1. (i) The magnitude of average velocity equals average speed when the object moves in a straight line without changing direction, making displacement equal to distance.
2. (ii) When an object returns to its starting point, its displacement becomes zero, so average velocity is zero. However, since it has covered some distance, its average speed remains non-zero.

1. Total distance travelled by Sarang in 50 s = 25 + 25 = 50 m. Displacement of Sarang in 50 s = 0 m (returns to start).
2. Average speed = total distance / time interval = 50 m / 50 s = 1 m s⁻¹.
3. Average velocity = displacement / time interval = 0 m / 50 s = 0 m s⁻¹.

1. An object kept on Earth cannot be considered truly at rest because motion is relative.
2. Even though the Earth moves around the Sun, the object shares the same motion as the Earth. So relative to the Earth's surface, it is at rest. However, relative to the Sun (or space), it is continuously moving along with the Earth's orbital motion.

1. The distance-time graph for uniform motion is a straight line not parallel to the time axis.
2. The distance-time graph for non-uniform motion is not a straight line — it can be a curve or a zigzag line not parallel to the time axis.

1. Yes, an object which has moved through a distance can have zero displacement.
2. Example: When a person walking along a circular path returns to the starting point after completing a circle, his displacement is zero, but he covers a distance of 2πr, where r is the radius of the circular path.
3. The displacement is zero because the shortest distance between the initial and final position of the person is zero.

1. Graphs allow easy comparison of two objects by showing their positions, speeds, or accelerations on the same axes.
2. By observing slopes, intersections, and curves, differences in their motion can be understood quickly, making graphical analysis useful in studying multiple moving objects together.

1. A reference point is important because it serves as a fixed point from which the position of an object is measured.
2. Without a reference point, it is impossible to state whether an object has moved or remained at rest. Position and motion are always described relative to it.

1. To determine the total distance travelled by a particle.
2. To determine the instantaneous velocity of the particle.
3. To determine the acceleration of the object.

1. (a) Retardation
2. (b) ms⁻²
3. (c) Circular

1. Convert speeds to m/s: Initial speed u = 54 km/h = 15 m/s. Final speed v = 36 km/h = 10 m/s. Time t = 36 s.
2. For constant acceleration, distance travelled: s = (u+v)/2 × t = (15+10)/2 × 36 = 12.5 × 36 = 450 m.

(i) When the driver presses the accelerator

1. u = 36 km/h = 10 m/s, v = 54 km/h = 15 m/s, t = 10 s.
2. a = (v−u)/t = (15−10)/10 = 0.5 m/s². Since velocity is increasing, acceleration acts in the direction of velocity.

(ii) When the driver presses the brake

1. u = 54 km/h = 15 m/s, v = 0 m/s, t = 5 s.
2. a = (0−15)/5 = −3 m/s². The minus sign indicates acceleration is opposite to the direction of velocity (velocity is decreasing).

1. Speed = Slope of the distance-time graph. The smaller the slope, the smaller is the speed.
2. From the graph, car D has the smallest (flattest) slope.

1. The total distance travelled and the magnitude of displacement are equal when an object moves along a straight line in one direction without turning back.
2. In this case, the actual path length is the same as the shortest distance between the starting and final positions.
3. If the object changes direction or follows a curved path, distance becomes greater than displacement.

1. (a) An object with constant acceleration can still have zero velocity at an instant. For example, an object at rest on the Earth's surface has zero velocity but is still acted upon by gravitational acceleration of 9.81 ms⁻² towards the centre. When it starts falling freely, it has constant acceleration but momentarily zero velocity.
2. (b) When an athlete moves with constant-magnitude velocity along a circular path, the only change in velocity is due to the change in direction. This is an example of accelerated motion where acceleration is always perpendicular to the direction of motion — hence possible in circular motion.

1. By making lines parallel to the axes, it is found that the displacement of object B is more than object A for the same time interval.
2. That is, the slope of the line for B is steeper than the slope for line A.

(i) Acceleration phase (0 to 20 m/s in 5 s)

1. s₁ = (u+v)/2 × t = (0+20)/2 × 5 = 10 × 5 = 50 m.

(ii) Constant speed phase (20 m/s for 10 s)

1. s₂ = v × t = 20 × 10 = 200 m.

(iii) Braking phase (20 to 0 m/s in 6 s)

1. s₃ = (20+0)/2 × 6 = 10 × 6 = 60 m.

1. Total distance = s₁ + s₂ + s₃ = 50 + 200 + 60 = 310 m.

1. Given: u = 0 m/s, v = 24 m/s, t = 6 s.
2. Average acceleration: a = (v−u)/t = (24−0)/6 = 4 m/s².
3. Distance travelled: s = ut + ½at² = 0 + ½(4)(6)² = 2 × 36 = 72 m.

1. Average acceleration is the rate of change of velocity of an object during a given time interval. It shows how quickly the velocity increases, decreases, or changes direction.
2. Mathematically, average acceleration is calculated by dividing the change in velocity by the time taken for that change.

