Chapter 2: Motion in a Straight Line Long Questions | physics Class 11 CBSE

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Chapter 2: Motion in a Straight Line Long Questions | physics Class 11 CBSE | ByteNotes.in

Long Answer

Medium difficulty • Structured explanation

Medium

Question 1

Long Form

Derive the three kinematic equations for uniformly accelerated motion using the graphical method.

For uniform acceleration, the v-t graph is a straight line starting at v0 at t = 0 and reaching v at time t. From the definition of constant acceleration: a = (v - v0)/t, giving the first equation v = v0 + at.
The displacement equals the area under the v-t curve (a trapezium OACB). Area = area of rectangle OACD + area of triangle ABC = v0·t + ½(v - v0)t = v0t + ½at², yielding x = v0t + ½at².
The displacement can also be written as x = [(v + v0)/2]·t, since average velocity for constant acceleration equals the arithmetic mean of initial and final velocities.
Substituting t = (v - v0)/a from the first equation into x = [(v + v0)/2]·t gives x = (v² - v0²)/(2a), or v² = v0² + 2ax as the third equation.
These three equations connect five quantities: initial velocity v0, final velocity v, acceleration a, time t, and displacement x; any three known quantities can be used to find the remaining two.
When the particle starts at position x0 (not zero), displacement x is replaced by (x - x0) in all equations, giving the more general forms used in practice.

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Long Answer

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Hard

Question 2

Long Form

Derive the kinematic equations for constant acceleration using the method of calculus and explain why this method is more powerful than the graphical approach.

Starting from the definition a = dv/dt, we write dv = a dt. Integrating both sides from v0 to v and 0 to t (a = constant), we get v - v0 = at, or v = v0 + at.
Using v = dx/dt, so dx = v dt = (v0 + at)dt. Integrating from x0 to x and 0 to t: x - x0 = v0t + ½at², giving x = x0 + v0t + ½at².
Writing a = v(dv/dx), we have v dv = a dx. Integrating from v0 to v and x0 to x: (v² - v0²)/2 = a(x - x0), giving v² = v0² + 2a(x - x0).
All three equations are consistent with each other and can be derived from one another by algebraic substitution — for instance, eliminating t from equations 1 and 2 gives equation 3.
The calculus method is more powerful because it can be extended directly to motion with non-uniform (variable) acceleration simply by integrating a(t) or a(v) as appropriate.
The graphical method is limited to cases where geometric shapes (rectangles, triangles) can represent the area under the curve; calculus handles all functional forms analytically and precisely.

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Long Answer

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Medium

Question 3

Long Form

Discuss the motion of an object under free fall. Derive expressions for velocity, position, and speed as functions of time, and describe the variation of these quantities graphically.

Free fall is motion under the influence of gravity alone with air resistance neglected. Taking upward as positive, the acceleration is a = -g = -9.8 m s⁻², and initial velocity v0 = 0 for an object released from rest.
Applying kinematic equations: velocity v = 0 - gt = -9.8t m s⁻¹, showing velocity increases in magnitude linearly with time in the downward direction.
Position: y = 0 - ½gt² = -4.9t² m, showing displacement grows as the square of time; the negative sign confirms downward motion from the origin.
The third equation gives v² = -2gy m² s⁻², linking speed to displacement; speed increases as the square root of distance fallen.
The a-t graph is a horizontal line at -9.8 m s⁻²; the v-t graph is a straight line with negative slope; the y-t graph is a downward-opening parabola.
If the height of fall is not small compared to Earth's radius, g cannot be treated as constant and the above equations become approximate; for everyday problems at small heights, this approximation is valid.

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Long Answer

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Hard

Question 4

Long Form

Compare and analyse the four different cases of velocity-time graphs for motion with constant acceleration as described in the chapter.

Case (a): Object moves in the positive direction with positive acceleration. The v-t graph is a straight line in the first quadrant with positive slope, showing speed increasing steadily over time.
Case (b): Object moves in the positive direction with negative acceleration. The v-t line starts in the first quadrant and slopes downward; the object decelerates, and if the line crosses the time axis, the object momentarily stops and then reverses direction.
Case (c): Object moves in the negative direction with negative acceleration. The v-t line lies in the third quadrant (negative velocity) and slopes further negatively, meaning the object speeds up in the negative direction.
Case (d): Object moves in the positive direction until time t1 (velocity decreases from v0 to 0), then reverses with the same negative acceleration. Between 0 and t1, the graph is in the first quadrant; beyond t1, it dips into the fourth quadrant (velocity becomes negative).
In all cases, the slope of the v-t line equals the constant acceleration; a steeper line indicates greater magnitude of acceleration regardless of direction.
The area under the v-t curve in all cases represents displacement — positive areas above the time axis indicate positive displacement, and negative areas below it indicate negative displacement.

