NCERT Solutions for Class 11 Physics Chapter 3: Motion in a Plane

Concept. A scalar has magnitude only and is added by ordinary algebra; a vector has both magnitude and direction and obeys the triangle (parallelogram) law of addition. To classify a quantity ask: ‘Does it need a direction to be completely specified?’ and ‘Is it defined by a vector operation (displacement, velocity-change, cross product)?’

1. Scalars: volume, mass, speed, density, number of moles, angular frequency.
2. Vectors: acceleration, velocity, displacement, angular velocity.

Pairs worth remembering: distance (scalar) – displacement (vector); speed (scalar) – velocity (vector); angular frequency (scalar) – angular velocity (vector). A vector’s magnitude is always a scalar: |velocity| = speed.

Extra depth. Angular velocity is a vector because for a rotating body we assign the direction along the axis of rotation using the right-hand rule (it is called an axial vector; Chapter 6). Angular frequency of a simple harmonic oscillator, in contrast, is only the number ω = 2π/T and is a scalar. Quantities with dimensions are not scalars merely because they have no direction in daily language: electric current has a sense but is a scalar since it does not obey the vector addition law (Ex 3.2).

Conclusion: Scalars: volume, mass, speed, density, number of moles, angular frequency. Vectors: acceleration, velocity, displacement, angular velocity.

Concept. Go through the list and test each quantity: force, linear momentum, electric field, average velocity and relative velocity are clearly vectors. Angular momentum (r × p) and magnetic moment are also vectors (axial). That leaves work and current.

Why is current a scalar? To be a vector a quantity must follow the parallelogram law. If two wires carrying currents I₁ and I₂ meet at a junction and one wire carries the combined current away, its value is I₁ + I₂ whatever the angle between the wires, which is not what the parallelogram law (R² = A² + B² + 2AB cosθ) would give. So current has a direction sign (+ / −) but is a scalar. Work is a scalar because it is the scalar (dot) product of force and displacement.

Extra depth. Mnemonic: a scalar results from dot product (work, power, flux) and from ratios of scalars; a vector results from cross product (torque, angular momentum, magnetic force). Work W = F·s is the dot product so it is a scalar even though F and s are vectors. Likewise power P = F·v. Electric current density J, in contrast, is a vector, although the current through a surface is a scalar (I = ∫J·dS).

More to know — classifying common quantities. Scalars: mass, length (as a magnitude), time, temperature, energy, work, power, charge, current, potential, density, pressure, speed. Vectors: displacement, velocity, acceleration, force, momentum, impulse, torque, angular momentum, electric and magnetic fields, area vector (for flux), magnetic moment. Remember that the sign of a scalar (e.g. charge ±, potential energy) is not a direction: scalars add algebraically, so 3 C + (−2 C) = 1 C regardless of any angle.

Conclusion: The two scalar quantities are work and current.

Concept. Temperature, pressure, time, power, total path length, energy, gravitational potential, coefficient of friction and charge are all scalars: they need only a magnitude (with sign where applicable). Impulse is the change in momentum, J = Δp = FΔt, and momentum is a vector, so impulse is a vector with the direction of the force.

A subtle case: pressure. Although pressure is defined with a force (a vector) it is a scalar: the force on an area is p A n̂ directed along the normal, but the pressure itself p = F_⊥/A does not carry direction and in a fluid it is the same in all directions. Similar subtle cases are surface tension (scalar, force per length) and current (scalar).

Extra depth. The direction of impulse is the direction of the change in momentum: when a cricket ball moving at 30 m s⁻¹ is hit back at 30 m s⁻¹, Δp = m(v_f − v_i) = −2mv_i… i.e. twice the initial momentum, opposite to the original motion, and so is the direction of the impulse. In Chapter 5 you will see that impulse equals the area under the force–time graph.

More to know — area as a vector. The area of a surface can be treated as a vector A = A n̂ directed along the outward normal; then flux = B·A is a scalar. The only vector in the given list is impulse, but note that many quantities that look like scalars in daily life (pressure, current, surface tension) are scalars even though they are linked to forces or flow. Torque τ = r × F, angular momentum, velocity and force are the standard NCERT vectors of mechanics.

Conclusion: The only vector quantity is impulse.

Concept. An operation is physically meaningful only if it gives a result with a definite physical meaning. Addition and subtraction make sense only between quantities of the same physical kind (same dimensions and units) and the same type (scalar with scalar, vector with vector). Multiplication of quantities of different kinds is allowed and gives a new kind of quantity.

• Meaningful ones: (a) with same dimensions, (c), (d), (e) with same kind, (f).
• Not meaningful: (b), and (a)/(e) when the quantities are of different kinds.

