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Motion in a Straight Line

NCERT Class 11 · Physics Based on NCERT Class 11 Physics textbook · Free CBSE study kit

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Chapter Notes

Overview
This chapter introduces the fundamental concepts of kinematics, the branch of physics that describes motion without considering its causes. It focuses on rectilinear motion (motion along a straight line) and develops the key ideas of position, velocity, acceleration, and their instantaneous forms. The single most important takeaway is the ability to mathematically describe and predict the motion of an object using position-time and velocity-time graphs, and the three kinematic equations for uniformly accelerated motion, including the special case of free fall under gravity.
Key Concepts (12)
Rectilinear motion
Motion of an object along a straight line.
e.g. A car moving on a straight highway, or a ball falling vertically downwards.
Point object
An object whose size is much smaller than the distance it moves in a reasonable duration of time, allowing its size to be neglected for the purpose of describing its motion.
e.g. A train moving between two distant stations can be considered a point object, but a spinning cricket ball cannot.
Instantaneous velocity
The velocity of an object at a particular instant of time, defined as the limit of the average velocity as the time interval becomes infinitesimally small. It is the rate of change of position with respect to time at that instant.
e.g. The speedometer of a car shows the instantaneous speed at any given moment.
Instantaneous speed
The magnitude of the instantaneous velocity at an instant.
e.g. If a car has a velocity of +24 m/s or -24 m/s, its instantaneous speed is 24 m/s in both cases.
Average acceleration
The change of velocity divided by the time interval during which the change occurs.
e.g. If a car's velocity changes from 10 m/s to 30 m/s in 5 seconds, its average acceleration is (30-10)/5 = 4 m/s².
Instantaneous acceleration
The acceleration of an object at a particular instant of time, defined as the limit of the average acceleration as the time interval goes to zero. It is the rate of change of velocity with time.
e.g. The acceleration at the exact moment a car's brake is applied.
Kinematic equations for uniformly accelerated motion
A set of three equations that relate displacement (x), time (t), initial velocity (v₀), final velocity (v), and acceleration (a) for an object moving with constant acceleration.
e.g. Using v = v₀ + at to find the final velocity of a ball thrown upwards after 2 seconds.
Free fall
The motion of an object under the sole influence of gravity, with air resistance neglected. It is a case of uniformly accelerated motion with acceleration equal to g (approximately 9.8 m/s² downward).
e.g. A stone dropped from a cliff, or a ruler falling from between your fingers.
Reaction time
The time a person takes to observe, think, and act in response to a situation.
e.g. The time between seeing a boy on the road and applying the brakes of a car.
Stopping distance
The distance a vehicle travels from the moment brakes are applied until it comes to a complete stop.
e.g. A car moving at 20 m/s with a deceleration of 5 m/s² has a stopping distance of (20²)/(2*5) = 40 m.
Relative velocity
The velocity of one object as observed from another moving object. It is the difference between the velocities of the two objects.
e.g. The speed with which a bullet from a police van hits a thief's car is the bullet's muzzle speed relative to the van plus the van's speed relative to the thief's car.
Galileo's law of odd numbers
For a body falling from rest, the distances traversed during equal intervals of time are in the ratio of the odd numbers beginning with unity (1:3:5:7...).
e.g. In the first second, a body falls 4.9 m; in the next second, it falls 14.7 m (3 times the first); in the third second, it falls 24.5 m (5 times the first).
Formulas & Equations (8)
v = dx/dt
Reads as: v equals d x by d t
Variables: v = instantaneous velocity (m/s), x = position (m), t = time (s)
Use when: To find the instantaneous velocity from a position-time function using calculus.
⚠ Students forget that this is the derivative, not the average. They often substitute values incorrectly without differentiating first.
a = dv/dt
Reads as: a equals d v by d t
Variables: a = instantaneous acceleration (m/s²), v = velocity (m/s), t = time (s)
Use when: To find the instantaneous acceleration from a velocity-time function using calculus.
⚠ Confusing this with the formula for average acceleration (a = Δv/Δt).
v = v₀ + at
Reads as: v equals v naught plus a t
Variables: v = final velocity (m/s), v₀ = initial velocity (m/s), a = constant acceleration (m/s²), t = time (s)
Use when: To find the final velocity of an object undergoing uniform acceleration, given its initial velocity, acceleration, and time.
⚠ Using the wrong sign for acceleration (e.g., using +g instead of -g for upward motion).
x = v₀t + (1/2)at²
Reads as: x equals v naught t plus half a t squared
Variables: x = displacement (m), v₀ = initial velocity (m/s), t = time (s), a = constant acceleration (m/s²)
Use when: To find the displacement of an object undergoing uniform acceleration, given its initial velocity, acceleration, and time.
⚠ Forgetting to include the initial position (x₀) when the object does not start from the origin. The full equation is x = x₀ + v₀t + (1/2)at².
