This MCQ module is based on: NCERT Exercises and Solutions: Motion in a Plane
TOPIC 13 OF 33
NCERT Exercises and Solutions: Motion in a Plane
🎓 Class 11
Physics
CBSE
Theory
Ch 3 – Motion in a Plane
⏱ ~8 min
🌐 Language: [gtranslate]
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NCERT Exercises and Solutions: Motion in a Plane
📋 Chapter Summary
Key Concepts of Motion in a Plane
- Scalars vs. Vectors: Scalars have only magnitude (mass, time, speed); vectors have both magnitude AND direction (velocity, acceleration, force) AND obey parallelogram law.
- Position vector: \(\vec{r} = x\hat{i} + y\hat{j}\); magnitude \(|\vec{r}| = \sqrt{x^2 + y^2}\); angle \(\theta = \tan^{-1}(y/x)\).
- Equality: Two vectors are equal only if BOTH magnitude AND direction match. Tail position is irrelevant.
- Vector addition: Triangle law (head-to-tail), or parallelogram law. Resultant: \(R = \sqrt{A^2 + B^2 + 2AB\cos\theta}\).
- Resolution: \(A_x = A\cos\theta\), \(A_y = A\sin\theta\). Reverse: \(A = \sqrt{A_x^2 + A_y^2}\), \(\tan\theta = A_y/A_x\).
- Displacement, Velocity, Acceleration in 2D: All defined by limits as in 1D — but as VECTORS. Velocity is tangent to trajectory.
- Constant acceleration in 2D: \(\vec{r} = \vec{r_0} + \vec{v_0}t + \frac{1}{2}\vec{a}t^2\) — solve component-wise.
- Projectile motion: Trajectory is a parabola. Range \(R = \frac{v_0^2 \sin 2\theta_0}{g}\); max range at θ = 45°. Max height \(h_m = \frac{v_0^2 \sin^2\theta_0}{2g}\). Time of flight \(T = \frac{2v_0 \sin\theta_0}{g}\).
- Uniform circular motion: Speed constant, direction changes. Centripetal acceleration \(a_c = v^2/r = \omega^2 r\), directed toward center. \(\omega = 2\pi/T\).
- Relative velocity: \(\vec{v_{AB}} = \vec{v_A} - \vec{v_B}\). Useful for problems involving boats in rivers, planes in wind, etc.
🔑 Key Terms & Formulas
ScalarQuantity with magnitude only (e.g., mass, temperature)
VectorQuantity with magnitude AND direction, obeying parallelogram law
Unit Vector\(\hat{a} = \vec{a}/|\vec{a}|\), magnitude 1, no units
Resultant\(R^2 = A^2 + B^2 + 2AB\cos\theta\)
Range (Projectile)\(R = v_0^2 \sin 2\theta_0 / g\); max at 45°
Max Height\(h_m = v_0^2 \sin^2\theta_0 / 2g\)
Time of Flight\(T = 2v_0\sin\theta_0/g\)
Centripetal Acceleration\(a_c = v^2/r = \omega^2 r\), towards center
Angular Speed\(\omega = 2\pi/T = 2\pi f\); v = ωr
Relative Velocity\(\vec{v}_{AB} = \vec{v}_A - \vec{v}_B\)
📝 NCERT Exercises (Worked Solutions)
Exercise 3.1
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.
Scalars: volume, mass, speed, density, number of moles, angular frequency.
Vectors: acceleration, velocity, displacement, angular velocity.
Note: Speed (scalar) vs. velocity (vector); angular frequency (rate, scalar) vs. angular velocity (rotation axis direction matters, vector).
Vectors: acceleration, velocity, displacement, angular velocity.
Note: Speed (scalar) vs. velocity (vector); angular frequency (rate, scalar) vs. angular velocity (rotation axis direction matters, vector).
Exercise 3.2
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.
Scalars: work and current.
Although electric current has a direction (along the wire), it does not obey the parallelogram law of addition, so it is a scalar. Work is force·displacement (a dot product), which yields a scalar.
Although electric current has a direction (along the wire), it does not obey the parallelogram law of addition, so it is a scalar. Work is force·displacement (a dot product), which yields a scalar.
