Exercises — Motion in a Plane

22 questions✓ Free · step-by-step

3.12 marksNCERT Cl-11 Physics Part I, Ch3 Exercises, Q3.1
State, for each of the following physical quantities, if it is a scalar or a vector: volume, mass, speed, acceleration, density, number of moles, velocity, angular frequency, displacement, angular velocity.
Hint. A quantity is a vector only if it needs a direction to be fully specified, not just a size.
Step 1 — Sort by whether direction is part of the quantity's definition. Volume, mass, speed, density, number of moles, and angular frequency are all fully described by a single number — no direction attaches to any of them.

Step 2 — The vector quantities. Acceleration, velocity, displacement, and angular velocity all inherently point somewhere, since each describes not just how much but also which way.

✦ Answer: Scalars — volume, mass, speed, density, number of moles, angular frequency. Vectors — acceleration, velocity, displacement, angular velocity.
Where students slip. Classifying angular frequency as a vector because it sounds related to angular velocity — angular frequency (ω, in rad/s) is just a rate, a single number with no attached direction, unlike angular velocity, which is conventionally given a direction along the rotation axis.
3.22 marksNCERT Cl-11 Physics Part I, Ch3 Exercises, Q3.2
Pick out the two scalar quantities in the following list: force, angular momentum, work, current, linear momentum, electric field, average velocity, magnetic moment, relative velocity.
Hint. Of the nine listed, all but two are defined with an inherent direction; find the pair that reduces to a plain number.
Step 1 — Check each quantity for direction. Force, angular momentum, linear momentum, electric field, average velocity, magnetic moment, and relative velocity are all vectors — each one describes a magnitude together with a specific direction.

Step 2 — The two that are not. Work is a scalar (it is the dot product of two vectors, force and displacement, and a dot product always yields a plain number). Electric current is likewise treated as a scalar in this introductory context, since charge flow rate through a wire has a magnitude without needing a full vector description.

✦ Answer: Work and current are the two scalar quantities in the list.
Where students slip. Assuming current must be a vector because it 'flows' in a direction along a wire — in this basic circuit-theory treatment, current is defined and added like a scalar (it does not obey the vector parallelogram law), even though charge clearly moves along a specific path.
3.31 markNCERT Cl-11 Physics Part I, Ch3 Exercises, Q3.3
Pick out the only vector quantity in the following list: Temperature, pressure, impulse, time, power, total path length, energy, gravitational potential, coefficient of friction, charge.
Hint. Impulse is defined as a change in momentum — check what kind of quantity momentum itself is.
Step 1 — Check the one flagged quantity. Impulse equals the change in a particle's momentum (impulse = FΔt = Δp), and since momentum is a vector, impulse inherits both a magnitude and a direction from it.

Step 2 — Confirm the rest are scalars. Temperature, pressure, time, power, total path length, energy, gravitational potential, coefficient of friction, and charge are each fully specified by a single number, with no direction attached to any of them.

✦ Answer: Impulse is the only vector quantity in the list.
Where students slip. Treating pressure as a vector because it involves force — pressure is force per unit area with the direction convention already built into its definition (always taken normal to the surface), leaving only a magnitude to specify at each point.
3.43 marksNCERT Cl-11 Physics Part I, Ch3 Exercises, Q3.4
State with reasons, whether the following algebraic operations with scalar and vector physical quantities are meaningful: (a) adding any two scalars, (b) adding a scalar to a vector of the same dimensions, (c) multiplying any vector by any scalar, (d) multiplying any two scalars, (e) adding any two vectors, (f) adding a component of a vector to the same vector.
Hint. For (a) and (e), ask whether the two quantities being combined represent the same kind of physical quantity, not just whether their arithmetic dimensions match.
Step 1 — (a) Adding any two scalars. Not always meaningful — only meaningful when both scalars represent the same physical quantity (e.g. adding two masses), since adding, say, a mass to a temperature produces a number with no physical sense even if you forced the arithmetic.

Step 2 — (b) Adding a scalar to a vector of the same dimensions. Never meaningful. A scalar and a vector are fundamentally different kinds of mathematical object — sharing dimensions (like both being a length) does not make them addable, since a vector carries direction and a scalar does not.