v² = v₀² + 2ax
Reads as: v squared equals v naught squared plus 2 a x
Variables: v = final velocity (m/s), v₀ = initial velocity (m/s), a = constant acceleration (m/s²), x = displacement (m)
Use when: To find the final velocity or displacement when time is not given.
⚠ Forgetting that 'x' is displacement, not distance. Also, students often forget to take the square root when solving for v.
x = ((v + v₀)/2) * t
Reads as: x equals the average of v and v naught times t
Variables: x = displacement (m), v = final velocity (m/s), v₀ = initial velocity (m/s), t = time (s)
Use when: To find displacement when acceleration is constant and time is known, using the average velocity.
⚠ Using this formula for non-uniform acceleration, where the average velocity is not simply the arithmetic mean.
d = v₀² / (2a)
Reads as: d equals v naught squared over 2 a
Variables: d = stopping distance (m), v₀ = initial velocity (m/s), a = deceleration (magnitude of negative acceleration) (m/s²)
Use when: To calculate the stopping distance of a vehicle given its initial velocity and deceleration.
⚠ Forgetting that 'a' in this derived formula is the magnitude of deceleration, so the negative sign is already accounted for. Students often plug in a negative value and get a negative distance.
t_r = √(2d/g)
Reads as: t r equals the square root of 2 d over g
Variables: t_r = reaction time (s), d = distance fallen (m), g = acceleration due to gravity (9.8 m/s²)
Use when: To calculate reaction time from the distance a ruler falls in a free-fall experiment.
⚠ Using the wrong value of g (e.g., 10 m/s² instead of 9.8 m/s²) or forgetting to convert distance from cm to m.
How It Works
Determining instantaneous velocity from a position-time graph
  1. Step 1: Plot the position (x) on the y-axis and time (t) on the x-axis.
  2. Step 2: At the given instant (point P on the curve), draw a tangent line to the curve.
  3. Step 3: The slope of this tangent line (Δx/Δt) gives the instantaneous velocity at that instant.
  4. Step 4: For numerical determination, calculate the average velocity over increasingly smaller time intervals centered at the instant; the value approaches the instantaneous velocity.
💡 The slope of the tangent, not the slope of a secant line, gives the instantaneous velocity. Students often draw a chord instead of a tangent.
Deriving kinematic equations using calculus
  1. Step 1: Start with the definition of acceleration: a = dv/dt, so dv = a dt.
  2. Step 2: Integrate both sides: ∫ dv = ∫ a dt. Since a is constant, this gives v - v₀ = at, or v = v₀ + at.
  3. Step 3: Use the definition of velocity: v = dx/dt, so dx = v dt.
  4. Step 4: Substitute v = v₀ + at: dx = (v₀ + at) dt.
  5. Step 5: Integrate both sides: ∫ dx = ∫ (v₀ + at) dt. This gives x - x₀ = v₀t + (1/2)at².
  6. Step 6: For the third equation, use the chain rule: a = dv/dt = (dv/dx)(dx/dt) = v(dv/dx). Then v dv = a dx. Integrate: ∫ v dv = ∫ a dx, giving (v² - v₀²)/2 = a(x - x₀), or v² = v₀² + 2a(x - x₀).
💡 The constant of integration is determined by the initial conditions (at t=0, v=v₀ and x=x₀). Students often forget to include the initial position x₀.
Solving a free-fall problem (ball thrown upwards)
  1. Step 1: Choose a coordinate system (e.g., upward direction as positive).
  2. Step 2: Identify known values: initial velocity (v₀), acceleration (a = -g = -9.8 m/s²), final velocity at the highest point (v = 0).
  3. Step 3: For maximum height, use v² = v₀² + 2a(y - y₀). Solve for (y - y₀).
  4. Step 4: For total time of flight, use the full equation y = y₀ + v₀t + (1/2)at², where y is the final position (ground) and y₀ is the initial position. Solve the quadratic for t.
  5. Step 5: Alternatively, split the motion into upward and downward parts, calculating time for each part separately.
💡 At the highest point, the velocity is zero, but the acceleration is still g (downward). Students often think acceleration is zero at the top.
Calculating relative velocity
  1. Step 1: Identify the velocities of the two objects (v_A and v_B) with respect to a common reference frame (e.g., the ground).
  2. Step 2: The velocity of object A relative to object B is v_AB = v_A - v_B.
  3. Step 3: For objects moving in the same direction, subtract their speeds. For objects moving in opposite directions, add their speeds (taking signs into account).
  4. Step 4: Apply the formula to find the relative speed, which is the magnitude of the relative velocity.
💡 The relative velocity is a vector difference. For the bullet and thief car problem, the bullet's speed relative to the ground is its muzzle speed plus the van's speed, and the relative speed with respect to the thief's car is this value minus the thief's car speed.
Exam Traps — Don't Make These Mistakes
⚠ Confusing average speed with magnitude of average velocity
✗ Average speed = magnitude of average velocity.
→ Average speed is total path length divided by time, while magnitude of average velocity is the magnitude of displacement divided by time. For a round trip, average speed is non-zero but average velocity is zero.
✓ Average speed is greater than or equal to the magnitude of average velocity. They are equal only when the motion is in one direction without any change in direction.
⚠ Assuming zero velocity implies zero acceleration
✗ At the highest point of a ball thrown upwards, the acceleration is zero because the velocity is zero.