Exercise 3.3
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.
Answer: Impulse.
Impulse = Force × time = change in linear momentum, which has direction (vector). All others are scalars: pressure (scalar in fluids), time, energy, charge, etc.
Impulse = Force × time = change in linear momentum, which has direction (vector). All others are scalars: pressure (scalar in fluids), time, energy, charge, etc.
Exercise 3.4
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.
(a) Meaningful only if both have same dimensions: e.g., 3 kg + 5 kg = 8 kg ✓; 3 kg + 5 m ✗.
(b) Not meaningful: a scalar and vector cannot be added even if dimensions match.
(c) Always meaningful: scaling a vector by any scalar gives another vector.
(d) Meaningful: result is a scalar (with combined units).
(e) Meaningful only if same dimensions: can't add velocity + force.
(f) Meaningful: a component of a vector IS a vector along an axis, with same dimensions, so addition is fine.
(b) Not meaningful: a scalar and vector cannot be added even if dimensions match.
(c) Always meaningful: scaling a vector by any scalar gives another vector.
(d) Meaningful: result is a scalar (with combined units).
(e) Meaningful only if same dimensions: can't add velocity + force.
(f) Meaningful: a component of a vector IS a vector along an axis, with same dimensions, so addition is fine.
Exercise 3.5
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.
(a) TRUE. Magnitude is always a non-negative number — by definition a scalar.
(b) FALSE. A component of a vector is itself a vector (in the direction of the corresponding axis). Its magnitude is a scalar, but the component as a directed quantity is a vector.
(c) FALSE. Path length ≥ |displacement|, with equality only for straight-line motion in a single direction. For curved paths, path length > |displacement|.
(b) FALSE. A component of a vector is itself a vector (in the direction of the corresponding axis). Its magnitude is a scalar, but the component as a directed quantity is a vector.
(c) FALSE. Path length ≥ |displacement|, with equality only for straight-line motion in a single direction. For curved paths, path length > |displacement|.
Exercise 3.6
Establish the following vector inequalities geometrically or otherwise:
(a) \(|\vec{a}+\vec{b}| \le |\vec{a}|+|\vec{b}|\); (b) \(|\vec{a}+\vec{b}| \ge ||\vec{a}|-|\vec{b}||\); (c) \(|\vec{a}-\vec{b}| \le |\vec{a}|+|\vec{b}|\); (d) \(|\vec{a}-\vec{b}| \ge ||\vec{a}|-|\vec{b}||\). When does equality sign hold?
By parallelogram law: \(|\vec{a} + \vec{b}|^2 = a^2 + b^2 + 2ab\cos\theta\).
Since \(-1 \le \cos\theta \le 1\):
- Maximum: when \(\theta = 0°\), \(|\vec{a}+\vec{b}| = a + b\) → inequality (a) becomes equality (vectors PARALLEL).
- Minimum: when \(\theta = 180°\), \(|\vec{a}+\vec{b}| = |a - b|\) → inequality (b) becomes equality (vectors ANTI-PARALLEL).
For (c) and (d), use \(\vec{a} - \vec{b} = \vec{a} + (-\vec{b})\) and apply the same reasoning.
Equality conditions:
(a), (d): when vectors are parallel (\(\theta = 0°\)).
(b), (c): when vectors are anti-parallel (\(\theta = 180°\)).
Since \(-1 \le \cos\theta \le 1\):
- Maximum: when \(\theta = 0°\), \(|\vec{a}+\vec{b}| = a + b\) → inequality (a) becomes equality (vectors PARALLEL).
- Minimum: when \(\theta = 180°\), \(|\vec{a}+\vec{b}| = |a - b|\) → inequality (b) becomes equality (vectors ANTI-PARALLEL).
For (c) and (d), use \(\vec{a} - \vec{b} = \vec{a} + (-\vec{b})\) and apply the same reasoning.
Equality conditions:
(a), (d): when vectors are parallel (\(\theta = 0°\)).
(b), (c): when vectors are anti-parallel (\(\theta = 180°\)).