Rule of thumb. ‘You cannot add apples and oranges’: addition needs the same kind of quantity; multiplication and division do not.

Extra depth. Dimensional homogeneity (Chapter 1) is the same principle: every term of a physical equation must have the same dimensions and the same tensor character (scalar with scalar, vector with vector). For instance, v = u + at is a vector equation: every term is a vector, and F = ma equates two vectors. An equation like E = mc² + v would be meaningless because it equates a scalar with a vector.

Conclusion: (a) yes if the same physical quantity; (b) no; (c) yes; (d) yes; (e) yes if same kind; (f) yes.

Concept. This question tests the definitions of scalars, vectors, components, displacement and average speed. Use the rules: the magnitude of a vector is a positive real number; a component A_x î (the resolved vector) is itself a vector; path length ≥ |displacement|; and a closed vector polygon needs vectors that lie in the same plane when only three vectors are used.

1. Example for (c): an athlete runs one lap of a 400 m track: path length 400 m, displacement 0.
2. Example for (e): take i, j and k: i + j + k has magnitude √3, not zero; in general a + b lies in the plane of a and b, while c has a part perpendicular to that plane which nothing can cancel. A null sum of three vectors is possible only when they are coplanar (they then form a triangle).

Note on (b). The textbook says ‘A_x is itself not a vector, but A_xî is a vector’. In the exercise the ‘component’ means the component vector, and the published answer is ‘false’; if by ‘component’ one means just the signed number A_x, it is a scalar. Be clear in the answer which meaning you use.

Extra depth. Part (e) generalises: n vectors can add to zero only if they form a closed polygon. Three vectors form a closed triangle, which lies in a plane; therefore three non-coplanar vectors cannot close. Four non-coplanar vectors can close (a skew quadrilateral in space), as in Ex 3.7 where a + b + c + d = 0 and the vectors need not be coplanar. Part (d) is the vector version of Ex 2.9 from Chapter 2.

Conclusion: (a) True; (b) False (the component vector is a vector); (c) False; (d) True; (e) True.

Concept. Draw vectors a and b head to tail. Their resultant a + b, together with a and b, forms the three sides of a triangle. The triangle inequalities of geometry then give all four results: any side of a triangle is less than the sum and greater than the difference of the other two sides. When the triangle collapses to a straight line (a and b collinear) the inequalities become equalities.

Proof by the triangle (geometrical)

1. (a) In triangle OPQ with OP = a, PQ = b, OQ = a + b: the side OQ is shorter than the sum of the other two sides, so |a + b| < |a| + |b|. If a and b point in the same direction the triangle degenerates into a straight line and |a + b| = |a| + |b|, i.e. |a + b| ≤ |a| + |b|.
2. (b) In the same triangle, any side is longer than the difference of the other two: |a + b| > ||a| − |b||. If a and b are antiparallel (opposite direction, collinear) the triangle collapses and |a + b| = ||a| − |b||; so |a + b| ≥ ||a| − |b||.
3. (c) Write a − b = a + (−b). Since |−b| = |b|, (a) gives |a − b| = |a + (−b)| ≤ |a| + |−b| = |a| + |b|. Equality holds when a and (−b) have the same direction, i.e. when a and b are antiparallel.
4. (d) From (b) with b replaced by −b: |a − b| ≥ ||a| − |−b|| = ||a| − |b||. Equality holds when a and (−b) are antiparallel, i.e. when a and b are parallel (same direction).

Algebraic check using the law of cosines

|a + b|² = a² + b² + 2ab cosθ and |a − b|² = a² + b² − 2ab cosθ, where θ is the angle between a and b. Since −1 ≤ cosθ ≤ 1: (a − b)² ≤ |a ± b|² ≤ (a + b)² for the sum, giving ||a| − |b|| ≤ |a + b| ≤ |a| + |b| (cosθ = 1 at θ = 0° for the upper limit; cosθ = −1 at θ = 180° for the lower limit), and similarly ||a| − |b|| ≤ |a − b| ≤ |a| + |b|.

Extra depth. Physical meaning: the magnitude of a vector sum lies between the sum and the difference of the magnitudes: two forces of 5 N and 3 N can have a resultant anywhere from 2 N (opposite) to 8 N (same direction), and the resultant for any other angle lies between these limits. This is why a boat can sometimes make progress against a current only if its own speed exceeds the current: |v_b − v_c| ≥ ||v_b| − |v_c||.

Conclusion: ||a| − |b|| ≤ |a ± b| ≤ |a| + |b|; equality when a and b are collinear (parallel for a + b upper and a − b lower; antiparallel for the others).