Step 3 — (c) and (d) Scalar multiplication and scalar-times-scalar. Both are always meaningful — multiplying a vector by a scalar is a standard, well-defined operation (it scales the vector's length, and flips its direction if the scalar is negative), and multiplying two scalars together always yields another well-defined scalar.

Step 4 — (e) Adding any two vectors. Not always meaningful — only meaningful if both vectors represent the same kind of physical quantity (e.g. two displacements, or two forces), since adding a velocity vector to a force vector, for instance, produces something with no physical interpretation.

Step 5 — (f) Adding a component of a vector to the same vector. Not meaningful — a component only captures part of a vector's information along one particular direction, so combining it back with the full vector does not correspond to any physically sensible operation, even where the dimensions happen to match.

✦ Answer: (a) meaningful only for like scalars (b) never meaningful (c) always meaningful (d) always meaningful (e) meaningful only for like vectors (f) not meaningful.
Where students slip. Assuming (a) and (e) are always fine just because the units cancel out correctly on paper — matching dimensions is necessary but not sufficient; the two quantities being combined must also be the same kind of physical quantity, not merely dimensionally compatible.
3.53 marksNCERT Cl-11 Physics Part I, Ch3 Exercises, Q3.5
Read each statement below carefully and state with reasons, if it is true or false: (a) The magnitude of a vector is always a scalar, (b) each component of a vector is always a scalar, (c) the total path length is always equal to the magnitude of the displacement vector of a particle, (d) the average speed of a particle is either greater or equal to the magnitude of average velocity of the particle over the same interval of time, (e) Three vectors not lying in a plane can never add up to give a null vector.
Hint. For (b), remember that an 'x-component' can be described either as a bare number along an axis, or as a full vector pointing along that axis — the standard convention in this chapter treats it as the latter.
Step 1 — (a) Magnitude of a vector. True — magnitude is defined as a non-negative number describing size alone, regardless of which vector it came from, so it is always a scalar.

Step 2 — (b) Each component of a vector. False — a component of a vector (say, its part along the x-axis) is itself a vector, since it still has a definite direction (along that axis), not merely a bare magnitude.

Step 3 — (c) Path length versus displacement. False — the two are equal only when the particle never changes direction; in general the path length is strictly greater whenever the motion involves any reversal or curving.

Step 4 — (d) Average speed versus average velocity magnitude. True — since path length is always at least as large as the displacement magnitude, dividing both by the same time interval preserves the inequality.

Step 5 — (e) Three non-coplanar vectors. True — for three vectors to sum to a null vector, they must form a closed triangle when placed head to tail, and a triangle is necessarily a flat, planar figure, so vectors that do not all lie in one plane can never close up into such a triangle.

✦ Answer: (a) True (b) False (c) False (d) True (e) True.
Where students slip. Marking (b) True by picturing a component as 'just a number' like 5 or -3 — in the vector-algebra convention this chapter uses, a component is treated as the vector 5î or -3î, which still points somewhere, not a direction-free scalar.
3.64 marksNCERT Cl-11 Physics Part I, Ch3 Exercises, Q3.6
Establish the following vector inequalities geometrically or otherwise: (a) |a+b| <= |a| + |b| (b) |a+b| >= ||a| - |b|| (c) |a-b| <= |a| + |b| (d) |a-b| >= ||a| - |b||. When does the equality sign above apply?
Hint. Draw a and b head to tail to form a triangle with a+b as the third side, then apply the two standard triangle facts: no side exceeds the sum of the other two, and no side is smaller than their difference.
Step 1 — (a) Triangle law setup. Placing a and b head to tail forms a triangle whose third side is a+b. Since one side of a triangle can never exceed the sum of the other two, |a+b| ≤ |a|+|b|, with equality exactly when the 'triangle' flattens into a straight line — that is, when a and b point in the same direction.

Step 2 — (b) The same triangle, the other inequality. A triangle's side is also never smaller than the difference of the other two sides, so |a+b| ≥ ||a|-|b||, with equality when the triangle again flattens out, this time with a and b pointing in exactly opposite directions.

Step 3 — (c) Replacing b with -b in result (a). Since |-b| = |b|, applying (a) to a and (-b) gives |a + (-b)| ≤ |a| + |b|, i.e. |a-b| ≤ |a|+|b|, with equality when a and -b point the same way — meaning a and b themselves point in opposite directions.

