MOTION IN A STRAIGHT LINE : A foundational concept in physics, linear motion, explores the movement of objects along straight paths and the relationships between displacement, velocity, and acceleration.”
*Key Concepts
Displacement: Displacement is the vector connecting an object’s starting point to its ending point.
Velocity: Velocity combines the scalar quantity of speed with a specified direction
Acceleration: An object accelerates when its velocity changes, either in magnitude or direction.
Equations of Motion: Equations of motion govern the relationship between displacement, velocity, acceleration, and time in uniformly accelerated motion.”
*Types of Motion
Uniform Motion: Motion with a constant velocity.
Uniformly Accelerated Motion: Motion with a constant rate of change of velocity.
*Graphs of Motion
Position-Time Graph: A plot of an object’s position over time.
Velocity-Time Graph: A graphical representation of an object’s speed and direction over time.
Acceleration-Time Graph: A plot that shows how an object’s rate of change of velocity varies with time.
*Important Formulas :
Average velocity:
v =
(s₂ – s₁)
———–
(t₂ – t₁)
Acceleration:
a =
(v₂ – v₁)
————
(t₂ – t₁)
Equations of Motion:
v = u + at
s = ut + (1/2)at²
v² = u² + 2as
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.
(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.
Ans : A body can be considered approximately a point object when its size is negligible compared to the distance it travels.
Based on this criterion, the following examples can be considered as approximately point objects:
(a) The size of a railway carriage is insignificant relative to the distance it covers between stations..
(b) A monkey sitting on top of a man cycling smoothly on a circular track: The size of the monkey is negligible compared to the circumference of the track.
However, the following examples cannot be considered as point objects:
(c) A spinning cricket ball that turns sharply on hitting the ground: The size of the cricket ball is significant compared to the distance it travels during its spin and bounce.
(d) A tumbling beaker that has slipped off the edge of a table: The size of the beaker is significant compared to the distance it falls.
Therefore, the correct options are (a) and (b).
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. 2.9. 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).
Ans : Analyzing the Position-Time Graph
Based on the provided graph (Fig. 2.9), here are the correct answers:
(a) A lives further from the school than B. This is because the graph shows that B reaches home at a shorter distance from the school compared to A.
(b) A departed from the school prior to B. This is evident from the fact that A’s graph begins at an earlier time on the x-axis.
(c)B is a faster walker than A.. The slope of a position-time graph represents velocity. B’s steeper slope indicates faster movement
(d) A and B reach home together. Both graphs end at the same time on the x-axis, indicating that they arrive home simultaneously.
(e) A overtakes B on the road once. This occurs where the two graphs intersect. At this point, A’s position becomes greater than B’s, indicating that A has overtaken B.
3. 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.
Ans : Analyzing the Woman’s Journey
Part 1: Walking to the Office
Distance: 2.5 km
Speed: 5 km/h
Time: Distance / Speed = 2.5 km / 5 km/h = 0.5 hours = 30 minutes
Therefore, she reaches the office at 9:30 AM.
Part 2: Staying at the Office
Time: 5:00 PM – 9:30 AM = 7.5 hours
Part 3: Returning Home by Auto
Distance: 2.5 km
Speed: 25 km/h
Time: Distance / Speed = 2.5 km / 25 km/h = 0.1 hours = 6 minutes
Therefore, she reaches home at 5:06 PM.
In summary:
The woman leaves home at 9:00 AM.
She arrives at the office at 9:30 AM.
She stays at the office until 5:00 PM.
She returns home at 5:06 PM.
Considering the origin (O) as the starting point for both time and distance:
At 9:00 AM (t = 0), the woman is at home (x = 0).
At 9:30 AM (t = 0.5 hours), she has walked 2.5 km to the office (x = 2.5 km).
The line segment OA on the x-t graph represents her journey to work.
She stays at the office from 9:30 AM to 5:00 PM (a total of 7.5 hours). This is shown as a horizontal line segment AB on the graph.
The return trip by auto takes 2.5 km / 25 km/h = 0.1 hours = 6 minutes.
Therefore, at 5:06 PM (t = 7.5 + 0.1 = 7.6 hours), she is back home (x = 0).
This return journey is represented by the line segment BC on the graph.
Note on the graph:
The time axis scale is 1 division = 1 hour.
The positive x-axis scale is 1 division = 0.5 km.
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.