Concept. If a + b + c + d = 0 the four vectors form a closed polygon (a, b, c, d placed head to tail return to the starting point). We rearrange the relation and use the triangle inequality to test each statement.

1. Rearranging: a + c = −(b + d) ⇒ (b) is true.
2. a = −(b + c + d): the magnitude of a sum of three vectors cannot exceed the sum of their magnitudes ⇒ (c) is true.
3. b + c = −(a + d): lies in the plane of a and d (any two non-collinear vectors define a plane; if a, d are collinear the sum is along that line) ⇒ (d) is true.
4. (a) is false (counter-example: a, b, c, d the four sides of a rhombus: each non-zero but sum zero).

Correct statements: (b), (c) and (d).

Extra depth. The four vectors need not all be in one plane: a + b + c + d = 0 can represent a closed skew quadrilateral in space. Only the pairs (a, d) and (b, c) are guaranteed coplanar with their sums. The result (c) is the generalisation of Ex 3.6(a) to many vectors: the magnitude of the resultant of many vectors never exceeds the sum of the magnitudes (maximum when all point the same way), and the magnitude of one vector in a closed polygon never exceeds the sum of the others (the longest side of a polygon is shorter than the sum of the others).

Conclusion: Statements (b), (c) and (d) are correct; (a) is wrong.

Concept. The displacement is the straight vector from the initial point P to the final point Q, independent of the path. Since Q is diametrically opposite P, PQ is a diameter of the circle of radius R = 200 m, so |PQ| = 2R for all three girls.

1. Radius of the ground R = 200 m, so the diameter PQ = 2R = 400 m.
2. Displacement of each girl = vector PQ: magnitude 400 m (the same for all three), direction from P to Q.
3. Girl A and girl C follow curved paths of length larger than 400 m (the straight line is the shortest path); so the actual path length for A and C is greater than the magnitude of displacement.
4. Girl B moves along the diameter, a straight line without any change of direction: her path length is 400 m = |displacement|.

Note. The semicircle PQ, if followed, would have length πR = 628 m; none of the paths drawn is a perfect semicircle, but each is longer than 400 m. The average velocity of each girl over her trip is displacement/time, same direction PQ, and the magnitude depends on the time taken.

Extra depth. This is a direct application of the property that |displacement| ≤ path length with equality only for straight-line motion without turning back (Ex 3.5(c) and Ex 2.9). The displacement depends on the end points only, which is why the three girls have the same displacement though their paths differ (Fig. 3.1b).

Conclusion: The displacement is 400 m for each girl; it equals the path length only for girl B.

Concept. The cyclist begins and ends at the same point O, so the net displacement is zero and hence so is the average velocity; but he has covered a definite path length (OP + arc PQ + QO), so the average speed is not zero.

1. (a) Net displacement = final position − initial position = O − O = 0.
2. (b) Average velocity = net displacement/time = 0/(10 min) = 0.
3. (c) Path length: OP = R = 1 km; arc PQ = a quarter of the circumference (P and Q are 90° apart in the figure) = (1/4)(2πR) = (π/2)(1 km) = 1.57 km; QO = 1 km. Total = 1 + 1.57 + 1 = 3.57 km.
4. Time = 10 min = 10/60 h = 1/6 h. Average speed = 3.57 km/(1/6 h) = 21.4 km/h (= 5.95 m/s).

Remark. If P and Q had been at a different angle φ on the circle the arc would be (φ/360°)(2πR); the figure shows P and Q at right angles to each other (OP is horizontal and OQ vertical), as assumed in the standard solution.

Extra depth. This exercise shows again why average speed is defined via path length: with a zero average velocity the cyclist would have ‘average speed zero’, which is absurd. Also compare the numerical value with typical cycling speeds: 21.4 km h⁻¹ is a brisk but plausible speed for a trained cyclist.

Conclusion: (a) 0; (b) 0; (c) 3.57 km/(1/6 h) = 21.4 km h⁻¹.

Concept. Turning left by 60° after every 500 m means that the motorist traces the sides of a regular hexagon of side 500 m (exterior angle 60°, interior angle 120°). The hexagon is completed after 6 sides (6 turns) and the motorist is back at the starting turn. Displacement = vector from the starting turn to the present position; the vertices are a distance of 500 m (the circumradius of the hexagon equals the side length) from the centre.

1. Third turn: after three sides the motorist is at the vertex opposite the start; the displacement is the diameter 2 × 500 m = 1000 m.
2. Sixth turn: after a full circuit he is back at the start: displacement 0.
3. Eighth turn: position is the same as after two sides. Chord of two sides: d² = 500² + 500² + 2(500)(500)cos60° (angle between the sides is 60° as direction vectors) = 3(500)², so d = 500√3 = 866 m, at 30° to the first side.