Step 4 — (d) Replacing b with -b in result (b). Similarly, applying (b) to a and (-b) gives |a-b| ≥ ||a|-|b||, with equality when a and -b point in opposite directions — meaning a and b themselves point in the same direction.

✦ Answer: All four follow from the triangle inequality applied to a, b (and to a, -b for (c), (d)). Equality in (a) and (d) holds when a and b are parallel (same direction); equality in (b) and (c) holds when a and b are antiparallel (opposite directions).
Where students slip. Mixing up which pair of inequalities needs 'same direction' versus 'opposite direction' for equality — (a) and (d) both need a, b parallel, while (b) and (c) both need them antiparallel; getting this backwards is the single most common slip on this question.
3.73 marksNCERT Cl-11 Physics Part I, Ch3 Exercises, Q3.7
Given a + b + c + d = 0, which of the following statements are correct: (a) a, b, c, and d must each be a null vector, (b) The magnitude of (a + c) equals the magnitude of (b + d), (c) The magnitude of a can never be greater than the sum of the magnitudes of b, c, and d, (d) b + c must lie in the plane of a and d if a and d are not collinear, and in the line of a and d, if they are collinear?
Hint. Rearrange the given equation to isolate the two vectors named in each statement, then reason directly from that rearranged form.
Step 1 — (a) Must every vector be null? False — the four vectors only need to sum to zero collectively; four equal-length vectors along the sides of a square, for instance, sum to zero without any one of them being null.

Step 2 — (b) Magnitude of (a+c) vs (b+d). True — rearranging the given equation gives a+c = -(b+d), and taking magnitudes of both sides, |a+c| = |-(b+d)| = |b+d|, since a vector and its negative always share the same magnitude.

Step 3 — (c) Magnitude of a vs sum of the others. True — rearranging gives a = -(b+c+d), so |a| = |b+c+d| ≤ |b|+|c|+|d| by repeated application of the triangle inequality, meaning |a| can never exceed that sum.

Step 4 — (d) Where b+c must lie. True — rearranging gives b+c = -(a+d), which is a linear combination of only a and d, so it must lie in whatever plane a and d define (or along their common line, if a and d happen to be collinear).

✦ Answer: (a) False (b) True (c) True (d) True.
Where students slip. Rejecting (b) by assuming a+c and b+d must be equal vectors, not just equal in magnitude — the equation only forces them to be negatives of each other, which guarantees matching magnitude but generally opposite direction.
3.83 marksNCERT Cl-11 Physics Part I, Ch3 Exercises, Q3.8
Three girls skating on a circular ice ground of radius 200 m start from a point P on the edge of the ground and reach a point Q diametrically opposite to P following different paths as shown in Fig. 3.19. What is the magnitude of the displacement vector for each? For which girl is this equal to the actual length of path skate?
Hint. Displacement depends only on the start and end points, never on the path taken between them — so the curvy paths change nothing about the answer to the first part.
Step 1 — Displacement magnitude for all three. Since P and Q are diametrically opposite points on a circle of radius 200 m, the straight-line distance between them is the diameter: 2 × 200 = 400 m. This value depends only on where P and Q are, not on the path skated, so it is identical for all three girls.

Step 2 — Which girl matches path length to displacement. Of the three paths shown, only the one girl who skates directly along the straight diameter from P to Q (labelled B in the figure) covers a path exactly 400 m long — equal to her displacement. The other two girls trace curved, wandering paths that are visibly longer than the straight-line distance.

✦ Answer: Displacement magnitude is 400 m for all three girls. Only the girl skating straight along the diameter (B) has a path length equal to her displacement; the two girls on curved paths cover more than 400 m.
Where students slip. Assuming the girl with the shortest-looking curved path must have the smallest displacement — displacement is fixed at 400 m for all three regardless of path shape; only the total distance skated (path length) differs between them.
3.94 marksNCERT Cl-11 Physics Part I, Ch3 Exercises, Q3.9
A cyclist starts from the centre O of a circular park of radius 1 km, reaches the edge P of the park, then cycles along the circumference, and returns to the centre along QO as shown in Fig. 3.20. If the round trip takes 10 min, what is the (a) net displacement, (b) average velocity, and (c) average speed of the cyclist?
Hint. The figure shows P and Q a quarter-circle apart (radii OP and OQ at right angles), so the middle leg of the trip is one-quarter of the full circumference.
Step 1 — (a) Net displacement. The cyclist starts at O and, after the full loop O → P → Q → O, ends back at O — so the net displacement is zero.