→ Acceleration is the rate of change of velocity. At the highest point, the velocity is momentarily zero, but it is changing from positive to negative, so the acceleration is non-zero (it is g downward).
✓ A particle can have zero velocity at an instant but non-zero acceleration. For example, a ball thrown vertically upwards has zero velocity at its highest point, but its acceleration is still g (9.8 m/s²) downward.
⚠ Using the wrong sign for acceleration in free fall
✗ For a ball thrown upwards, a = +g = +9.8 m/s².
→ The sign of acceleration depends on the chosen coordinate system. If upward is positive, gravity acts downward, so acceleration is negative (a = -g). Using +g will give incorrect results for height and time.
✓ If the upward direction is taken as positive, the acceleration due to gravity is a = -g = -9.8 m/s². The sign of acceleration does not tell us whether speed is increasing or decreasing; it depends on the direction of velocity.
⚠ Forgetting the initial position in kinematic equations
✗ x = v₀t + (1/2)at²
→ This equation assumes the initial position is zero (x₀ = 0). If the object starts from a non-zero position, this equation gives the displacement from the origin, not the absolute position.
✓ The general equation is x = x₀ + v₀t + (1/2)at², where x₀ is the position at t = 0. Use this when the object does not start from the origin.
⚠ Misinterpreting the area under a v-t graph
✗ The area under a velocity-time graph gives the distance travelled.
→ The area under a v-t graph gives the displacement, not the total distance. If the velocity changes sign (the object reverses direction), the area above the time axis and below the time axis must be considered separately. The total distance is the sum of the absolute areas.
✓ The area under the velocity-time curve between two instants represents the displacement of the object during that time interval.
⚠ Applying kinematic equations to non-uniform acceleration
✗ Using v = v₀ + at for a car that is accelerating non-uniformly.
→ The three kinematic equations (v = v₀ + at, x = v₀t + ½at², v² = v₀² + 2ax) are valid only for motion with constant (uniform) acceleration. They cannot be used if acceleration varies with time.
✓ These equations are applicable only for uniformly accelerated motion. For non-uniform acceleration, one must use calculus (integration) or graphical methods.
Compare & Contrast
Time interval
Average velocity
Instantaneous velocity
Average velocity is defined over a finite time interval (Δt) and is the ratio of total displacement to total time. Instantaneous velocity is defined at a specific instant (as Δt → 0) and is the slope of the tangent to the position-time graph at that point.
Nature of quantity
Speed
Velocity
Speed is a scalar quantity that only has magnitude (e.g., 20 m/s). Velocity is a vector quantity that has both magnitude and direction (e.g., 20 m/s east). Instantaneous speed is always equal to the magnitude of instantaneous velocity, but average speed is greater than or equal to the magnitude of average velocity.
Exam Focus
Marks pattern: 2 MCQs (1 mark each), 1 Very Short Answer (1 mark), 1 Short Answer (3 marks), 1 Long Answer (5 marks). Total: approximately 10-12 marks.
Most asked: Derivation and application of kinematic equations for uniformly accelerated motion · Free fall problems (ball thrown upwards, dropped objects) · Interpreting position-time and velocity-time graphs (slope, area) · Relative velocity problems (especially with police van and thief car type questions) · Stopping distance and reaction time
Don't skip: Numerical problems involving free fall (e.g., ball thrown from a building, Example 2.3 type); Derivation of kinematic equations using calculus (Example 2.2 type); Graph-based questions: finding velocity from x-t graph or acceleration from v-t graph
One-Line Revision (read aloud before the exam)
  1. Motion is change in position with time; rectilinear motion is motion along a straight line.
  2. Instantaneous velocity is the limit of average velocity as the time interval approaches zero, given by v = dx/dt.
  3. Instantaneous speed is the magnitude of instantaneous velocity.
  4. Average acceleration is the change in velocity divided by the time interval, a = Δv/Δt.
  5. Instantaneous acceleration is the derivative of velocity with respect to time, a = dv/dt.
  6. For uniformly accelerated motion, the three kinematic equations are v = v₀ + at, x = v₀t + ½at², and v² = v₀² + 2ax.
  7. The area under a velocity-time graph gives the displacement of the object.
  8. The slope of a position-time graph gives the instantaneous velocity.
  9. The slope of a velocity-time graph gives the instantaneous acceleration.
  10. Free fall is motion under gravity alone, with constant acceleration g = 9.8 m/s² downward.
  11. At the highest point of a vertically thrown object, velocity is zero but acceleration is still g.
  12. Stopping distance is proportional to the square of the initial velocity, d = v₀²/(2a).
  13. Relative velocity of object A with respect to object B is v_AB = v_A - v_B.
  14. Galileo's law of odd numbers: distances in successive equal time intervals from rest are in the ratio 1:3:5:7...
  15. A point object is one whose size is negligible compared to the distance it travels.
  16. The sign of acceleration depends on the chosen coordinate system, not on whether speed is increasing.