Check of the hexagon geometry. The two sides after the first turn make the direction vectors differ by 60°; the vector sum has magnitude 2 × 500 cos30° = 866 m, along the bisector, i.e. 30° from the first side (towards the left, the direction of turning). After four, five sides the displacement magnitudes are 866 m and 500 m respectively.

Extra depth. General result: for a regular polygon track, the displacement after n sides is 2R sin(nπ/N) where N is the number of sides and R the circumradius (here N = 6, R = 500 m): n = 3 gives 2(500) sin90° = 1000 m; n = 6 gives 0; n = 8 (equivalent to n = 2) gives 2(500) sin60° = 866 m. The path length always exceeds or equals the displacement, with strict inequality whenever the path turns.

Conclusion: 3rd turn: 1000 m (path 1500 m); 6th turn: 0 (path 3000 m); 8th turn: 866 m at 30° to the initial direction (path 4000 m).

Concept. The station and the hotel are 10 km apart in a straight line, so the displacement of the taxi is 10 km, but the path length is 23 km. Average speed uses path length, average velocity uses displacement; both are divided by the same time interval, 28 min = 28/60 h = 0.467 h.

1. Time t = 28 min = 28/60 h = 0.4667 h.
2. (a) Average speed = total path length / time = 23 km / 0.4667 h = 49.3 km h⁻¹ (13.7 m s⁻¹).
3. (b) Magnitude of average velocity = |displacement| / time = 10 km / 0.4667 h = 21.4 km h⁻¹ (5.95 m s⁻¹).
4. The two are not equal: average speed (49.3 km h⁻¹) > |average velocity| (21.4 km h⁻¹), because the path length (23 km) is greater than the displacement (10 km).

Meaning. The ratio 2.3 is exactly the ratio of path length to displacement; the cabman’s detour made the journey 2.3 times longer than necessary. A taximeter measures path length (the fare is charged on it), while the passenger’s displacement is what matters for reaching the hotel.

Extra depth. The two quantities would be equal only if the cab had gone straight along the 10 km road without reversing; then both would be 10/0.467 = 21.4 km h⁻¹. The inequality average speed ≥ |average velocity| holds always (Ex 2.9 and Ex 3.5d). Note also that the time of 28 min for 23 km corresponds to a speed of 49.3 km h⁻¹, a plausible taxi speed, while the same time for 10 km would give only 21.4 km h⁻¹, which the cabman’s detour disguises.

More to know — average speed versus average velocity in everyday life. A car’s odometer measures path length; a GPS ‘as the crow flies’ distance measures displacement. For a round trip the average velocity is zero but the average speed equals the total distance divided by the total time. Many exam questions (Ex 2.10, 3.9, 3.11) test this distinction: always identify whether the quantity asked is a speed (uses path length, scalar) or a velocity (uses displacement, vector).

Conclusion: (a) 49.3 km h⁻¹; (b) 21.4 km h⁻¹; the two are not equal.

Concept. The ball must stay below the ceiling, so its maximum height must not exceed 25 m. For a projectile, h_m = (v₀ sinθ)²/2g and R = v₀² sin2θ/g. We find the largest angle allowed by the ceiling and then compute the range for that angle. (Range increases with θ up to 45°, so we need the biggest θ that clears the ceiling, if it is below 45°.)

1. Condition: h_m = v₀² sin²θ/(2g) = 25 m with v₀ = 40 m s⁻¹, g = 9.8 m s⁻².
2. sin²θ = 2g h_m/v₀² = 2(9.8)(25)/(40)² = 490/1600 = 0.3063 ⇒ sinθ = 0.5534 ⇒ θ = 33.6°.
3. cosθ = √(1 − 0.3063) = 0.8329; sin2θ = 2 sinθ cosθ = 2 × 0.5534 × 0.8329 = 0.9219.
4. R = v₀² sin2θ/g = (1600)(0.9219)/9.8 = 150.5 m ≈ 150 m.

Why not 45°? At 45° the height would be (40 sin45°)²/(2 × 9.8) = 800/19.6 = 40.8 m, more than 25 m: the ball would hit the ceiling. At the smaller angle 33.6° the ball just grazes the ceiling and the range, 150 m, is the largest possible without hitting it (a smaller angle gives smaller range; the maximum range with no ceiling would be v₀²/g = 163 m).

Extra depth. Alternative approach: use the relation between R and h_m, h_m/R = tanθ/4 (since tanθ = 4h_m/R): tan 33.6° = 0.665 and R = 4 h_m/tanθ = 100/0.665 = 150 m, which confirms the answer. The same method answers questions about the clearance of walls or goalposts. Note that the problem implicitly assumes a ceiling of the long hall extends beyond the range and that the ball is thrown from ground level.