Step 2 — (b) Average velocity. Average velocity = net displacement / time = 0 / 10 min = 0, regardless of how far the cyclist actually travelled.

Step 3 — (c) Average speed. Total path length = OP (a straight radius, 1000 m) + the quarter-circle arc PQ (¼ × 2π × 1000 ≈ 1570.8 m) + QO (another straight radius, 1000 m) = 2000 + 1570.8 ≈ 3570.8 m. Dividing by the 10 min (600 s) taken: average speed ≈ 3570.8/600 ≈ 5.95 m s⁻¹, which is about 21.4 km h⁻¹.

✦ Answer: Net displacement = 0; average velocity = 0; average speed ≈ 5.95 m s⁻¹ (≈ 21.4 km h⁻¹).
Where students slip. Reporting average speed as 0 just because average velocity is 0 — the cyclist still covered roughly 3.57 km of actual path, and average speed depends on that total distance, not on where the cyclist ends up relative to the start.
3.104 marksNCERT Cl-11 Physics Part I, Ch3 Exercises, Q3.10
On an open ground, a motorist follows a track that turns to his left by an angle of 60 degrees after every 500 m. Starting from a given turn, specify the displacement of the motorist at the third, sixth and eighth turn. Compare the magnitude of the displacement with the total path length covered by the motorist in each case.
Hint. A constant 60-degree left turn after every equal-length segment traces out the sides of a regular hexagon — use that shape's known symmetry instead of adding vectors from scratch each time.
Step 1 — Recognising the hexagon. Since the exterior angle of a regular hexagon is exactly 60°, six consecutive 500 m segments, each turning 60° left from the last, trace out a closed regular hexagon with 500 m sides.

Step 2 — After the 3rd turn. Three sides of the hexagon carry the motorist to the vertex diametrically opposite the start — the hexagon's long diagonal, which for a regular hexagon of side 500 m equals 2 × 500 = 1000 m. Path length covered = 3 × 500 = 1500 m.

Step 3 — After the 6th turn. Six segments complete the full hexagon and return the motorist exactly to the starting point, so displacement = 0. Path length covered = 6 × 500 = 3000 m.

Step 4 — After the 8th turn. Since 8 = 6 + 2, this is one complete hexagon loop (net effect: back to start) followed by 2 more segments, so the net displacement equals whatever it would be after just 2 segments from the start: two 500 m sides of a hexagon meeting at 60° give a resultant of 500√3 ≈ 866 m. Path length covered = 8 × 500 = 4000 m.

✦ Answer: 3rd turn: displacement 1000 m, path length 1500 m. 6th turn: displacement 0, path length 3000 m. 8th turn: displacement ≈ 866 m, path length 4000 m. In every case the path length is greater, since the motorist keeps changing direction.
Where students slip. Trying to recompute the 8th-turn displacement as if it were a fresh 8-sided figure — the key shortcut is that 8 segments equal one full closed hexagon (net zero) plus 2 extra segments, so the 8th-turn displacement is identical to the 2nd-turn displacement, not some new shape.
3.113 marksNCERT Cl-11 Physics Part I, Ch3 Exercises, Q3.11
A passenger arriving in a new town wishes to go from the station to a hotel located 10 km away on a straight road from the station. A dishonest cabman takes him along a circuitous path 23 km long and reaches the hotel in 28 min. What is (a) the average speed of the taxi, (b) the magnitude of average velocity? Are the two equal?
Hint. Average speed uses the distance actually driven; average velocity uses only the straight-line distance between start and end.
Step 1 — (a) Average speed. Average speed = total path length / time = 23 km / (28/60 h) = 23/0.4667 ≈ 49.3 km h⁻¹.

Step 2 — (b) Average velocity magnitude. Since the hotel is 10 km away in a straight line, average velocity magnitude = 10 km / (28/60 h) ≈ 21.4 km h⁻¹.

Step 3 — Comparing the two. The two are clearly not equal (49.3 km h⁻¹ vs 21.4 km h⁻¹), since the cabman's circuitous 23 km route is far longer than the straight 10 km distance the hotel actually sits at.

