\[
v = u + at
\]

3. Substituting the First Equation of Motion into the Velocity Definition

Substitute \( v = u + at \) into \( v = \frac{ds}{dt} \):

\[
\frac{ds}{dt} = u + at
\]

4. Separating Variables

We can separate the variables to integrate both sides with respect to their respective variables.

\[
ds = (u + at) \, dt
\]

5. Integrating Both Sides

Integrate both sides with respect to their respective variables. For the left side, integrate with respect to \( s \), and for the right side, integrate with respect to \( t \).

\[
\int ds = \int (u + at) \, dt
\]

6. Evaluating the Integrals

• The left side integrates to \( s \) (displacement).
• The right side integrates to \( ut + \frac{1}{2}at^2 \).

\[
s = ut + \frac{1}{2}at^2
\]

7. Final Equation

Thus, the second equation of motion relating displacement, initial velocity, acceleration, and time is:

\[
s = ut + \frac{1}{2}at^2
\]

3. \( v^2 = u^2 + 2as \)

4. \( s = \frac{u+v}{2} \cdot t \)

Third Equation: \( v^2 = u^2 + 2as \)

Variables

• \( u \): Initial velocity
• \( v \): Final velocity
• \( a \): Constant acceleration
• \( s \): Displacement

1. Using the First Equation of Motion

From the first equation of motion, we know that:

\[
v = u + at
\]

2. Expressing Time (\( t \))

We can solve the above equation for time \( t \):

\[
t = \frac{v – u}{a}
\]

3. Using the Definition of Average Velocity

Average velocity (\( \bar{v} \)) can be expressed as:

\[
\bar{v} = \frac{u + v}{2}
\]

4. Using the Definition of Displacement

Displacement (\( s \)) is given by the product of average velocity and time:

\[
s = \bar{v} \cdot t
\]

Substitute the expressions for \( \bar{v} \) and \( t \):

\[
s = \left( \frac{u + v}{2} \right) \left( \frac{v – u}{a} \right)
\]

5. Simplifying the Expression

Simplify the right-hand side of the equation:

\[
s = \frac{(u + v)(v – u)}{2a}
\]

6. Expanding the Numerator

Expand the numerator using the difference of squares:

\[
s = \frac{v^2 – u^2}{2a}
\]

7. Solving for \( v^2 \)

Multiply both sides by \( 2a \) to isolate \( v^2 \) on one side of the equation:

\[
2as = v^2 – u^2
\]

Rearrange to solve for \( v^2 \):

\[
v^2 = u^2 + 2as
\]

Free Fall

Free fall refers to the motion of an object where gravity is the only force acting upon it. This means that air resistance is negligible, and the object accelerates downward due to gravity.

The acceleration due to gravity (\( g \)) is a constant near the Earth’s surface and has a value of approximately \( 9.8 \, \text{m/s}^2 \). This acceleration acts downward toward the center of the Earth.

Equations of Motion for Free Fall

The motion of an object in free fall can be described by the equations of motion, with the acceleration \( a \) replaced by \( g \).

\[
v = u + gt
\]

\[
s = ut + \frac{1}{2}gt^2
\]

\[
v^2 = u^2 + 2gs
\]

• \( u \): Initial velocity (usually 0 if the object is dropped)
• \( v \): Final velocity
• \( g \): Acceleration due to gravity (\( 9.8 \, \text{m/s}^2 \))
• \( t \): Time
• \( s \): Displacement

Initial Conditions

• If an object is dropped from rest, its initial velocity (\( u \)) is 0.
• If an object is thrown upwards or downwards, the initial velocity (\( u \)) is the velocity at which it is thrown.

Upward and Downward Motion

• When an object is thrown upwards, it will decelerate at a rate of \( g \) until it reaches its maximum height, where its velocity becomes 0.
• After reaching the maximum height, it will start to fall back down, accelerating at a rate of \( g \).

Time to Reach Maximum Height

The time to reach the maximum height (\( t_{\text{up}} \)) can be found using:

\[
t_{\text{up}} = \frac{u}{g}
\]

Maximum Height

The maximum height (\( h \)) reached by the object can be calculated using:

\[
h = \frac{u^2}{2g}
\]

Total Time of Flight

The total time of flight (\( T \)) for an object thrown upwards and then falling back to the same height can be found using:

\[
T = \frac{2u}{g}
\]

 

Questions

Question 3. 1. 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.

Explanation: A railway carriage is relatively large compared to the distance between two stations. However, for the purposes of analyzing its motion over long distances, its size can be neglected, and it can be treated as a point object.

Conclusion: The railway carriage can be considered approximately a point object.

(b) A monkey sitting on top of a man cycling smoothly on a circular track.

Explanation: The combined system of the man and the monkey can be treated as a single object. If the size of this system is small compared to the radius of the circular track, it can be considered a point object.