Conclusion: The maximum horizontal distance is about 150 m (angle of projection 33.6°).

Concept. The cricketer can give the ball a fixed maximum speed u (the same for any direction). The horizontal range is maximum when the ball is thrown at 45°, R_max = u²/g = 100 m. The greatest height is attained by throwing the ball vertically upward (θ = 90°) with the same speed u, giving H = u²/2g.

1. Maximum range: R_max = u²/g = 100 m ⇒ u² = 100g = 980 m² s⁻² (u = 31.3 m s⁻¹).
2. For vertical projection (θ = 90°) the maximum height H = u²/(2g).
3. H = (100 g)/(2g) = 50 m.
4. Hence the cricketer can throw the ball to a height of 50 m, i.e. half the maximum range.

Note that the result is independent of g: H_max = R_max/2 for any planet, because both are proportional to u²/g. On the Moon (g smaller) both would be six times larger, but H would still be half of R.

Extra depth. The speed of throw is the same in both cases: u = √(100 g) = 31.3 m s⁻¹ (about 113 km h⁻¹), realistic for a fast bowler’s arm. For other angles H = u² sin²θ/2g = 50 sin²θ m and R = 100 sin2θ m, so, e.g. at 30° the height is 12.5 m and the range 86.6 m. The relation H_max = R_max/2 is a handy result for numerical questions.

More to know — the general range and height relations. For a throw of speed u at angle θ: R = u² sin2θ/g and H = u² sin²θ/2g, so H/R = tanθ/4. At θ = 45° H = R/4 = 25 m (for R = 100 m) while the maximum height (θ = 90°) is 50 m. The ball thrown at 45° stays in the air for T = 2u sin45°/g = 4.5 s and in a vertical throw for 2u/g = 6.4 s. These relations are useful for every sport involving throws (javelin, shot put, cricket).

Conclusion: The cricketer can throw the ball to a maximum height of 50 m.

Concept. The stone performs uniform circular motion of radius R = 80 cm = 0.80 m. Its acceleration is the centripetal acceleration a_c = ω² R = 4π² ν² R, with magnitude constant, and directed along the string towards the centre of the circle at every instant.

1. Frequency ν = number of revolutions/time = 14/25 s = 0.56 s⁻¹; period T = 25/14 = 1.786 s.
2. Angular speed ω = 2πν = 2 × 3.14 × 0.56 = 3.52 rad s⁻¹.
3. Speed v = ωR = 3.52 × 0.80 = 2.81 m s⁻¹.
4. Acceleration a_c = ω² R = (3.52)² × 0.80 = 12.38 × 0.80 = 9.9 m s⁻² (equivalently v²/R = 7.9/0.8 = 9.9 m s⁻²).
5. Direction: along the radius, from the stone towards the centre of the circle (i.e. along the string towards the hand), in the horizontal plane; it changes continuously as the stone moves.

Interesting numerical coincidence: a_c = 9.9 m s⁻² is almost exactly g; the tension in the string is therefore about equal to the weight of the stone (T = m a_c = mg approximately) if the string were horizontal (in practice it must slope slightly because gravity must be balanced).

Extra depth. The acceleration vector has a constant magnitude but a changing direction (always pointing to the centre), so it is not a constant vector, and so the kinematic equations of constant acceleration do not apply to circular motion. The direction of the velocity is tangent to the circle, perpendicular to the acceleration. If the string breaks, the stone flies off along the tangent with speed 2.8 m s⁻¹ (no centripetal force remains).

More to know — tension in the string. The centripetal force that produces the acceleration is supplied by the tension in the string: T = m a_c = mω² R. For a stone of 100 g, T = 0.1 × 9.9 = 0.99 N. If the whirling speed doubles, the acceleration becomes four times (39.6 m s⁻²) and the tension also four times; this is why strings break at high speeds. Also note the period T_period = 1.79 s and that the stone covers 2πR = 5.03 m per revolution, i.e. 14 revolutions cover 70.4 m in 25 s (speed 2.8 m s⁻¹).

Conclusion: The acceleration is 9.9 m s⁻², directed towards the centre of the circle along the string.

Concept. In a horizontal loop at steady speed v the aircraft undergoes uniform circular motion with centripetal acceleration a_c = v²/R. We convert the speed to SI units (m s⁻¹) and compare a_c with g = 9.8 m s⁻².