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Long Answer

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Medium

Question 5

Long Form

Explain the concept of relative velocity for two objects moving along the same straight line. Analyse how relative velocity differs when the objects move in the same direction versus opposite directions.

Relative velocity of object A with respect to object B is defined as vAB = vA - vB; it represents the velocity of A as seen by an observer sitting on (moving with) B.
Similarly, relative velocity of B with respect to A is vBA = vB - vA = -vAB; these two relative velocities are always equal in magnitude but opposite in direction.
When A and B move in the same direction (say both positive), |vAB| = |vA - vB|, which is less than either individual speed. If they have equal speeds, relative velocity is zero and they appear stationary to each other.
When A and B move in opposite directions, vAB = vA - (-vB) = vA + vB; the magnitude of relative velocity exceeds either individual speed and equals their sum.
A practical example: a police van at 30 km h⁻¹ fires a bullet (muzzle speed 150 m s⁻¹) at a thief's car at 192 km h⁻¹ in the same direction. The bullet's speed relative to the thief's car determines the impact.
The concept is essential for problems involving two trains, two swimmers, or any pair of moving objects — it reduces a two-body problem to an equivalent single-body problem by switching to the reference frame of one object.

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Long Answer

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Hard

Question 6

Long Form

Analyse the distinction between average velocity and instantaneous velocity, and explain using numerical and graphical methods how instantaneous velocity can be determined.

Average velocity is the ratio of total displacement to total time (v̄ = Δx/Δt) and gives information about motion over a finite interval, not at a specific moment.
Instantaneous velocity is the limit of average velocity as Δt → 0: v = dx/dt. It captures how fast an object is moving at a single instant rather than over an interval.
Graphically, instantaneous velocity at a point on the x-t curve equals the slope of the tangent at that point; this is found by decreasing Δt until the chord P1P2 becomes tangent to the curve.
Numerically, for x = 0.08t³, the average velocity (Δx/Δt) was computed for Δt = 2.0, 1.0, 0.5, 0.1, 0.01 s centred at t = 4 s. The values approach 3.84 m s⁻¹, confirming the limiting value.
Using calculus: v = dx/dt = 3 × 0.08 × t² = 0.24t². At t = 4 s, v = 0.24 × 16 = 3.84 m s⁻¹, agreeing precisely with the numerical limit.
For uniform motion, instantaneous velocity equals average velocity at all instants; for non-uniform motion, they differ, and only the instantaneous velocity fully describes motion at a given point in time.

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Long Answer

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Hard

Question 7

Long Form

Prove Galileo's law of odd numbers for a body falling freely from rest. What does this law reveal about the nature of free fall?

For a body falling from rest under free fall, y = -½gt². Dividing time into equal intervals τ, the position at time nτ is yn = -½g(nτ)² = -n²(½gτ²).
Taking y0 = -(½)gτ² (magnitude of position after first interval τ), we can write positions as 0, -y0, -4y0, -9y0, -16y0, -25y0, -36y0... at t = 0, τ, 2τ, 3τ...
Distance covered in first interval: y0; second: 4y0 - y0 = 3y0; third: 9y0 - 4y0 = 5y0; fourth: 7y0; fifth: 9y0; sixth: 11y0.
The ratio of distances in successive equal time intervals is 1:3:5:7:9:11, confirming Galileo's law of odd numbers directly from kinematics.
This law reveals that free fall has uniformly increasing distance increments per time interval — the body gains equal increments of velocity every equal time interval, which is the hallmark of uniform acceleration.
Galileo Galilei (1564-1642) established this law experimentally and thus empirically proved that free fall is uniformly accelerated motion, a cornerstone of classical mechanics.

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Long Answer

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Hard

Question 8

Long Form

Examine the Points to Ponder from the chapter: discuss why the sign of acceleration alone does not determine whether a particle is speeding up or slowing down.