Conclusion: The monkey and the man can be considered approximately a point object.

(c) A spinning cricket ball that turns sharply on hitting the ground.

Explanation: A spinning cricket ball has a significant size compared to the distance it travels after turning sharply. Its rotational motion and the effect of spin cannot be neglected.

Conclusion: The spinning cricket ball cannot be considered a point object.

(d) A tumbling beaker that has slipped off the edge of a table.

Explanation: A tumbling beaker exhibits complex motion, including rotation and translation. Its size and shape are significant in determining its motion.

Conclusion: The tumbling beaker cannot be considered a point object.

Question 3. 2. The position-time (x -t) graphs for two children A and B returning from their school O to their homes P and Q respectively are shown in Fig. Choose the correct entries in the brackets below:
(a) (A/B) lives closer to the school than (B/A).
(b) (A/B) starts from the school earlier than (B/A).
(c) (A/B) walks faster than (B/A).
(d) A and B reach home at the (same/different) time.
(e) (A/B) overtakes (B/A) on the road (once/twice).

Solution: 

(a) A lives closer to the school as the distance covered by A is OP which is lesser than the distance travelled by B which is OQ.

(b) A starts at the origin whereas B starts after a definite time gap.

(c) As the slope of B is steeper than that of A, B walks faster.

(d) A and B reached the home at the same time.

(e) B overtakes A only once, at the X position in the graph.

Question 3. 3. A woman starts from her home at 9.00 am, walks with a speed of 5 Km/h 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 2.5 km/h. Choose suitable scales and plot the x-t graph of her motion.

Solution:

Step 1: Calculate Times

• Walking time: Distance = 2.5 km, Speed = 5 km/h, Time = Distance/Speed = 0.5 hours = 30 minutes.
• Total time at office: 9:30 AM to 5:00 PM
• Return journey time: Distance = 2.5 km, Speed = 25 km/h, Time = Distance/Speed = 0.1 hour = 6 Minutes

Step 2: Plot the Graph

1. From 9:00 AM to 9:30 AM: The woman walks from home (0 km) to office (2.5 km) at a constant speed. This is represented by a straight line with a positive slope.
2. From 9:30 AM to 5:00 PM: The woman stays at the office (2.5 km). This is represented by a horizontal line at y = 2.5 km.
3. From 5:00 PM to 5:06 PM: The woman returns home (0 km) at a constant speed. This is represented by a straight line with a negative slope.

Question 3.4. 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.

Solution: 

1. First Cycle (0 to 8 seconds):
   • 5 seconds forward: 5 meters
   • 3 seconds backward: \(5 – 3 = 2\) meters
2. Second Cycle (9 to 16 seconds):
   • 5 seconds forward: \(2 + 5 = 7\) meters
   • 3 seconds backward: \(7 – 3 = 4\) meters
3. Third Cycle (17 to 24 seconds):
   • 5 seconds forward: \(4 + 5 = 9\) meters
   • 3 seconds backward: \(9 – 3 = 6\) meters
4. Fourth Cycle (25 to 32 seconds):
   • 5 seconds forward: \(6 + 5 = 11\) meters
   • 3 seconds backward: \(11 – 3 = 8\) meters
5. Fifth Cycle (33 to 40 seconds):
   • 5 seconds forward: \(8 + 5 = 13\) meters

Thus, after 32 seconds, the drunken will be 8 meters away from the pit. As he takes five more steps forward, he will fall down. So, in total it will take 32 seconds and 5 seconds.  In the next second (37th second), he takes one more step forward and reaches 13 meters, falling into the pit.

Question 3.5. A jet airplane travelling at the speed of 500 km/h ejects its products of combustion at the speed of 1500 km/h relative to the jet plane. What is the speed of the latter with respect to an observer on the ground?

Question 3. 6. A car moving along a straight highway with speed of 126 km h-1 is brought to a stop within a distance of 200 m. What is the retardation of the car (assumed uniform), and how long does it take for the car to stop?

Solution: We’re asked to find the deceleration (retardation) of a car and the time it takes to stop, given its initial speed and stopping distance.

Question 3. 7. Two trains A and B of length 400 m each are moving on two parallel tracks with a uniform speed of 71 km h-1 in the same direction, with A ahead of B. The driver of B decides to overtake A and accelerates by 1 ms-1. If after 50 s, the guard of B just brushes past the driver of A, what was the original distance between them?

Solution: 

Initial speed of both trains = 71 km/h = (71 * 1000 m) / (3600 s) = 20 m/s

• Since both trains are moving in the same direction with the same speed, their initial relative velocity is 0 m/s.
• Relative acceleration (a): This is the acceleration of train B with respect to train A, which is 1 m/s².
• Time (t): 50 seconds

Let d be the distance between the guard of Train A and the Driver of Train B.