✦ Answer: Average speed ≈ 49.3 km h⁻¹; average velocity magnitude ≈ 21.4 km h⁻¹ — not equal, because the path taken was not a straight line.
Where students slip. Using the 23 km circuitous distance to compute average velocity too — average velocity only ever depends on the net straight-line displacement (10 km here), never on how winding the actual path was.
3.124 marksNCERT Cl-11 Physics Part I, Ch3 Exercises, Q3.12
The ceiling of a long hall is 25 m high. What is the maximum horizontal distance that a ball thrown with a speed of 40 m/s can go without hitting the ceiling of the hall?
Hint. For a fixed launch speed, range keeps increasing with angle right up to 45 degrees — so the best strategy is to use the steepest angle that still just grazes the 25 m ceiling, not to guess an angle directly.
Step 1 — Set up the height constraint. Maximum height reached = u²sin²θ / (2g). Setting this equal to the ceiling height: 25 = (40)²sin²θ / (2×9.8), so sin²θ = 25×2×9.8/1600 = 490/1600 = 0.30625, giving sinθ ≈ 0.5534 and θ ≈ 33.6°.

Step 2 — Why this boundary angle gives the maximum range. Range R = u²sin(2θ)/g keeps increasing with θ for any θ below 45°, and every angle allowed by the ceiling constraint here is below 45° (since the unconstrained 45° throw would reach u²/(4g) ≈ 40.8 m, well above the 25 m ceiling) — so the largest range under the constraint is reached exactly at the steepest angle the ceiling still allows.

Step 3 — Compute the range at that angle. cosθ = √(1−0.30625) ≈ 0.8329. R = u²×2sinθcosθ/g = 1600×2×0.5534×0.8329/9.8 ≈ 1475/9.8 ≈ 150.5 m.

✦ Answer: The ball can travel about 150 m horizontally without striking the 25 m ceiling.
Where students slip. Assuming the 45° angle (which gives maximum range in open space) is still the right choice here — at 45° this ball would rise to about 40.8 m, well above the 25 m ceiling, so a shallower angle is required, even though it means giving up the usual 'best' angle.
3.133 marksNCERT Cl-11 Physics Part I, Ch3 Exercises, Q3.13
A cricketer can throw a ball to a maximum horizontal distance of 100 m. How much high above the ground can the cricketer throw the same ball?
Hint. A 'maximum horizontal distance' is achieved specifically at a 45-degree launch angle — use that fact to first pin down the ball's launch speed before switching to a straight-up throw.
Step 1 — Extract the launch speed from the maximum range. Maximum range occurs at θ = 45°, where R = u²/g. So 100 = u²/g, meaning u²/g = 100 (a fixed property of how hard this cricketer can throw, independent of angle).

Step 2 — Apply the same speed to a straight-up throw. Thrown straight up (θ = 90°), the maximum height reached is H = u²sin²90°/(2g) = u²/(2g) = (u²/g)/2.

Step 3 — Substitute the known value. H = 100/2 = 50 m.

✦ Answer: The cricketer can throw the same ball to a maximum height of 50 m.
Where students slip. Trying to use 100 m directly as a height or as u² without dividing by g first — the useful fact carried over between the two throws is the ratio u²/g = 100, not the raw range value itself.
3.143 marksNCERT Cl-11 Physics Part I, Ch3 Exercises, Q3.14
A stone tied to the end of a string 80 cm long is whirled in a horizontal circle with a constant speed. If the stone makes 14 revolutions in 25 s, what is the magnitude and direction of acceleration of the stone?
Hint. Work out the angular speed from the revolution rate first, then use a = omega squared times r for centripetal acceleration.
Step 1 — Angular speed. Frequency = 14 revolutions / 25 s = 0.56 rev/s, so angular speed ω = 2π × 0.56 ≈ 3.52 rad/s.

Step 2 — Centripetal acceleration. a = ω²r = (3.52)² × 0.80 ≈ 12.39 × 0.80 ≈ 9.90 m s⁻².

✦ Answer: The acceleration is about 9.90 m s⁻² in magnitude, directed radially inward — that is, always along the string, toward the centre of the circle.
Where students slip. Reporting the direction as 'forward' or 'along the motion' — for circular motion at constant speed, acceleration always points toward the centre (perpendicular to the velocity), never along the direction of travel itself.
3.153 marksNCERT Cl-11 Physics Part I, Ch3 Exercises, Q3.15
An aircraft executes a horizontal loop of radius 1.00 km with a steady speed of 900 km/h. Compare its centripetal acceleration with the acceleration due to gravity.
Hint. Convert the speed to m/s before applying a = v squared over r, since mixing km/h with a radius in metres would give a meaningless result.
Step 1 — Convert the speed. 900 km h⁻¹ = 900 × 1000/3600 = 250 m s⁻¹.