Exercise 3.7
Given \(\vec{a} + \vec{b} + \vec{c} + \vec{d} = 0\), which of the following statements are correct: (a) \(\vec{a}, \vec{b}, \vec{c}, \vec{d}\) must each be a null vector; (b) the magnitude of \(\vec{a}+\vec{c}\) equals the magnitude of \(\vec{b}+\vec{d}\); (c) the magnitude of \(\vec{a}\) can never be greater than the sum of the magnitudes of \(\vec{b}, \vec{c}\) and \(\vec{d}\); (d) \(\vec{b}+\vec{c}\) must lie in the plane of \(\vec{a}\) and \(\vec{d}\) if \(\vec{a}\) and \(\vec{d}\) are not collinear.
(a) FALSE. Sum can be zero without all vectors being null. Example: 4 sides of a closed quadrilateral.
(b) TRUE. Rearranging: \(\vec{a}+\vec{c} = -(\vec{b}+\vec{d})\). Magnitudes equal (sign just opposite direction).
(c) TRUE. \(\vec{a} = -(\vec{b}+\vec{c}+\vec{d})\), so \(|\vec{a}| \le |\vec{b}|+|\vec{c}|+|\vec{d}|\) (triangle inequality).
(d) TRUE. \(\vec{b}+\vec{c} = -(\vec{a}+\vec{d})\), and \(\vec{a}+\vec{d}\) is in the plane of \(\vec{a}\) and \(\vec{d}\).
(b) TRUE. Rearranging: \(\vec{a}+\vec{c} = -(\vec{b}+\vec{d})\). Magnitudes equal (sign just opposite direction).
(c) TRUE. \(\vec{a} = -(\vec{b}+\vec{c}+\vec{d})\), so \(|\vec{a}| \le |\vec{b}|+|\vec{c}|+|\vec{d}|\) (triangle inequality).
(d) TRUE. \(\vec{b}+\vec{c} = -(\vec{a}+\vec{d})\), and \(\vec{a}+\vec{d}\) is in the plane of \(\vec{a}\) and \(\vec{d}\).
Exercise 3.8
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 figure. What is the magnitude of the displacement vector for each? For which girl is this equal to the actual length of path skate?
Displacement for each girl = straight line from P to Q = diameter = 2 × 200 = 400 m.
The girl who skates along the diameter (straight line P to Q) has her path length = displacement = 400 m. For the others (curved paths), path length > 400 m.
Insight: Displacement depends only on initial and final positions, not the path. Path length depends on the actual route.
The girl who skates along the diameter (straight line P to Q) has her path length = displacement = 400 m. For the others (curved paths), path length > 400 m.
Insight: Displacement depends only on initial and final positions, not the path. Path length depends on the actual route.
Exercise 3.9
A cyclist starts from the center 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 center along QO as shown. If the round trip takes 10 min, what is the (a) net displacement, (b) average velocity, and (c) average speed of the cyclist?
The cyclist's path: O → P (along radius, 1 km) → Q (along quarter-arc of circumference) → O (along radius, 1 km). Quarter-arc length = (πr)/2 = π/2 km ≈ 1.57 km.
(a) Net displacement = 0 (returns to starting point O).
(b) Average velocity = displacement/time = 0/10 = 0 km/min.
(c) Average speed: total path = 1 + 1.57 + 1 = 3.57 km. Time = 10 min = 1/6 h. \[\text{Avg. speed} = \frac{3.57}{1/6} = 21.42 \text{ km/h} \approx \boxed{21.42 \text{ km/h}}\]
(a) Net displacement = 0 (returns to starting point O).
(b) Average velocity = displacement/time = 0/10 = 0 km/min.
(c) Average speed: total path = 1 + 1.57 + 1 = 3.57 km. Time = 10 min = 1/6 h. \[\text{Avg. speed} = \frac{3.57}{1/6} = 21.42 \text{ km/h} \approx \boxed{21.42 \text{ km/h}}\]
Exercise 3.10
On an open ground, a motorist follows a track that turns to his left by an angle of 60° 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.
The motorist traces a regular hexagon (since 60° × 6 = 360°).