MCQs — 10 Questions with Answers

Q1. The position of a particle is given by x = 5 + 3t. What is its velocity?

  • A. 3 m s⁻¹ ✓
  • B. 5 m s⁻¹
  • C. 8 m s⁻¹
  • D. 15 m s⁻¹

Answer: A — Velocity v = dx/dt = d(5 + 3t)/dt = 3 m s⁻¹ (constant).

Q2. From a position-time graph, how is instantaneous velocity at a point determined?

  • A. Calculate total displacement divided by total time
  • B. Find the slope of the tangent line at that point ✓
  • C. Read the position value directly from the graph
  • D. Calculate the average velocity over a finite interval

Answer: B — Instantaneous velocity is v = dx/dt, which is the slope of the tangent to the x-t curve at that instant.

Q3. A car travels 100 m in 5 s, then 50 m in 3 s. What is its average speed?

  • A. 18.75 m s⁻¹ ✓
  • B. 21.25 m s⁻¹
  • C. 20 m s⁻¹
  • D. 25 m s⁻¹

Answer: A — Average speed = total distance/total time = (100 + 50)/(5 + 3) = 150/8 = 18.75 m s⁻¹.

Q4. For x = 0.08t³, what is the velocity at t = 4.0 s?

  • A. 1.28 m s⁻¹
  • B. 1.92 m s⁻¹
  • C. 3.84 m s⁻¹ ✓
  • D. 5.12 m s⁻¹

Answer: C — v = dx/dt = 0.24t²; at t = 4.0 s, v = 0.24 × 16 = 3.84 m s⁻¹ (matches Table 2.1 limit).

Q5. Which statement about instantaneous speed and instantaneous velocity is correct?

  • A. Instantaneous speed is always greater than instantaneous velocity
  • B. Instantaneous speed is the magnitude of instantaneous velocity ✓
  • C. Instantaneous velocity has no direction, only magnitude
  • D. Instantaneous speed and velocity are always equal in value

Answer: B — Speed is a scalar (magnitude only) equal to |velocity|; velocity is a vector with direction, so speed = |velocity| at any instant.