Ans : The man’s consistent forward movement over time can be represented by a linear graph. He travels 5 meters in the first 5 seconds and then retreats 3 meters in the following 3 seconds.
Therefore, in a total of 8 seconds, he has only covered 2 meters. As depicted in the graph, he will fall into the pit after approximately 37 seconds, based on his consistent rate of movement.
The man covers 2 meters in 8 seconds. Therefore, he will cover 8 meters in 32 seconds (4 times the distance).
However, he needs to cover an additional 5 meters to reach the pit. This will take an extra 5 seconds.
So, in total, he will fall into the pit after 32 seconds + 5 seconds = 37 seconds.
5. 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 ?
Ans : To determine the retardation of the car and the time taken to stop, we can use the equations of motion:
v² = u² + 2as
v = u + at
where:
v is the final velocity (0 m/s, as the car comes to a stop)
u is the initial velocity (126 km/h = 35 m/s)
a is the acceleration (retardation in this case)
s is the distance traveled (200 m)
t is the time taken
1. Finding the retardation:
Using the equation v² = u² + 2as, we can rearrange it to solve for a:
a = (v² – u²) / (2s)
Substituting the given values:
a = (0² – (35 m/s)²) / (2 * 200 m)
a ≈ -3.06 m/s²
Hence, The car’s speed decreases by 3.06 meters per second every second.
2. Finding the time taken:
Using the equation v = u + at, we can rearrange it to solve for t:
t = (v – u) / a
Substituting the given values:
t = (0 – 35 m/s) / (-3.06 m/s²)
t ≈ 11.4 seconds
Therefore, it takes the car approximately 11.4 seconds to stop.
6. A player throws a ball upwards with an initial speed of 29.4 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 the x = 0 m and t = 0 s to be the location and time of the ball at its highest point, vertically downward direction to be the positive direction of x-axis, and give the signs of position, velocity and acceleration of the ball during its 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 m s–2 and neglect air resistance).
Ans : Analyzing the Ball’s Motion
(a) Direction of acceleration during upward motion
The direction of acceleration due to gravity is always downward, regardless of the direction of the object’s motion. So, even when the ball is moving upwards, the acceleration due to gravity is acting downwards.
(b)The ball’s upward motion comes to a brief halt at its maximum height before reversing direction. Therefore, its velocity is 0 m/s. However, the acceleration due to gravity continues to act downwards, so the ball’s acceleration remains -9.8 m/s².
(c) Signs of position, velocity, and acceleration
Upward motion:
Position (x): Positive
Velocity (v): Positive
Acceleration (a): Negative
Downward motion:
Position (x): Negative
Velocity (v): Negative
Acceleration (a): Negative
(d) Maximum height and time to return
To find the maximum height and time taken to return, we can use the equations of motion:
v² = u² + 2as
v = u + at
where:
The velocity of the object reaches zero at the highest point of its trajectory.
u is the initial velocity (29.4 m/s)
a is the acceleration (-9.8 m/s²)
s is the displacement (maximum height)
t is the time
Finding the maximum height (s):
0² = (29.4 m/s)² + 2 * (-9.8 m/s²) * s
s ≈ 44.1 m
Finding the time to return:
Since the time taken to reach the highest point is equal to the time taken to fall back down, we can double the time it takes to reach the maximum height.
v = u + at
0 = 29.4 m/s + (-9.8 m/s²) * t
t ≈ 3 seconds
Therefore:
The ball rises to a maximum height of approximately 44.1 meters.
The ball returns to the player’s hands after approximately 6 seconds (double the time it takes to reach the maximum height).
7. 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.
Ans : (a) True.
A ball thrown upward has zero speed at its highest point. However, due to the constant force of gravity pulling it downward, it has a non-zero acceleration.
(b) False.
Speed is the magnitude of velocity. If a particle has non-zero velocity, it means it is moving at a certain speed in a specific direction.
(c) True.
Instantaneous rebound with the same speed implies an infinite change in velocity over zero time, which is physically impossible. This would require an infinite force.
(d) False.
The statement is only true when the chosen position direction aligns with the direction of motion. If the position direction is perpendicular to the motion, the velocity would be zero even if the particle is moving.
8. 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 s.
Ans :Given:
Initial velocity (u) = 0 m/s
Acceleration (a) = 10 m/s²
Distance (s) = 90 m
Time (t) = ?
Final velocity (v) = ?