3rd turn: Displacement = diameter = 2 × 500 = 1000 m; path length = 3 × 500 = 1500 m. Ratio = 0.667.
6th turn: Returns to start. Displacement = 0; path length = 6 × 500 = 3000 m. Ratio = 0.
8th turn: Has gone around once + 2 more sides. Displacement = 500 m (since after 6 it's at start, then 2 more sides at 60° gives same as 2nd-turn position). Direction: 60° from initial direction. Path length = 4000 m. Ratio = 0.125.
3rd turn: Displacement = diameter = 2 × 500 = 1000 m; path length = 3 × 500 = 1500 m. Ratio = 0.667.
6th turn: Returns to start. Displacement = 0; path length = 6 × 500 = 3000 m. Ratio = 0.
8th turn: Has gone around once + 2 more sides. Displacement = 500 m (since after 6 it's at start, then 2 more sides at 60° gives same as 2nd-turn position). Direction: 60° from initial direction. Path length = 4000 m. Ratio = 0.125.
Exercise 3.11
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?
(a) Average speed:
\[\text{Avg. speed} = \frac{23 \text{ km}}{28 \text{ min}} = \frac{23}{28/60} = \frac{23 \times 60}{28} \approx \boxed{49.3 \text{ km/h}}\]
(b) Average velocity magnitude: \[|\vec{v}|_{avg} = \frac{|\Delta \vec{r}|}{\Delta t} = \frac{10}{28/60} \approx \boxed{21.4 \text{ km/h}}\]
Are they equal? NO — they differ because path length (23 km) ≠ displacement (10 km). Avg speed depends on path; avg velocity only on net displacement.
(b) Average velocity magnitude: \[|\vec{v}|_{avg} = \frac{|\Delta \vec{r}|}{\Delta t} = \frac{10}{28/60} \approx \boxed{21.4 \text{ km/h}}\]
Are they equal? NO — they differ because path length (23 km) ≠ displacement (10 km). Avg speed depends on path; avg velocity only on net displacement.
Exercise 3.12
Rain is falling vertically with a speed of 30 m/s. A woman rides a bicycle with a speed of 10 m/s in the north-to-south direction. What is the direction in which she should hold her umbrella?
Setup: +y up, +x north. Woman moves south: \(\vec{v_w} = -10\hat{x}\). Rain (ground): \(\vec{v_r} = -30\hat{y}\).
Rain relative to woman: \(\vec{v_{r,w}} = \vec{v_r} - \vec{v_w} = -30\hat{y} + 10\hat{x}\).
The rain appears to come from the north (positive x direction) and from above. Angle from vertical: \[\tan\theta = \frac{10}{30} = 0.333 \Rightarrow \theta = \tan^{-1}(0.333) \approx \boxed{18.4°}\] She should tilt umbrella ≈ 18.4° from vertical, towards the NORTH (the direction she came from).
Rain relative to woman: \(\vec{v_{r,w}} = \vec{v_r} - \vec{v_w} = -30\hat{y} + 10\hat{x}\).
The rain appears to come from the north (positive x direction) and from above. Angle from vertical: \[\tan\theta = \frac{10}{30} = 0.333 \Rightarrow \theta = \tan^{-1}(0.333) \approx \boxed{18.4°}\] She should tilt umbrella ≈ 18.4° from vertical, towards the NORTH (the direction she came from).
Exercise 3.13
A man can swim with a speed of 4.0 km/h in still water. How long does he take to cross a river 1.0 km wide if the river flows steadily at 3.0 km/h and he makes his strokes normal to the river current? How far down the river does he go when he reaches the other bank?
The width of the river is 1 km. Swimmer's velocity perpendicular to current = 4 km/h.
Time to cross: \[t = \frac{1 \text{ km}}{4 \text{ km/h}} = 0.25 \text{ h} = \boxed{15 \text{ min}}\]
Distance carried downstream: \[d = v_{current} \times t = 3 \times 0.25 = \boxed{0.75 \text{ km} = 750 \text{ m}}\] The swimmer reaches the opposite bank 750 m downstream from his target.