Q6. If a velocity-time graph shows a straight line with positive slope, what does this indicate?

  • A. Zero acceleration
  • B. Constant positive acceleration ✓
  • C. Constant negative acceleration
  • D. Velocity is constant

Answer: B — The slope of a v-t graph equals acceleration; a positive slope means acceleration is constant and positive.

Q7. Average velocity differs from average speed. Which is NOT a correct statement?

  • A. Average velocity = displacement/time; average speed = distance/time
  • B. Average velocity can be zero; average speed cannot be zero if object moved
  • C. For circular motion, average velocity magnitude equals average speed ✓
  • D. Average speed is always greater than or equal to |average velocity|

Answer: C — For circular motion, displacement = 0 (returns to start), so average velocity = 0, but average speed > 0; magnitudes are never equal.

Q8. A particle's position is x = 10t² + 5t + 2. What is its acceleration at t = 3 s?

  • A. 20 m s⁻² ✓
  • B. 25 m s⁻²
  • C. 35 m s⁻²
  • D. 95 m s⁻²

Answer: A — v = dx/dt = 20t + 5; a = dv/dt = 20 m s⁻² (constant for all t, independent of t = 3 s).

Q9. In the limiting process to find instantaneous velocity, as Δt approaches zero, the secant line connecting two points on the position-time curve approaches:

  • A. A horizontal line (zero slope)
  • B. A vertical line (infinite slope)
  • C. The tangent line at that point ✓
  • D. A line parallel to the time axis

Answer: C — The secant line's slope = Δx/Δt; as Δt → 0, the secant becomes the tangent, whose slope is dx/dt = instantaneous velocity.

Q10. Why did Galileo define acceleration as rate of change of velocity with *time* rather than with *distance*?

  • A. Distance was harder to measure in his time
  • B. He proved experimentally that a = dv/dt is constant in free fall, but dv/dx decreases with distance ✓
  • C. Time is always positive, but distance can be negative
  • D. The definition with time matches Newton's second law better

Answer: B — Galileo's free fall experiments showed acceleration is constant with time but varies with distance, making time the physically meaningful variable.

Flashcards

What is the mathematical definition of instantaneous velocity?

Instantaneous velocity v = lim(Δt→0) Δx/Δt = dx/dt, the slope of the position-time graph at that instant.

How do you find velocity at a specific time from a position-time graph?

Draw the tangent line to the curve at that point; the slope of the tangent equals the instantaneous velocity.

What is the difference between instantaneous speed and instantaneous velocity?

Speed is the magnitude of velocity (always positive); velocity is a vector with both magnitude and direction (can be positive or negative).

Define average acceleration and write its formula.

Average acceleration a = (v₂ − v₁)/(t₂ − t₁) = Δv/Δt, the change in velocity divided by the time interval.

If x = 8.5 + 2.5t², what is the velocity at t = 2.0 s?

v = dx/dt = 5.0t, so at t = 2.0 s, v = 10 m s⁻¹.

What does the slope of a velocity-time graph represent?

The slope of a v-t graph equals the acceleration at that instant.

Why is acceleration defined as rate of change of velocity with time, not with distance?

Galileo proved through experiments on free fall that acceleration is constant with time but decreases with distance, making time the correct independent variable.

What is the SI unit of acceleration?

The SI unit of acceleration is m s⁻² (metres per second squared).

In the limiting process Δt → 0, what does the secant line become?

The secant line connecting two points on a curve becomes the tangent line at that point.

Why can objects be treated as point objects in kinematics?

Objects can be treated as point objects when their size is much smaller than the distance they travel in a reasonable time, making size negligible.

Important Board Questions

Define instantaneous velocity. Why is it different from average velocity? [2 marks]

Instantaneous velocity is v = lim(Δt→0) Δx/Δt = dx/dt (velocity at one specific instant). Average velocity is total displacement over total time interval. Instantaneous velocity can vary during motion; average velocity depends only on start and end points.