Calculation:
Finding final velocity (v):
Using the equation of motion:
v² = u² + 2as
Substituting the values:
v² = 0² + 2 × 10 × 90
Solving for v:
v = 30√2 m/s
Finding time taken to reach the ground (t):
Using another equation of motion:
s = ut + 1/2 at²
Substituting the values:
90 = 0 × t + 1/2 × 10 × t²
Solving for t:
t = 3√2 s
Finding rebound velocity:
The ball rebounds with 9/10th of its impact velocity.
Rebound velocity = (9/10) × 30√2 m/s
= √2 m/s
Finding time taken to reach the highest point after rebound:
Using the same equation of motion as in step 2, but with initial velocity as √2 m/s and final velocity as 0 m/s:
0² = (√2)² + 2 × (-10) × t²
Solving for t:
t = 2.7√2 s
Finding total time:
Total time = Time to fall + Time to reach highest point after rebound
Total time = (3√2 + 2.7√2) s
= 5.7√2 s
Graph Interpretation:
OA shown downward motion of the ball.
AB represents the upward motion after the rebound.
The ball hits the ground with a velocity of 30√2 m/s after 3√2 seconds.
It rebounds with a velocity of √2 m/s and reaches the highest point in 2.7√2 seconds.
The total time of motion is 5.7√2 seconds.
Note: The graph shows speed vs. time. Speed is always positive, so the graph will be above the time axis
Analyzing the Ball’s Motion
Initial phase (AB): The ball moves downward for 3√2 seconds. The vertical line AB represents the loss of 1/10th of its speed due to the first collision.
Second phase (BC): The ball rebounds upward and reaches its highest point in 2.7√2 seconds.
Total time for the first bounce: 3√2 + 2.7√2 = 5.7√2 seconds.
Third phase (CD): The ball falls back down, represented by the vertical line CD.
Fourth phase (DE): The ball hits the ground again, losing another 1/10th of its speed (represented by the vertical line DE).
9. 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 : (a) Magnitude of displacement vs. total path length
Magnitude of displacement: The shortest distance between the initial and final positions. It can take on values greater than(+), less than(-), or equal to zero
Total path length: The actual distance traveled, including any changes in direction. It is always positive.
Example:
A car travels 10 km north and then 5 km south.
Displacement: 5 km north (since the car ended up 5 km north of its starting point)
Total path length: 15 km (10 km north + 5 km south)
In general, total path length is always greater than or equal to the magnitude of displacement. The equality holds when the motion is in a straight line without changing direction.
(b) Magnitude of average velocity vs. average speed
Magnitude of average velocity: Displacement divided by time interval. It can take on values greater than(+), less than(-), or equal to zero
Average speed: Total path length divided by time interval. It is always positive.
Example:
A car travels 10 km north in 2 hours, then 5 km south in 1 hour.
Average velocity: (5 km north) / 3 hours ≈ 1.67 km/h north
Average speed: (15 km) / 3 hours = 5 km/h
In general, average speed is always greater than or equal to the magnitude of average velocity. The equality holds when the motion is in a straight line without changing direction
10. 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 ?
[Note: You will appreciate from this exercise why it is better to define average speed as total path length divided by time, and not as magnitude of average velocity. You would not like to tell the tired man on his return home that his average speed was zero !]
Ans : Analyzing the Man’s Journey
Time to reach the market: 2.5 km / 5 km/h = 0.5 hours
Time to return: 2.5 km / 7.5 km/h = 0.333 hours
Average velocities:
(i) 0-30 minutes: 2.5 km / 0.5 h = 5 km/h
(ii) 0-50 minutes: (2.5 km + 2.5 km) / (0.5 h + 0.333 h) = 8 km/h
(iii) 0-4 minutes: (2.5 km – 1.25 km) / (40/60 h) = 1.875 km/h
Calculating Average Speed (0-40 minutes)
During the first 30 minutes, the man covers 2.5 km.
In the next 10 minutes, he covers 1.25 km (half of the remaining distance).
Therefore, the total distance traveled is 2.5 km + 1.25 km = 3.75 km.
The total time is 40 minutes, which is 2/3 hours.
Average speed = Total distance / Total time
= 3.75 km / (2/3) h
= 5.625 km/h
11. 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?
Ans : The reason instantaneous speed is always equal to the magnitude of instantaneous velocity is because at an infinitesimally small instant of time, the direction of an object’s motion can be considered essentially constant.