Time to cross: \[t = \frac{1 \text{ km}}{4 \text{ km/h}} = 0.25 \text{ h} = \boxed{15 \text{ min}}\]
Distance carried downstream: \[d = v_{current} \times t = 3 \times 0.25 = \boxed{0.75 \text{ km} = 750 \text{ m}}\] The swimmer reaches the opposite bank 750 m downstream from his target.
Exercise 3.14
In a harbour, wind is blowing at the speed of 72 km/h and the flag on the mast of a boat anchored in the harbour flutters along the N-E direction. If the boat starts moving at a speed of 51 km/h to the north, what is the direction of the flag on the mast of the boat?
Wind velocity (ground): 72 km/h NE = \(72\cos45°\hat{i} + 72\sin45°\hat{j} = 50.91\hat{i} + 50.91\hat{j}\) (i = east, j = north).
Boat velocity: \(\vec{v_b} = 51\hat{j}\) km/h.
Wind relative to boat: \[\vec{v_{w,b}} = \vec{v_w} - \vec{v_b} = 50.91\hat{i} + (50.91 - 51)\hat{j} \approx 50.91\hat{i} - 0.09\hat{j}\] This is approximately 50.91 km/h due East. So the flag flutters very nearly due EAST.
Insight: The boat's northward motion almost exactly cancels the northward component of wind.
Boat velocity: \(\vec{v_b} = 51\hat{j}\) km/h.
Wind relative to boat: \[\vec{v_{w,b}} = \vec{v_w} - \vec{v_b} = 50.91\hat{i} + (50.91 - 51)\hat{j} \approx 50.91\hat{i} - 0.09\hat{j}\] This is approximately 50.91 km/h due East. So the flag flutters very nearly due EAST.
Insight: The boat's northward motion almost exactly cancels the northward component of wind.
Exercise 3.15
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?
Constraint: Max height ≤ 25 m. Find the angle that gives \(h_m = 25\) m, then compute range.
\[h_m = \frac{v_0^2 \sin^2\theta_0}{2g} = 25\] \[\sin^2\theta_0 = \frac{25 \times 2 \times 9.8}{40^2} = \frac{490}{1600} = 0.306\] \[\sin\theta_0 = 0.553 \Rightarrow \theta_0 \approx 33.6°\]
Maximum range with this angle: \[R = \frac{v_0^2 \sin 2\theta_0}{g} = \frac{1600 \times \sin 67.2°}{9.8} = \frac{1600 \times 0.921}{9.8} \approx \boxed{150.4 \text{ m}}\]
\[h_m = \frac{v_0^2 \sin^2\theta_0}{2g} = 25\] \[\sin^2\theta_0 = \frac{25 \times 2 \times 9.8}{40^2} = \frac{490}{1600} = 0.306\] \[\sin\theta_0 = 0.553 \Rightarrow \theta_0 \approx 33.6°\]
Maximum range with this angle: \[R = \frac{v_0^2 \sin 2\theta_0}{g} = \frac{1600 \times \sin 67.2°}{9.8} = \frac{1600 \times 0.921}{9.8} \approx \boxed{150.4 \text{ m}}\]
Exercise 3.16
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?
Step 1: Maximum range = 100 m occurs at θ = 45°. So:
\[R_{max} = \frac{v_0^2}{g} = 100 \Rightarrow v_0^2 = 100g\]
Step 2: Maximum vertical height = thrown straight up (θ = 90°). Using \(v^2 = u^2 - 2gh\) at top \(v = 0\): \[h = \frac{v_0^2}{2g} = \frac{100g}{2g} = \boxed{50 \text{ m}}\]
Insight: Max vertical height = R_max / 2.
Step 2: Maximum vertical height = thrown straight up (θ = 90°). Using \(v^2 = u^2 - 2gh\) at top \(v = 0\): \[h = \frac{v_0^2}{2g} = \frac{100g}{2g} = \boxed{50 \text{ m}}\]
Insight: Max vertical height = R_max / 2.
Exercise 3.17
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?
Given: r = 0.80 m; 14 revolutions in 25 s.