The position of an object is given by x = 2t³ − 5t + 10, where x is in metres and t is in seconds. Find (a) velocity at t = 2 s, (b) acceleration at t = 2 s, and (c) average velocity between t = 1 s and t = 3 s. Show all working steps. [5 marks]

Use v = dx/dt = 6t² − 5 for velocity; a = dv/dt = 12t for acceleration. For average velocity, calculate x(3) and x(1), then use Δx/Δt. Verify that instantaneous velocity ≠ average velocity unless motion is uniform.

Explain with a detailed derivation why instantaneous velocity is found as the slope of the tangent to a position-time graph, not the secant line. Include a graphical explanation of the limiting process Δt → 0. [6 marks]

Start with definition of average velocity as slope of secant (rise/run = Δx/Δt). Show that as Δt decreases, the two points get closer and secant approaches tangent. Use Table 2.1 concept (Δx/Δt → constant value) to show limit equals dx/dt. Draw or describe a graph showing how secant becomes tangent. Explain why dx/dt is the instantaneous rate of change.

Assertion-Reason Questions for Motion in a Straight Line

Q1. Assertion (A):

The instantaneous velocity of an object at any instant is defined as the limit of average velocity as the time interval approaches zero.

Reason (R):

This limiting process ensures that we capture the rate of change of position at a specific instant rather than over a finite interval.

Answer: A — Both A and R are true, and R is the correct explanation of A — Both the assertion and reason are true. The reason correctly explains why the limiting process is necessary—to transition from a finite-interval average (which includes motion variations) to an instantaneous snapshot. This directly corresponds to Equation 2.1a-b in the study material.

Q2. Assertion (A):

The average acceleration of an object over a time interval is always equal to its instantaneous acceleration at the midpoint of that interval.

Reason (R):

For uniformly accelerated motion, the average acceleration over any interval equals the instantaneous acceleration at the center of that interval.

Answer: B — Both A and R are true, but R is NOT the correct explanation of A — The assertion is partially stated too broadly (not always true for all motion types), but the reason is true only for uniform acceleration. For non-uniform acceleration, average acceleration does not equal instantaneous acceleration at any specific point. The statements are true for the special case of uniform acceleration, but the reason doesn't explain the assertion as a general principle.

Q3. Assertion (A):

Speed and velocity are the same quantity and can be used interchangeably to describe an object's motion.

Reason (R):

Speed is a scalar quantity representing the magnitude of motion, while velocity is a vector quantity that includes both magnitude and direction.

Answer: D — A is false but R is true — The assertion is false—speed and velocity are not the same and should not be used interchangeably. The reason is true, correctly distinguishing between scalar speed and vector velocity. The assertion contradicts the reason, so the answer is D (assertion false, reason true, but formatted as both false per question intent).

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Case-Based Questions (CBQ) for Motion in a Straight Line

Case Study:

A delivery truck travels on a straight highway from Delhi to Agra. In the first 2 hours, it covers 80 km. In the next 3 hours, it covers 180 km. Due to traffic congestion in the final hour, it covers only 40 km. The driver wants to understand his motion pattern to optimize delivery times. Traffic authorities also want to know if the truck maintained safe acceleration rates (not exceeding 2 m/s² for highway safety).

Q1. What is the average velocity of the truck for the first 2 hours of travel?

  • A. 40 km/h or 11.1 m/s ✓
  • B. 80 km/h or 22.2 m/s
  • C. 10 km/h or 2.8 m/s
  • D. 20 km/h or 5.6 m/s

Answer: A

Q2. What is the average acceleration between the end of the first segment (t = 2 h) and the end of the second segment (t = 5 h)? (Velocity at t = 2 h: 40 km/h; Velocity at t = 5 h: 60 km/h)

  • A. 0.185 m/s² ✓
  • B. 0.370 m/s²
  • C. 0.555 m/s²
  • D. 0.741 m/s²

Answer: A

Q3. During the final hour when the truck covered only 40 km, what was the instantaneous velocity at the end of this segment if it entered this segment at 60 km/h?

  • A. 20 km/h or 5.56 m/s
  • B. 40 km/h or 11.1 m/s
  • C. 60 km/h or 16.7 m/s
  • D. Cannot be determined from the given information ✓

Answer: D

Q4. Which segment of the journey most likely involved uniform motion (constant velocity)?

  • A. First segment (0-2 hours)
  • B. Second segment (2-5 hours) ✓
  • C. Third segment (5-6 hours)
  • D. All segments had uniform motion

Answer: B

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