1. v = 900 km h⁻¹ = 900 × (5/18) m s⁻¹ = 250 m s⁻¹.
2. R = 1.00 km = 1000 m.
3. a_c = v²/R = (250)²/1000 = 62500/1000 = 62.5 m s⁻².
4. Ratio a_c/g = 62.5/9.8 = 6.4.

Interpretation. The centripetal acceleration is about 6.4 times g, i.e. the pilot experiences a force of about 6.4 times his weight directed towards the centre of the loop (because gravity acts perpendicular to the horizontal loop, his apparent weight is about √(6.4² + 1) ≈ 6.5 times his normal weight). Such accelerations are close to the limits that a trained pilot can withstand without blacking out (typical limit 5–9 g with special suits).

Extra depth. Because a_c ∝ v²/R, doubling the speed multiplies the acceleration by four; halving the radius doubles it. In fighter-jet manoeuvres engineers quote the load factor in ‘g’; tight turns at high speeds quickly lead to loads > 5 g, which is why pilots wear anti-g suits and why aircraft wings are designed for such loads.

More to know — g-force and human tolerance. A load of 6.4 g means the seat pushes on the pilot with a force 6.4 times his weight (plus 1 g for gravity). Untrained people begin to lose vision (grey-out) at about 4–5 g sustained; fighter pilots use g-suits and breathing techniques to tolerate up to about 9 g briefly. Roller-coaster loops are designed for about 3–4 g. The formula a_c = v²/R shows why a tight loop (small R) at high v is dangerous: halving the radius at the same speed doubles the g-load.

Conclusion: The centripetal acceleration is 62.5 m s⁻², about 6.4 times the acceleration due to gravity.

Concept. In circular motion the velocity is always tangent to the circle. The acceleration has a centripetal component v²/R (changing the direction of velocity) and, if the speed changes, a tangential component dv/dt (changing the magnitude of velocity). Only in uniform circular motion (constant speed) is the tangential component zero.

1. (a) a = a_c + a_t; a_c = v²/R (towards the centre), a_t = dv/dt (along the tangent). The net direction is towards the centre only if a_t = 0.
2. (b) Use v = dr/dt: the velocity is directed along the instantaneous direction of motion, i.e. along the tangent.
3. (c) ā = Δv/Δt with Δv = v(T) − v(0) = 0 for a full period T. Also, summing the inward vectors −ω² r over a full circle gives zero by symmetry.

Examples. (a) false: a swing at the bottom of its arc speeding up; (b) true for a car taking a bend: the headlight points along the tangent; (c) true for the Moon or an electron in a circular orbit.

Extra depth. Part (c) is a vector statement; the magnitude of the acceleration averaged over a cycle is not zero: the average of |a| equals v²/R. The average of the vector is zero because the direction rotates through 360°. A similar vector argument shows that the average displacement of a particle in uniform circular motion over one cycle is zero, and so is its average velocity. This is a good example of how a quantity can have non-zero magnitude at each instant but zero vector average.

Conclusion: (a) False (only for uniform motion); (b) True; (c) True.

Concept. When position is a function of time, velocity v = dr/dt and acceleration a = dv/dt are found by differentiating each component with respect to time (the unit vectors î, ĵ, k̂ are constants). The magnitude of a vector is the square root of the sum of the squares of its components and the direction is given by tanθ = v_y/v_x.

1. (a) Velocity: v = dr/dt = d(3.0t)/dt î + d(−2.0t²)/dt ĵ + d(4.0)/dt k̂ = (3.0 î − 4.0 t ĵ) m s⁻¹ (the k term is a constant so its derivative is zero).
2. Acceleration: a = dv/dt = −4.0 ĵ m s⁻² (constant, directed along the negative y-axis).
3. (b) At t = 2.0 s: v = 3.0 î − 4.0(2.0) ĵ = 3.0 î − 8.0 ĵ m s⁻¹.
4. Magnitude |v| = √(3.0² + 8.0²) = √73 = 8.5 m s⁻¹.
5. Direction: tanθ = v_y/v_x = −8.0/3.0 = −2.67 ⇒ θ = tan⁻¹(−2.67) = −69.4°, i.e. 69.4° below the +x axis (in the x–y plane, fourth quadrant).

Note. Because the acceleration is constant, the motion is projectile-like: along x uniform motion at 3.0 m s⁻¹, along y uniformly accelerated with a = −4.0 m s⁻², and z fixed at 4.0 m (the particle moves in the plane z = 4.0 m).

Extra depth. In Example 3.4 of the text the position r = 3.0t î + 2.0t² ĵ + 5.0 k̂ is handled in the same way. The path is the parabola y = −(2/9) x² (eliminate t between x = 3t and y = −2t²), and the velocity at any time is tangent to it. At t = 0 the velocity is purely along x (3.0 m s⁻¹) and the speed increases with time as √(9 + 16t²).