So total distance travelled by the driver of Train B will be d + 400 m + 400 m = d+800 m

Using the equation of motion:

• d + 800 = ut + (1/2)at²
• d + 800 = 0 * 50 + (1/2) * 1 * 50²
• d = 450 m

Hence, initial distance between the guard of train A and the driver of train B will be 450 + 400 + 400 = 1250 Meter.

Question 3.8. On a two-lane road, car A is travelling with a speed of 36 km/h. Two cars B and C approach car A in opposite directions with a speed of 54 km/h each. At a certain instant, when the distance AB is equal to AC, both being 1 km, B decides to overtake A before C does. What minimum acceleration of car B is required to avoid an accident?

Solution: We have three cars, A, B, and C on a two-lane road. Car A is moving in one direction, while B and C are moving in the opposite direction. Initially, cars A and B are 1 km apart, as are cars A and C. Car B needs to overtake car A before car C does. We need to find the minimum acceleration for car B to achieve this.

Question 3.13. 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? [For simplicity, consider one dimensional motion only]

Ans:  Distinction between Displacement and Distance, Velocity and Speed

Solution:

Given Data:

• Speed of the child with respect to the belt (\(v_{\text{child, belt}}\)) = 9 km/h
• Speed of the belt (\(v_{\text{belt}}\)) = 4 km/h
• Distance between the father and mother on the belt = 50 m (0.05 km)

(a) Speed of the child running in the direction of motion of the belt:

When the child runs in the direction of the belt’s motion, the speeds add up:

\[
v_{\text{child, direction}} = v_{\text{child, belt}} + v_{\text{belt}} = 9 \text{ km/h} + 4 \text{ km/h} = 13 \text{ km/h}
\]

(b) Speed of the child running opposite to the direction of motion of the belt:

When the child runs opposite to the belt’s motion, the speeds subtract:

\[
v_{\text{child, opposite}} = v_{\text{child, belt}} – v_{\text{belt}} = 9 \text{ km/h} – 4 \text{ km/h} = 5 \text{ km/h}
\]

(c) Time taken by the child in (a) and (b):

To find the time taken, we need to convert the distance to the same unit as the speed (i.e., kilometers).

\[
\text{Distance} = 50 \text{ m} = 0.05 \text{ km}
\]

Time taken when running in the direction of the belt

\[
t_{\text{direction}} = \frac{\text{Distance}}{v_{\text{child, direction}}} = \frac{0.05 \text{ km}}{9 \text{ km/h}} = \frac{0.05}{9} \text{ hours} \approx 0.0055 \text{ hours} \approx 20 \text{ seconds}
\]

Time taken when running opposite to the direction of the belt

\[
t_{\text{direction}} = \frac{\text{Distance}}{v_{\text{child, direction}}} = \frac{0.05 \text{ km}}{9 \text{ km/h}} = \frac{0.05}{9} \text{ hours} \approx 0.0055 \text{ hours} \approx 20 \text{ seconds}
\]

Which of the answers alter if motion is viewed by one of the parents?

If the motion is viewed by one of the parents standing on the belt, they would see the child’s speed as simply 9 km/h. both (a) and (b) will alter but (c) will remain unchanged.

Question 3.26. Two stones are thrown up simultaneously from the edge of a cliff 200 m high with initial speeds of 15 m/s and 30 m/s. Verify that the graph shown in Fig. correctly represents the time variation of the relative position of the second stone with respect to the first. Neglect air resistance and assume that the stones do not rebound after hitting the ground. Take g = 10 m/s2. Give the equations for the linear and curved parts of the plot.

Solution: 

Given Data:

• Height of the cliff = 200 m
• Initial speed of stone \( S_1 \) = 15 m/s
• Initial speed of stone \( S_2 \) = 30 m/s
• Acceleration due to gravity \( g \) = 10 m/s²

Equations of Motion

For a stone thrown upwards with an initial velocity \( u \) from a height \( h \), its position \( y \) at time \( t \) is given by:

\[
y = h + ut – \frac{1}{2}gt^2
\]

For Stone \( S_1 \):

\[
y_1 = 200 + 15t – 5t^2
\]

For Stone \( S_2 \):

\[
y_2 = 200 + 30t – 5t^2
\]

Relative Position

The relative position of \( S_2 \) with respect to \( S_1 \) is:

\[
\Delta y = y_2 – y_1 = (200 + 30t – 5t^2) – (200 + 15t – 5t^2)
\]
\[
\Delta y = 30t – 15t = 15t
\]

So, the relative position of \( S_2 \) with respect to \( S_1 \) is a linear function given by:

\[
\Delta y = 15t
\]

Time to Reach the Ground

To find the time at which each stone hits the ground, we set \( y = 0 \).