Step 2 — Centripetal acceleration. a = v²/r = 250²/1000 = 62500/1000 = 62.5 m s⁻².

Step 3 — Compare with g. 62.5 / 9.8 ≈ 6.38, so the centripetal acceleration is about 6.4 times the acceleration due to gravity.

✦ Answer: Centripetal acceleration ≈ 62.5 m s⁻², which is about 6.4g.
Where students slip. Plugging in 900 (km/h) directly against a radius of 1.00 (km) without converting either to base SI units first — the two must share a consistent unit system (metres and seconds here) before dividing.
3.163 marksNCERT Cl-11 Physics Part I, Ch3 Exercises, Q3.16
Read each statement below carefully and state, with reasons, if it is true or false: (a) The net acceleration of a particle in circular motion is always along the radius of the circle towards the centre, (b) The velocity vector of a particle at a point is always along the tangent to the path of the particle at that point, (c) The acceleration vector of a particle in uniform circular motion averaged over one cycle is a null vector.
Hint. Part (a) is only about circular motion in general — check whether it still holds when the particle's speed is allowed to change along the circle, not just its direction.
Step 1 — (a) Net acceleration always radial? False in general — this only holds for uniform circular motion (constant speed). If the speed is also changing, there is an additional tangential acceleration component along the direction of motion, so the net acceleration is not purely radial.

Step 2 — (b) Velocity always tangential. True — this holds for any curved path whatsoever, not just circles: velocity at a point is always directed along the tangent to the trajectory at that instant.

Step 3 — (c) Acceleration averaged over one cycle of uniform circular motion. True — in uniform circular motion the (centripetal) acceleration vector continuously rotates in direction, always pointing toward the centre, and over one complete revolution it points equally in every direction around the circle, so its components cancel out symmetrically and the average over the full cycle is zero.

✦ Answer: (a) False (b) True (c) True.
Where students slip. Assuming (a) must be true simply because the question is inside a section about circular motion — 'circular motion' alone doesn't guarantee constant speed; only the qualifier 'uniform' (present in part (c) but absent in part (a)) guarantees a purely radial acceleration.
3.174 marksNCERT Cl-11 Physics Part I, Ch3 Exercises, Q3.17
The position of a particle is given by r = 3.0t i - 2.0t^2 j + 4.0 k m, where t is in seconds and the coefficients have the proper units for r to be in metres. (a) Find the v and a of the particle. (b) What is the magnitude and direction of velocity of the particle at t = 2.0 s?
Hint. Differentiate the position vector term by term with respect to t to get velocity, then differentiate again for acceleration — the constant 4.0 k term contributes nothing to either.
Step 1 — (a) Velocity by differentiating r(t). v(t) = dr/dt = 3.0 î − 4.0t ĵ m/s (the constant 4.0 k̂ term vanishes on differentiating, since a constant has zero rate of change).

Step 2 — (a) Acceleration by differentiating v(t). a(t) = dv/dt = −4.0 ĵ m/s², a constant vector — the particle's acceleration never changes with time here, since v's x-component is itself constant.

Step 3 — (b) Velocity at t = 2.0 s. v(2.0) = 3.0 î − 4.0(2.0) ĵ = 3.0 î − 8.0 ĵ m/s. Magnitude = √(3.0² + 8.0²) = √73 ≈ 8.54 m/s.

Step 4 — (b) Direction. Since the x-component is positive and the y-component is negative, the velocity points below the positive x-axis, at angle = tan⁻¹(8.0/3.0) ≈ 69.4° below it.