Frequency: f = 14/25 Hz. Angular speed: \[\omega = 2\pi f = \frac{2\pi \times 14}{25} = \frac{28\pi}{25} \approx 3.52 \text{ rad/s}\]
Centripetal acceleration: \[a_c = \omega^2 r = (3.52)^2 \times 0.80 = 12.39 \times 0.80 \approx \boxed{9.91 \text{ m/s}^2}\]
Direction: always pointing towards the center of the circle (along the string, toward the hand).
Frequency: f = 14/25 Hz. Angular speed: \[\omega = 2\pi f = \frac{2\pi \times 14}{25} = \frac{28\pi}{25} \approx 3.52 \text{ rad/s}\]
Centripetal acceleration: \[a_c = \omega^2 r = (3.52)^2 \times 0.80 = 12.39 \times 0.80 \approx \boxed{9.91 \text{ m/s}^2}\]
Direction: always pointing towards the center of the circle (along the string, toward the hand).
Exercise 3.18
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.
Convert: v = 900 km/h = 250 m/s; r = 1000 m.
\[a_c = \frac{v^2}{r} = \frac{(250)^2}{1000} = \frac{62500}{1000} = 62.5 \text{ m/s}^2\]
Ratio: \[\frac{a_c}{g} = \frac{62.5}{9.8} \approx \boxed{6.38}\] The centripetal acceleration is ~6.4 times g — pilots experience strong "g-forces" during such turns.
\[a_c = \frac{v^2}{r} = \frac{(250)^2}{1000} = \frac{62500}{1000} = 62.5 \text{ m/s}^2\]
Ratio: \[\frac{a_c}{g} = \frac{62.5}{9.8} \approx \boxed{6.38}\] The centripetal acceleration is ~6.4 times g — pilots experience strong "g-forces" during such turns.
Exercise 3.19
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 center; (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.
(a) FALSE. Only true for UNIFORM circular motion. In non-uniform circular motion, there is also a tangential component (changing speed).
(b) TRUE. Instantaneous velocity is always tangent to the path (proven via limit argument).
(c) TRUE. Over one full revolution, the net change in velocity is zero (start = end), so average acceleration = Δv/Δt = 0.
(b) TRUE. Instantaneous velocity is always tangent to the path (proven via limit argument).
(c) TRUE. Over one full revolution, the net change in velocity is zero (start = end), so average acceleration = Δv/Δt = 0.
Exercise 3.20
The position of a particle is given by \(\vec{r} = 3.0t\hat{i} - 2.0t^2\hat{j} + 4.0\hat{k}\) m, where t is in seconds. (a) Find v and a of the particle. (b) What is the magnitude and direction of velocity of the particle at t = 2.0 s?
(a)
\[\vec{v} = \frac{d\vec{r}}{dt} = 3.0\hat{i} - 4.0t\hat{j} \text{ m/s}\]
\[\vec{a} = \frac{d\vec{v}}{dt} = -4.0\hat{j} \text{ m/s}^2 \text{ (constant)}\]
(b) At t = 2.0 s: \[\vec{v}(2) = 3.0\hat{i} - 8.0\hat{j} \text{ m/s}\] \[|\vec{v}| = \sqrt{9 + 64} = \sqrt{73} \approx \boxed{8.54 \text{ m/s}}\]
Direction (below x-axis): \[\tan\theta = \frac{8}{3} = 2.667 \Rightarrow \theta \approx 69.4°\] The velocity makes ~ 69.4° below the +x-axis (since v_y is negative).
(b) At t = 2.0 s: \[\vec{v}(2) = 3.0\hat{i} - 8.0\hat{j} \text{ m/s}\] \[|\vec{v}| = \sqrt{9 + 64} = \sqrt{73} \approx \boxed{8.54 \text{ m/s}}\]
Direction (below x-axis): \[\tan\theta = \frac{8}{3} = 2.667 \Rightarrow \theta \approx 69.4°\] The velocity makes ~ 69.4° below the +x-axis (since v_y is negative).
Exercise 3.21
A particle starts from the origin at t = 0 with a velocity of \(10.0\hat{j}\) m/s and moves in the x-y plane with a constant acceleration of \((8.0\hat{i} + 2.0\hat{j})\) m/s². (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?