The sign of acceleration depends on the choice of the positive direction of the coordinate axis, which is arbitrary. It does not by itself tell whether speed is increasing or decreasing.
If the vertically upward direction is chosen as positive, acceleration due to gravity is negative (-9.8 m s⁻²) for all objects near Earth's surface regardless of their motion.
A particle falling downward (velocity negative) with this negative acceleration is speeding up — acceleration and velocity are in the same direction (both negative), so |v| increases.
A particle thrown upward (velocity positive) with the same negative acceleration is slowing down — acceleration is opposite to velocity, so |v| decreases until it reaches zero.
The rule is: if acceleration and velocity are in the same direction (same sign), speed increases; if they are in opposite directions (opposite signs), speed decreases.
This subtle point is important for interpreting motion problems correctly: a 'negative acceleration' in one coordinate choice might mean speeding up in the negative direction, not necessarily decelerating.

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Long Answer

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Medium

Question 9

Long Form

A ball is thrown vertically upward at 20 m s⁻¹ from the top of a 25 m building. Analyse its complete motion to find how high it rises and total time before it hits the ground. Take g = 10 m s⁻².

Taking upward as positive with origin at the ground, initial position y0 = 25 m, initial velocity v0 = +20 m s⁻¹, acceleration a = -10 m s⁻².
At the highest point, v = 0. Using v² = v0² + 2a(y - y0): 0 = 400 + 2(-10)(y - 25), giving y - 25 = 20 m, so the ball rises 20 m above the launch point, reaching y = 45 m above ground.
Method 1 (two stages): Time to reach top: 0 = 20 - 10t1, so t1 = 2 s. Free fall from 45 m: 0 = 45 + 0·t2 + ½(-10)t2², giving t2 = 3 s. Total time = t1 + t2 = 5 s.
Method 2 (single equation): Using y = y0 + v0t + ½at², with y = 0: 0 = 25 + 20t - 5t². Simplifying: 5t² - 20t - 25 = 0, or t² - 4t - 5 = 0, giving (t-5)(t+1) = 0, so t = 5 s (taking positive value).
The second method is preferred as it directly gives the answer without splitting the problem into sub-paths, reducing the risk of sign errors.
The problem illustrates how kinematic equations handle the complete trajectory under constant acceleration regardless of direction changes mid-flight.

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Long Answer

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Medium

Question 10

Long Form

Critically compare average speed and average velocity using the example of a woman who walks to her office and returns home by auto. Explain when they are equal.

Average velocity = total displacement / total time. If the woman returns to her home (starting point), displacement = 0, so average velocity for the entire trip = 0 m/s.
Average speed = total distance / total time = (2 × 2.5 km) / total time. Since she walks 5 km h⁻¹ for 2.5 km (takes 0.5 h) and returns at 25 km h⁻¹ (takes 0.1 h), total time = 0.6 h. Average speed = 5/0.6 ≈ 8.33 km h⁻¹.
This example demonstrates that average speed is not zero even when average velocity is zero — because path length is not the same as displacement magnitude.
The relationship is: average speed ≥ |average velocity|, with equality when the object moves in one direction without reversing, so that path length equals displacement magnitude.
For instantaneous quantities, no such distinction exists: instantaneous speed always equals the magnitude of instantaneous velocity at that instant, because over an infinitesimally small interval, path length and displacement magnitude are identical.
This is why it is more physically meaningful to define average speed as total path divided by time rather than as magnitude of average velocity — the latter would give a misleading zero for a round trip.

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Chapter 2: Motion in a Straight Line | Long Questions

Derive the three kinematic equations for uniformly accelerated motion using the graphical method.

Derive the kinematic equations for constant acceleration using the method of calculus and explain why this method is more powerful than the graphical approach.

Discuss the motion of an object under free fall. Derive expressions for velocity, position, and speed as functions of time, and describe the variation of these quantities graphically.

Compare and analyse the four different cases of velocity-time graphs for motion with constant acceleration as described in the chapter.

Explain the concept of relative velocity for two objects moving along the same straight line. Analyse how relative velocity differs when the objects move in the same direction versus opposite directions.

Analyse the distinction between average velocity and instantaneous velocity, and explain using numerical and graphical methods how instantaneous velocity can be determined.

Prove Galileo's law of odd numbers for a body falling freely from rest. What does this law reveal about the nature of free fall?

Examine the Points to Ponder from the chapter: discuss why the sign of acceleration alone does not determine whether a particle is speeding up or slowing down.

A ball is thrown vertically upward at 20 m s⁻¹ from the top of a 25 m building. Analyse its complete motion to find how high it rises and total time before it hits the ground. Take g = 10 m s⁻².

Critically compare average speed and average velocity using the example of a woman who walks to her office and returns home by auto. Explain when they are equal.

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