When we talk about instantaneous quantities, we’re referring to values at a specific moment in time. At such a moment, the direction of an object’s motion is essentially fixed. This means that the displacement of the object in that infinitesimally small time interval is also in the same direction as its velocity.
Therefore, the ratio of displacement to time (which defines instantaneous velocity) will have the same magnitude as the ratio of distance traveled to time (which defines instantaneous speed). In essence, when dealing with infinitesimally small time intervals, the distinction between displacement (a vector) and distance (a scalar) becomes less relevant.
To summarize:
Instantaneous velocity considers both magnitude (speed) and direction.
At an infinitesimally small time interval, the direction is essentially constant.
This leads to the magnitude of instantaneous velocity being equal to instantaneous speed.
12. 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.
Ans :
Ans : None of the provided graphs (a), (b), (c), or (d) accurately depict one-dimensional motion.
Graphs (a) and (b): These graphs clearly represent two-dimensional motion.
Graph (a) shows two positions at the same time, which is physically impossible.
Graph (b) indicates simultaneous motion in opposite directions, also impossible in one dimension.
Graph (c): This graph is incorrect because it shows negative speed. Speed is always a positive quantity.
Graph (d): This graph suggests that the path length is both increasing and decreasing. This is not possible in one dimension, as the path length can only increase or remain constant.
Therefore, none of the given graphs can represent a particle’s motion in one dimension.
13. Figure 2.11shows the x-t plot of onedimensional 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.
Ans : The statement “the particle moves in a straight line for t < 0 and on a parabolic path for t > 0” is incorrect.
Here’s why:
The x-t graph is a parabola for all values of t. This indicates that the acceleration of the particle is constant, as a parabolic path is characteristic of constant acceleration motion.
A straight line on an x-t graph would imply constant velocity. If the particle were moving in a straight line, the slope of the graph would be constant. In this case, the slope is changing, indicating a changing velocity.
A suitable physical context for this graph could be:
A ball thrown vertically upwards:
The initial part of the graph (t < 0) could represent the ball being held at rest.
The parabolic part (t > 0) could represent the ball’s upward motion, influenced by gravity.
In this scenario, the constant acceleration due to gravity causes the parabolic path. The straight line portion before t = 0 simply indicates that the ball was at rest before being thrown.
14. 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 that speed which is relevant for damaging the thief’s car).
Ans : To determine the speed at which the bullet hits the thief’s car, we need to consider the relative velocities of the bullet and the thief’s car.
1. Relative velocity of the bullet with respect to the police van:
Since the bullet is fired from the moving police van, its velocity relative to the ground is the sum of its muzzle speed and the speed of the police van.
Relative velocity of bullet = Muzzle speed + Speed of police van
Relative velocity of bullet = 150 m/s + (30 km/h * 1000 m/km * 1 h/3600 s)
Relative velocity of bullet = 150 m/s + 8.33 m/s ≈ 158.33 m/s
2.The velocity of the bullet as observed from the perspective of the thief’s car
Since the thief’s car is moving in the same direction as the bullet, we need to subtract the speed of the thief’s car from the relative velocity of the bullet with respect to the police van.
Relative velocity of bullet with respect to thief’s car =
Relative velocity of bullet – Speed of thief’s car
Relative velocity of bullet with respect to thief’s car = 158.33 m/s – (192 km/h * 1000 m/km * 1 h/3600 s)
Relative velocity of bullet with respect to thief’s car = 158.33 m/s – 53.33 m/s ≈
105 m/s
Therefore, the bullet hits the thief’s car with a speed of approximately 105 m/s.
15. Suggest a suitable physical situation for each of the following graphs (Fig 2.12):
Ans : (a) Position-Time Graph
Possible Physical Situation: A ball being kicked against a wall. The ball initially moves forward, hits the wall, and rebounds backward.
The initial increase in position represents the ball moving forward.
The sudden decrease in position represents the ball hitting the wall and rebounding.
The final constant position indicates the ball coming to rest.
(b) Velocity-Time Graph
Possible Physical Situation: A car accelerating from rest, then maintaining a constant velocity, and finally decelerating to a stop.
The initial increase in velocity represents the car accelerating.
The constant velocity is shown by the horizontal line.
The final decrease in velocity represents the car decelerating to a stop.