Position:
\[x = 0 + 0 \cdot t + \frac{1}{2}(8.0)t^2 = 4t^2\]
\[y = 0 + 10t + \frac{1}{2}(2)t^2 = 10t + t^2\]
(a) When x = 16 m: \[4t^2 = 16 \Rightarrow t = 2 \text{ s}\] \[y(2) = 10(2) + (2)^2 = 20 + 4 = \boxed{24 \text{ m}}\]
(b) Speed at t = 2 s: \[v_x = 0 + 8(2) = 16 \text{ m/s}\] \[v_y = 10 + 2(2) = 14 \text{ m/s}\] \[|\vec{v}| = \sqrt{16^2 + 14^2} = \sqrt{256+196} = \sqrt{452} \approx \boxed{21.26 \text{ m/s}}\]
(a) When x = 16 m: \[4t^2 = 16 \Rightarrow t = 2 \text{ s}\] \[y(2) = 10(2) + (2)^2 = 20 + 4 = \boxed{24 \text{ m}}\]
(b) Speed at t = 2 s: \[v_x = 0 + 8(2) = 16 \text{ m/s}\] \[v_y = 10 + 2(2) = 14 \text{ m/s}\] \[|\vec{v}| = \sqrt{16^2 + 14^2} = \sqrt{256+196} = \sqrt{452} \approx \boxed{21.26 \text{ m/s}}\]
Exercise 3.22
\(\hat{i}\) and \(\hat{j}\) are unit vectors along x- and y-axis respectively. What is the magnitude and direction of the vectors \(\hat{i}+\hat{j}\) and \(\hat{i}-\hat{j}\)? What are the components of a vector \(\vec{A}=2\hat{i}+3\hat{j}\) along the directions of \(\hat{i}+\hat{j}\) and \(\hat{i}-\hat{j}\)?
Magnitudes:
\[|\hat{i}+\hat{j}| = \sqrt{1^2 + 1^2} = \sqrt{2}\]
\[|\hat{i}-\hat{j}| = \sqrt{1^2 + (-1)^2} = \sqrt{2}\]
Directions: \(\hat{i}+\hat{j}\) is at 45° above x-axis; \(\hat{i}-\hat{j}\) is at 45° below x-axis (i.e., −45°).
Unit vectors along these directions: \[\hat{n_1} = \frac{\hat{i}+\hat{j}}{\sqrt{2}}, \quad \hat{n_2} = \frac{\hat{i}-\hat{j}}{\sqrt{2}}\]
Component of \(\vec{A} = 2\hat{i}+3\hat{j}\) along \(\hat{n_1}\): \[A_{n_1} = \vec{A} \cdot \hat{n_1} = \frac{(2)(1) + (3)(1)}{\sqrt{2}} = \frac{5}{\sqrt{2}} \approx \boxed{3.54}\]
Along \(\hat{n_2}\): \[A_{n_2} = \vec{A} \cdot \hat{n_2} = \frac{(2)(1) + (3)(-1)}{\sqrt{2}} = \frac{-1}{\sqrt{2}} \approx \boxed{-0.71}\]
Directions: \(\hat{i}+\hat{j}\) is at 45° above x-axis; \(\hat{i}-\hat{j}\) is at 45° below x-axis (i.e., −45°).
Unit vectors along these directions: \[\hat{n_1} = \frac{\hat{i}+\hat{j}}{\sqrt{2}}, \quad \hat{n_2} = \frac{\hat{i}-\hat{j}}{\sqrt{2}}\]
Component of \(\vec{A} = 2\hat{i}+3\hat{j}\) along \(\hat{n_1}\): \[A_{n_1} = \vec{A} \cdot \hat{n_1} = \frac{(2)(1) + (3)(1)}{\sqrt{2}} = \frac{5}{\sqrt{2}} \approx \boxed{3.54}\]
Along \(\hat{n_2}\): \[A_{n_2} = \vec{A} \cdot \hat{n_2} = \frac{(2)(1) + (3)(-1)}{\sqrt{2}} = \frac{-1}{\sqrt{2}} \approx \boxed{-0.71}\]
Exercise 3.23
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°, what is the speed of the aircraft?