✦ Answer: v(t) = (3.0 î − 4.0t ĵ) m/s; a = −4.0 ĵ m/s² (constant). At t = 2.0 s: |v| ≈ 8.54 m/s, directed about 69.4° below the positive x-axis.
Where students slip. Forgetting that acceleration here is constant (independent of t) and instead trying to plug t = 2.0 s into a — differentiating −4.0t ĵ a second time leaves no t at all, since it's already linear in t after the first derivative.
3.184 marksNCERT Cl-11 Physics Part I, Ch3 Exercises, Q3.18
A particle starts from the origin at t = 0 s with a velocity of 10.0 j m/s and moves in the x-y plane with a constant acceleration of (8.0 i + 2.0 j) m/s^2. (a) At what time is the x-coordinate of the particle 16 m? What is the y-coordinate of the particle at that time? (b) What is the speed of the particle at the time?
Hint. Write x(t) and y(t) separately using s = ut + half a t squared along each axis on its own, since the x and y motions here don't interact.
Step 1 — Set up x(t) and y(t). With x0 = y0 = 0, initial velocity purely in y (10.0 m/s), and acceleration (8.0, 2.0) m/s², since x starts with zero initial velocity: x(t) = ½(8.0)t² = 4.0t², while y(t) = 10.0t + ½(2.0)t² = 10.0t + 1.0t².

Step 2 — (a) Solve for t when x = 16 m. 4.0t² = 16 → t² = 4 → t = 2.0 s.

Step 3 — (a) Find y at that time. y(2.0) = 10.0(2.0) + 1.0(2.0)² = 20.0 + 4.0 = 24.0 m.

Step 4 — (b) Find the speed at t = 2.0 s. vx(t) = 8.0t → vx(2.0) = 16.0 m/s. vy(t) = 10.0 + 2.0t → vy(2.0) = 10.0 + 4.0 = 14.0 m/s. Speed = √(16.0² + 14.0²) = √452 ≈ 21.3 m/s.

✦ Answer: x = 16 m at t = 2.0 s, when y = 24.0 m; speed at that time ≈ 21.3 m/s.
Where students slip. Forgetting the initial 10.0 m/s in the y-direction when computing y(t) — y's motion is not starting from rest the way x's is, since the particle already has an initial velocity entirely along y.
3.194 marksNCERT Cl-11 Physics Part I, Ch3 Exercises, Q3.19
i and j are unit vectors along x- and y- axis respectively. What is the magnitude and direction of the vectors i+j, and i-j? What are the components of a vector A = 2i + 3j along the directions of i+j and i-j?
Hint. To find a vector's component along a given direction, first turn that direction into a unit vector, then take the dot product.
Step 1 — Magnitude and direction of î+ĵ. |î+ĵ| = √(1²+1²) = √2. Since both components are equal and positive, it points at 45° from the x-axis, into the first quadrant.

Step 2 — Magnitude and direction of î−ĵ. |î−ĵ| = √(1²+(−1)²) = √2. With a positive x-component and negative y-component, it points at 45° below the x-axis (−45°), into the fourth quadrant.

Step 3 — Component of A along î+ĵ. The unit vector along î+ĵ is (î+ĵ)/√2. Component = A·(î+ĵ)/√2 = (2×1 + 3×1)/√2 = 5/√2 ≈ 3.54.

Step 4 — Component of A along î−ĵ. The unit vector along î−ĵ is (î−ĵ)/√2. Component = A·(î−ĵ)/√2 = (2×1 + 3×(−1))/√2 = −1/√2 ≈ −0.71.

✦ Answer: Both î+ĵ and î−ĵ have magnitude √2, at +45° and −45° from the x-axis respectively. A's component along î+ĵ is 5/√2 ≈ 3.54, and along î−ĵ is −1/√2 ≈ −0.71.
Where students slip. Skipping the division by √2 when finding A's components — the dot product with the un-normalised vector (î+ĵ) gives A's projection scaled by that vector's own length, not the true component, unless it's first turned into a unit vector.
3.203 marksNCERT Cl-11 Physics Part I, Ch3 Exercises, Q3.20
For any arbitrary motion in space, which of the following relations are true: (a) v_average = (1/2)(v(t1) + v(t2)) (b) v_average = [r(t2) - r(t1)]/(t2 - t1) (c) v(t) = v(0) + at (d) r(t) = r(0) + v(0)t + (1/2)at^2 (e) a_average = [v(t2) - v(t1)]/(t2 - t1). (The 'average' stands for average of the quantity over the time interval t1 to t2).
Hint. Ask which of these five are true purely by definition, holding for any motion whatsoever, and which secretly assume the acceleration stays constant.
Step 1 — Identify the pure definitions. Average velocity is, by definition, the total displacement divided by the total time, and average acceleration is, by definition, the total change in velocity divided by the total time — these definitions hold for absolutely any motion, however complicated.