For Stone \( S_1 \):

\[
0 = 200 + 15t – 5t^2
\]

Solving the quadratic equation:

\[
5t^2 – 15t – 200 = 0
\]
\[
t^2 – 3t – 40 = 0
\]
\[
t = \frac{3 \pm \sqrt{9 + 160}}{2} = \frac{3 \pm 13}{2}
\]
\[
t = 8 \text{ s} \quad (\text{since time cannot be negative})
\]

For Stone \( S_2 \):

\[
0 = 200 + 30t – 5t^2
\]

Solving the quadratic equation:

\[
5t^2 – 30t – 200 = 0
\]
\[
t^2 – 6t – 40 = 0
\]
\[
t = \frac{6 \pm \sqrt{36 + 160}}{2} = \frac{6 \pm 14}{2}
\]
\[
t = 10 \text{ s} \quad (\text{since time cannot be negative})
\]

Hence,

– Stone \( S_1 \) takes 8 seconds to hit the ground.

– Stone \( S_2 \) takes 10 seconds to hit the ground.

Equations for the Linear and Curved Parts of the Plot

Linear Part:

The relative position \( \Delta y = 15t \) is valid until \( t = 8 \) seconds when \( S_1 \) hits the ground.

Curved Part:

After \( S_1 \) hits the ground (at \( t = 8 \) seconds), \( S_1 \) remains at 0 height, and \( S_2 \) continues its motion until it hits the ground (at \( t = 10 \) seconds).

For \( t > 8 \):

– The position of \( S_2 \) can be given by:

\[
y_2 = 200 + 30t – 5t^2
\]

– The relative position after \( t = 8 \) is:

\[
\Delta y = y_2 – 0 = 200 + 30t – 5t^2
\]

So, for \( 8 \leq t \leq 10 \):

\[
\Delta y = 200 + 30t – 5t^2
\]

Summary: 

• Relative position (linear part): \( \Delta y = 15t \) for \( t \leq 8 \) s
• Relative position (curved part): \( \Delta y = 200 + 30t – 5t^2 \) for \( 8 \leq t \leq 10 \) s

Question 3. 27. The speed-time graph of a particle moving along a fixed direction is shown in Fig. Obtain the distance traversed by the particle between
(a) t = 0 s to 10 s. (b) t = 2 s to 6 s.
What is the average speed of the particle over the intervals in (a) and (b)?

Solution: 

(a) From \( t = 0 \) s to \( t = 10 \) s

We need to find the area under the speed-time graph from \( t = 0 \) s to \( t = 10 \) s, which is divided into two triangles:

• First triangle (from \( t = 0 \) s to \( t = 5 \) s):

\[
\text{Area}_1 = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 5 \times 12 = 30 \text{ m}
\]

• Second triangle (from \( t = 5 \) s to \( t = 10 \) s):

\[
\text{Area}_2 = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 5 \times 12 = 30 \text{ m}
\]

Total distance traversed from \( t = 0 \) s to \( t = 10 \) s:

\[
\text{Total distance} = \text{Area}_1 + \text{Area}_2 = 30 + 30 = 60 \text{ m}
\]

(b) From \( t = 2 \) s to \( t = 6 \) s

We need to find the area under the speed-time graph from \( t = 2 \) s to \( t = 6 \) s. This section is divided into two parts:

• Trapezoid (from \( t = 2 \) s to \( t = 5 \) s):

\[
\text{Area}_{\text{trapezoid}} = \frac{1}{2} \times (\text{base}_1 + \text{base}_2) \times \text{height} = \frac{1}{2} \times (4.8 + 12) \times 3 = \frac{1}{2} \times 16.8 \times 3 = 25.2 \text{ m}
\]

• Trapezoid (from \( t = 5 \) s to \( t = 6 \) s):

\[
\text{Area}_{\text{trapezoid}} = \frac{1}{2} \times (\text{base}_1 + \text{base}_2) \times \text{height} = \frac{1}{2} \times (12 + 9.6) \times 1 = \frac{1}{2} \times 21.6 \times 1 = 10.8 \text{ m}
\]

Total distance traversed from \( t = 2 \) s to \( t = 6 \) s:

\[
\text{Total distance} = \text{Area}_{\text{trapezoid1}} + \text{Area}_{\text{trapezoid2}} = 25.2 + 10.8 = 36 \text{ m}
\]

Average Speed

The average speed is the total distance divided by the time interval.

(a) From \( t = 0 \) s to \( t = 10 \) s

\[
\text{Average speed} = \frac{\text{Total distance}}{\text{Time interval}} = \frac{60 \text{ m}}{10 \text{ s}} = 6 \text{ m/s}
\]

(b) From \( t = 2 \) s to \( t = 6 \) s

\[
\text{Average speed} = \frac{\text{Total distance}}{\text{Time interval}} = \frac{36 \text{ m}}{4 \text{ s}} = 9 \text{ m/s}
\]