Setup: The aircraft is at height h = 3400 m. Two positions 10 s apart subtend angle 30° at observer. The horizontal distance covered = chord length on the path.
At the midpoint, the line from observer to aircraft is perpendicular (closest). For symmetric angle subtension, the chord length is: \[d = 2 \times h \tan(15°) = 2 \times 3400 \times 0.2679 \approx 1822 \text{ m}\]
Speed: \[v = \frac{d}{t} = \frac{1822}{10} \approx \boxed{182.2 \text{ m/s}} \approx 656 \text{ km/h}\]
At the midpoint, the line from observer to aircraft is perpendicular (closest). For symmetric angle subtension, the chord length is: \[d = 2 \times h \tan(15°) = 2 \times 3400 \times 0.2679 \approx 1822 \text{ m}\]
Speed: \[v = \frac{d}{t} = \frac{1822}{10} \approx \boxed{182.2 \text{ m/s}} \approx 656 \text{ km/h}\]
🎓 Chapter 3 Complete! You've now mastered the fundamentals of motion in two dimensions: vectors, projectile motion, and circular motion. These are the building blocks for analysing forces (Chapter 5), gravitational orbits, and rotational dynamics. Keep practicing problems — each NCERT exercise builds your physical intuition!
Frequently Asked Questions - NCERT Exercises and Solutions: Motion in a Plane
What are the key NCERT exercise types in Chapter 3 Motion in a Plane?
NCERT Class 11 Physics Chapter 3 Motion in a Plane exercises cover conceptual MCQs, short numerical problems (2-3 marks), long numerical derivations (5 marks), and assertion-reason questions. The MyAiSchool solution set provides step-by-step worked solutions for every NCERT exercise question in the chapter, aligned with the CBSE board exam pattern. Students should focus on dimensional analysis, formula application, and unit consistency to score full marks.
How should students approach numerical problems in Motion in a Plane?
For numerical problems in NCERT Class 11 Physics Chapter 3 Motion in a Plane: (1) list all given data with units, (2) write the relevant formula(e), (3) substitute values carefully, (4) calculate with proper significant figures, (5) state the final answer with the correct unit. Always draw a free-body or schematic diagram where relevant. The MyAiSchool solutions follow this 5-step CBSE-aligned format consistently.
What are the most-asked CBSE board questions from Chapter 3?
From NCERT Class 11 Physics Chapter 3 (Motion in a Plane), the most-asked CBSE board questions test conceptual understanding of fundamental principles, application of derived formulas to standard scenarios, and proof-style derivations. 5-mark questions usually combine derivation + application. The MyAiSchool exercise set tags each question by board frequency so students can prioritize high-yield problems before exams.
How do I check the dimensional correctness of my answer?
Dimensional correctness in NCERT Class 11 Physics is verified by ensuring the LHS and RHS of any equation have the same dimensional formula (M^a L^b T^c). For Chapter 3 Motion in a Plane problems, write the dimensions of each quantity, substitute, and simplify. If dimensions match the equation is dimensionally valid (necessary but not sufficient). The MyAiSchool solutions include dimensional checks in every numerical answer.
What are common mistakes students make in Chapter 3 exercises?
Common mistakes in NCERT Class 11 Physics Chapter 3 Motion in a Plane exercises include: (1) unit conversion errors (CGS vs SI), (2) sign convention mistakes for vectors, (3) forgetting to consider all forces in free-body diagrams, (4) algebra errors during derivations, and (5) misreading the problem statement. The MyAiSchool solutions highlight these traps with red-flag annotations so students learn to avoid them.
How does the MyAiSchool solution differ from other NCERT solution sets?
MyAiSchool Class 11 Physics Chapter 3 Motion in a Plane solutions use NEP 2024-aligned pedagogy with Bloom Taxonomy tagged questions (L1 Remember to L6 Create), step-by-step working with reasoning, dimensional checks, interactive simulations, and Competency-Based Questions (CBQs) for board exam practice. Each solution is verified against NCERT and CBSE marking schemes for accuracy.
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