Step 2 — Identify the assumptions built into the others. (a), (c), and (d) are all only valid for motion with constant (uniform) acceleration — (a) assumes velocity changes linearly with time, while (c) and (d) are the standard kinematic equations that were derived specifically under that assumption, and none of them need hold for a general, arbitrarily varying acceleration.

✦ Answer: Only (b) and (e) are true in general, since they are the actual definitions of average velocity and average acceleration; (a), (c), and (d) hold only for the special case of uniform (constant) acceleration.
Where students slip. Assuming all five must be true since they 'look like' standard kinematics formulas — (a), (c), (d) are specifically the constant-acceleration equations, which is a real restriction the question is testing, not a formality.
3.213 marksNCERT Cl-11 Physics Part I, Ch3 Exercises, Q3.21
Read each statement below carefully and state, with reasons and examples, if it is true or false: A scalar quantity is one that (a) is conserved in a process (b) can never take negative values (c) must be dimensionless (d) does not vary from one point to another in space (e) has the same value for observers with different orientations of axes.
Hint. Test each proposed property against a real, everyday scalar quantity that clearly breaks it — temperature is a useful one to try against several of these at once.
Step 1 — (a) Conserved in a process. False — many scalars are not conserved (distance travelled, temperature), and conversely some conserved quantities (momentum) are vectors, not scalars, so conservation has nothing to do with being scalar.

Step 2 — (b) Can never take negative values. False — many scalars can be negative, such as temperature in Celsius, electric charge, or work done against a force; only a few special scalars like mass or path length are inherently non-negative.

Step 3 — (c) Must be dimensionless. False — mass, energy, and temperature are all scalars but carry real physical dimensions; 'dimensionless' is a separate, stricter category (pure numbers and ratios), not a requirement for being scalar.

Step 4 — (d) Does not vary from point to point. False — a scalar can still vary across space, such as temperature or pressure at different points in a room; that variation makes it a scalar field, not a vector.

Step 5 — (e) Same value regardless of axis orientation. True — this is the actual defining property of a scalar: rotating the coordinate axes used to measure it changes nothing about its value, unlike a vector's individual components, which do change when the axes are rotated.

✦ Answer: (a) False (b) False (c) False (d) False (e) True — only (e) captures what actually makes a quantity a scalar.
Where students slip. Assuming mass or path length (both non-negative) as proof that (b) is generally true — a single well-behaved example doesn't establish a universal property; charge and temperature are enough to break the claim.
3.224 marksNCERT Cl-11 Physics Part I, Ch3 Exercises, Q3.22
An aircraft is flying at a height of 3400 m above the ground. If the angle subtended at a ground observation point by the aircraft positions 10.0 s apart is 30 degrees, what is the speed of the aircraft?
Hint. Picture the ground point directly beneath the midpoint of the aircraft's short flight path — the 30-degree angle then splits evenly into two right triangles, each with the height as one leg.
Step 1 — Set up the symmetric right triangle. With the ground observer directly below the midpoint of the short segment flown, the 30° angle splits into two equal 15° halves, each in a right triangle with height 3400 m and half of the horizontal distance d travelled as the opposite side.

Step 2 — Solve for the horizontal distance. tan(15°) = (d/2)/3400, so d = 2 × 3400 × tan(15°) ≈ 6800 × 0.2679 ≈ 1821.9 m.

Step 3 — Convert to speed. This distance was covered in 10.0 s, so speed = 1821.9/10.0 ≈ 182.2 m/s.

✦ Answer: The aircraft's speed is approximately 182 m/s.
Where students slip. Using the full 30° angle directly in a single right triangle against the full height — the height forms a right triangle with only half of the horizontal distance, matched against half of the given angle (15°), not the full 30°.

Solutions written by the tuition.in editorial team and checked against the NCERT Class 11 Physics Part I textbook, Reprint 2026-27 (keph103.pdf) — one end-of-chapter Exercises set (22 questions, 3.1-3.22); figures 3.19 and 3.20 and the exact vector-equation coefficients in Q3.17 and Q3.18 were rendered directly from the PDF at up to 400dpi and read visually/textually to verify before answering. Questions are referenced from the NCERT textbook for identification.

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