Conclusion: v = (3.0î − 4.0tĵ) m s⁻¹, a = −4.0ĵ m s⁻²; at t = 2 s speed 8.5 m s⁻¹ at 69.4° below the +x axis.

Concept. The motion in a plane with constant acceleration separates into two independent one-dimensional motions with constant acceleration: along x with v_0x = 0, a_x = 8.0 m s⁻² and along y with v_0y = 10.0 m s⁻¹, a_y = 2.0 m s⁻². We use x = x₀ + v_0xt + ½a_x t² and similarly for y, v = v₀ + at.

1. x-motion: x = 0 + 0 + ½(8.0)t² = 4.0 t². Setting x = 16 m: 4.0 t² = 16 ⇒ t² = 4 ⇒ t = 2.0 s (negative root rejected).
2. y-motion: y = v_0y t + ½ a_y t² = 10.0(2.0) + ½(2.0)(2.0)² = 20 + 4 = 24 m.
3. Velocity at t = 2.0 s: v_x = v_0x + a_x t = 0 + 8.0 × 2.0 = 16 m s⁻¹; v_y = v_0y + a_y t = 10.0 + 2.0 × 2.0 = 14 m s⁻¹.
4. Speed = √(v_x² + v_y²) = √(16² + 14²) = √(256 + 196) = √452 = 21.3 m s⁻¹ (≈ 21 m s⁻¹).

Direction of the velocity at that instant: tanθ = 14/16 = 0.875 ⇒ θ = 41.2° with the x-axis. The vector form: r(2) = 16î + 24ĵ m, v(2) = 16î + 14ĵ m s⁻¹.

Extra depth. Compare with Example 3.5 in the text (v₀ = 5.0 î, a = 3.0 î + 2.0 ĵ), which is solved in the same way: first find t from the x-equation (a quadratic there), then substitute in the y-equation. In general the ‘independent motions’ principle is the key to every problem on two-dimensional motion with constant acceleration, including projectile motion where a_x = 0 and a_y = −g.

More to know — projectile-motion connection. Here the acceleration (8.0î + 2.0ĵ) is constant, so the path is a parabola, but the axis of the parabola is along the acceleration vector, which is not vertical. For a true projectile, a = −gĵ. The speed increases with time from 10 m s⁻¹ at t = 0 to 21.3 m s⁻¹ at t = 2 s, because the acceleration has a component along the velocity throughout. The distance from the origin at t = 2 s is |r| = √(16² + 24²) = 28.8 m.

Conclusion: (a) t = 2.0 s and y = 24 m; (b) speed = √452 ≈ 21 m s⁻¹.

Concept. A vector’s magnitude is √(sum of squares of components); its direction is the angle with the x-axis, tanθ = A_y/A_x. To find the component of A along a direction, resolve A along the unit vector n̂ of that direction using the dot product, A · n̂; it equals the projection of A on that line. Since î + ĵ and î − ĵ are perpendicular, they form a rotated pair of axes.

(i) Magnitude and direction of î + ĵ and î − ĵ

1. |î + ĵ| = √(1² + 1²) = √2 = 1.41; tanθ = 1/1 = 1 ⇒ θ = 45° with the +x axis (towards +y).
2. |î − ĵ| = √(1² + (−1)²) = √2; tanθ = −1 ⇒ θ = −45° (i.e. 315°): 45° below the +x axis.
3. The two vectors are perpendicular: (î + ĵ)·(î − ĵ) = 1 − 1 = 0.

(ii) Components of A = 2î + 3ĵ along the two directions

1. Unit vectors: n̂₁ = (î + ĵ)/√2 and n̂₂ = (î − ĵ)/√2.
2. Component along (î + ĵ): A · n̂₁ = (2 × 1 + 3 × 1)/√2 = 5/√2 = 3.54 (as a vector: (5/2)(î + ĵ)).
3. Component along (î − ĵ): A · n̂₂ = (2 × 1 + 3 × (−1))/√2 = −1/√2 = −0.71 (as a vector: −(1/2)(î − ĵ)).
4. Check: (5/2)(î + ĵ) − (1/2)(î − ĵ) = (2.5 − 0.5)î + (2.5 + 0.5)ĵ = 2î + 3ĵ = A ✓. Also |A|² = 13 = (5/√2)² + (1/√2)² = 12.5 + 0.5 ✓.