(c) Acceleration-Time Graph
Possible Physical Situation: A ball being thrown upward, reaching its maximum height, and then falling back down.
The initial spike in acceleration represents the force of the throw.
The constant negative acceleration represents the force of gravity pulling the ball downward.
The spike in acceleration at the end could represent the ball hitting the ground.
16. 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 Chapter13). Give the signs of position, velocity and acceleration variables of the particle at t = 0.3 s, 1.2 s, – 1.2 s.
Ans : To analyze the signs of position, velocity, and acceleration at the given times, let’s examine the x-t graph:
t = 0.3 s:
Position (x): Positive
Velocity (v): Negative (since the slope of the x-t plot is negative at this point)
Acceleration (a): Positive (since the slope of the v-t plot is positive at this point)
t = 1.2 s:
Position (x): Negative
Velocity (v): Positive (since the slope of the x-t plot is positive at this point)
Acceleration (a): Negative (since the slope of the v-t plot is negative at this point)
t = -1.2 s:
Position (x): Negative
Velocity (v): Positive (since the slope of the x-t plot is positive at this point)
Acceleration (a): Positive (since the slope of the v-t plot is negative at this point)
Summary:
Time (s) Position (x) Velocity (v) Acceleration (a)
0.3 Positive Negative Positive
1.2 Negative Positive Negative
-1.2 Negative Positive Positive
17. Figure 2.14 gives the x-t plot of a particle in one-dimensional motion. Three different equal intervals of time are shown. In which interval is the average speed greatest, and in which is it the least ? Give the sign of average velocity for each interval.
Ans : Understanding the Graph:
The graph shows the position (x) of a particle over time (t). The shaded regions represent equal intervals of time.
Average Speed:
To find average speed, divide the total distance traveled by the elapsed time. In this case, since we don’t have the exact values of distance and time, we can qualitatively
analyze the graph.
Interval 1: The slope of the graph is steepest in this interval, indicating a large change in position over a short time. This suggests the greatest average speed.
Interval 2: The slope is less steep than in Interval 1, indicating a smaller change in position over the same time. This suggests a lower average speed than Interval 1.
Interval 3: The slope is the least steep, indicating the smallest change in position over the same time. This suggests the least average speed.
Average Velocity:
To find average velocity, divide the change in position by the time taken.
Interval 1: The displacement is positive, and the time is positive. Therefore, the average velocity is positive.
Interval 2: The displacement is negative, and the time is positive. Therefore, the average velocity is negative.
Interval 3: The displacement is positive, and the time is positive. Therefore, the average velocity is positive.
Summary:
Greatest average speed: Interval 1
Least average speed: Interval 3
Sign of average velocity:
Interval 1: Positive
Interval 2: Negative
Interval 3: Positive
18. Figure 2.15 gives a speed-time graph of a particle in motion along a constant direction. Three equal intervals of time are shown. In which interval is the average acceleration greatest in magnitude?
In which interval is the average speed greatest ? Choosing the positive direction as the constant direction of motion, give the signs of v and a in the three intervals. What are the accelerations at the points A, B, C and D ?
Ans : Analyzing the Speed-Time Graph
Average Acceleration:
Average acceleration is the slope of the line connecting the starting and ending points of a time interval on the speed-time graph.
The steeper the slope, the greater the magnitude of the average acceleration.
Average Speed:
Average speed is calculated by dividing the total distance covered by the time elapsed.
Since the intervals are equal in time, the average speed is directly proportional to the total change in speed during the interval.
Analyzing the Intervals:
Interval 1 (0 to 1):
Average acceleration: Greatest magnitude (steepest slope)
Average speed: Greatest (largest change in speed)
v: Positive
a: Positive
Interval 2 (1 to 2):
Average acceleration: Smaller magnitude (less steep slope)
Average speed: Intermediate
v: Positive
a: Negative (since the speed is decreasing)
Interval 3 (2 to 3):
Average acceleration: Greatest magnitude (steepest slope, but negative)
Average speed: Least (smallest change in speed)
v: Positive
a: Negative
Summary:
Greatest average acceleration magnitude: Interval 1 and Interval 3
Greatest average speed: Interval 1
Signs of v and a:
Interval 1: v positive, a positive
Interval 2: v positive, a negative
Interval 3: v positive, a negative
Note: At points A, B, C, and D, the slope of the speed-time graph is zero, which means the acceleration is zero at these points.