Extra depth. Geometrically, the problem rotates the coordinate axes by 45°: in the new axes A has components (3.54, −0.71). The negative sign means A has a projection opposite to î − ĵ. Components depend on the choice of axes but the magnitude |A| = √13 = 3.61 does not; this is the idea behind invariance of vectors under rotation of axes. You may check graphically by drawing A on graph paper and dropping perpendiculars on the two diagonals.

Conclusion: |î + ĵ| = |î − ĵ| = √2 at +45° and −45°; components of A: 5/√2 (3.54) along î + ĵ and −1/√2 (−0.71) along î − ĵ.

Concept. Relations (b) and (e) are definitions of average velocity and average acceleration and hold for any motion. Relations (a), (c) and (d) are consequences of constant acceleration (uniformly accelerated motion); they fail when the acceleration changes with time, as in an arbitrary motion.

1. For arbitrary motion (variable a) use the calculus forms: v(t) = v(0) + ∫₀^t a dt, r(t) = r(0) + ∫₀^t v dt.
2. The ‘true’ relations are exactly the definitions that involve no assumption about the motion.

Example showing (a) false. Let v(t) = 3t² m s⁻¹ from t = 0 to t = 2 s: average velocity = (1/2)∫₀² 3t² dt = (1/2)(8) = 4 m s⁻¹, but (1/2)[v(0) + v(2)] = (1/2)(0 + 12) = 6 m s⁻¹, different. With constant acceleration (v = 3t) both give 3 m s⁻¹.

Extra depth. This exercise reinforces point 3 of ‘Points to ponder’: the vector equations (3.33a) and (3.34a) hold only for constant acceleration. The definitions (b) and (e), and the definitions v = dr/dt and a = dv/dt, are always valid. In projectile motion a = −g is constant, so (a), (c), (d) are valid there; in circular motion the acceleration is not constant (its direction changes), so they are not.

Conclusion: (b) and (e) are true; (a), (c) and (d) are true only for constant acceleration, hence false for arbitrary motion.

Concept. A scalar is a quantity specified by a magnitude alone and whose value does not change when the coordinate axes are rotated. It may carry a unit and a sign; it may vary from place to place and from time to time, and it need not be conserved. Only statement (e) states the defining property of a scalar.

1. Example for (e): speed |v| = √(v_x² + v_y²) has the same value whatever axes are used, though v_x and v_y change when the axes are rotated.
2. Counter-example for (a)–(d): temperature (a scalar) is not conserved, can be negative on the Celsius scale, has dimensions and varies from point to point.

Answer in one line: (a) false, (b) false, (c) false, (d) false, (e) true.

Extra depth. The property in (e) is also the mathematical definition used in advanced physics: a scalar is a quantity that transforms trivially (unchanged) under rotation of coordinates, a vector transforms like the coordinates of a point, and so on. This is why the dot product A·B = AB cosθ is a scalar (invariant under rotation) while the cross product A × B is a (pseudo)vector.

Conclusion: Only (e) is true: a scalar has the same value for observers with different orientations of axes.

Concept. The aircraft flies in a horizontal straight line at constant height h = 3400 m. The observer is at O on the ground, and the two aircraft positions A and B, 10.0 s apart, subtend an angle of 30° at O. Take the observation point such that the perpendicular from O to the flight path bisects the angle (the symmetrical situation of the standard solution); then the speed = AB/t.

1. Let the vertical through O meet the flight path at C. By symmetry C is the midpoint of AB and angle AOC = angle COB = 30°/2 = 15°.
2. In right triangle OCA: AC = OC tan15° = 3400 × tan15° = 3400 × 0.2679 = 911 m.
3. AB = 2 AC = 2 × 911 = 1822 m (= 2 h tan15°).
4. Speed v = AB/t = 1822 m/10.0 s = 182 m s⁻¹ (≈ 656 km h⁻¹).

Common mistake. Using AB = h tan30° = 3400 × 0.577 = 1963 m (speed 196 m s⁻¹) treats the whole 30° angle as measured from the vertical; that is wrong because the vertical lies between the two lines of sight. The correct statement is AB = h(tanα₁ + tanα₂) where α₁ and α₂ are the angles of A and B from the vertical, here 15° each.

Extra depth. The problem has a geometrical assumption (the symmetric observation point directly under the midpoint of the arc). If the observer is elsewhere the distance AB would differ. In the general case with angles α₁ and α₂ on opposite sides of the vertical: AB = h(tanα₁ + tanα₂) and with α₁ + α₂ = 30° the answer varies between 1822 m (symmetric) and h tan30° = 1963 m (when one line of sight is vertical), i.e. speed between 182 and 196 m s⁻¹. The standard answer 182 m s⁻¹ corresponds to the symmetric case.

Conclusion: The speed of the aircraft is about 182 m s⁻¹ (656 km h⁻¹).