1. Its SI unit is metre per second squared (m/s²).

1. (a) Origin
2. (b) Uniform acceleration
3. (c) Same
4. (d) Retardation

1. (i) The displacement is 0 as the body returns to its original position, therefore average velocity is 0 m/min.
2. (ii) If a body is moving with uniform velocity, the velocity-time graph is a straight line parallel to the time axis.
3. (iii) When an object is moving along a straight path (one direction), the magnitude of velocity equals the speed. Therefore, the ratio of speed to velocity magnitude is 1.

OR

1. The slope of the distance-time graph is constant, showing the rate of change of distance is constant — so this represents uniform speed.

1. (i) Kinematic equations are applicable when an object moves with constant acceleration in a straight line. They cannot be used if acceleration is changing or motion is not linear.
2. (ii) Kinematic equations relate displacement, initial velocity, final velocity, acceleration, and time. These quantities are interconnected and help describe motion under constant acceleration.
3. (iii) Kinematic equations are important because they help analyse motion and predict future positions or velocities of objects. They are used in solving practical problems such as vehicle motion, falling objects, and sports activities involving motion under constant acceleration.

OR

1. Direction is important because quantities like velocity, displacement, and acceleration are vectors. The sign (+ or −) indicates direction, and incorrect signs can lead to wrong results when solving motion problems.

1. (i) Linear motion is the motion of an object along a straight line. The object moves in a single direction, either forward or backward, and its position changes with respect to a reference point over time.
2. (ii) Displacement is the shortest distance between the initial and final position of an object, along with the direction. It is a vector quantity because it requires both magnitude and direction.
3. (iii) Distance is the total path travelled by an object and has only magnitude, so it is a scalar quantity. Displacement is the shortest straight-line distance between two points and includes direction, making it a vector quantity. Distance is always positive, while displacement can be positive, negative, or zero.

OR

1. Distance and displacement are equal when an object moves in a straight line without changing direction. In such cases, the total path travelled is the same as the shortest distance between the starting and ending points.

1. (i) Motion in a plane is motion in two dimensions where an object moves along a path that lies on a flat surface. Examples include a ball being thrown or a vehicle taking a turn.
2. (ii) Acceleration is present because even though speed is constant, the direction of velocity keeps changing continuously. Any change in velocity, whether in magnitude or direction, results in acceleration.
3. (iii) The displacement after one complete revolution is zero because the object returns to its starting point. Although it travels a certain distance, its initial and final positions are the same.

OR

1. The velocity in circular motion is always directed along the tangent to the circle at any given point. This means the direction of motion is constantly changing as the object moves along the circular path.

1. (a) It is clear from the graph that B covers more distance in less time. Therefore, B is the fastest.
2. (b) All of them never come at the same point at the same time.
3. (c) Each small division shows about 0.57 km. A is passing B at a point in line with about 9.14 km on the distance axis. At this point, C travels about 9.14 − (0.57×3.75) km = 9.14 − 2.1375 ≈ 7 km.
4. (d) B passes C at a point ≈ 4 km + 0.57 km × 2.25 = 5.28 km.

Time-displacement graph plotted from the given data

1. Average velocity for first 4s = (4−0)/4 = 1 ms⁻¹.
2. Average velocity for next 4s = (4−4)/(8−4) = 0/4 = 0 ms⁻¹.
3. Average velocity for last 6s = (0−6)/(16−10) = −6/6 = −1 ms⁻¹.

1. (i) The displacement of an object after one full revolution in a circular path is zero. This is because the object returns to its starting point, so the initial and final positions are the same — there is no net change in position.
2. (ii) The average speed of an object for one revolution in a circular path is total distance divided by total time. Since the distance for one revolution is the circumference of the circle, average speed = 2πr/t, where r is the radius and t is the time taken.
3. (iii) A key characteristic of uniform circular motion is that the object moves with constant speed while its direction of velocity continuously changes. Thus, even though speed remains the same, the motion is accelerated because velocity changes at every point.

1. (i) The distance traveled and the magnitude of displacement are equal when an object moves in a straight line and in one direction only. In this scenario, there are no changes in the direction of motion, ensuring their congruence.
2. (ii) A curved position-time graph indicates that an object is undergoing accelerated motion. This shape signifies that the velocity is not constant but is continuously changing over time.
3. (iii) For an object moving with constant velocity, its velocity-time graph is a straight line parallel to the time axis. In contrast, for an object undergoing constant acceleration, the velocity-time graph is a straight line with a non-zero, constant slope, indicating a steady increase or decrease in velocity.

1. (i) In situation (iii), the distance between them will remain the same (parallel lines).
2. (ii) In situation (i), they are moving in opposite directions (one line has positive slope, the other negative).
3. (iii) Velocity of object A is positive in situation (i) because the slope of the s-t graph for A is positive.
4. (iv) Yes, in situations (i) and (ii), they are crossing each other because in these two cases, the s-t graphs of A and B intersect each other.

(a) From A to B

1. Total distance covered = displacement = AB = 300 m. Total time = 2 min 30 s = 150 s.
2. Average speed = 300/150 = 2 ms⁻¹. Average velocity = 300/150 = 2 ms⁻¹.

(b) From A to C

1. Total distance covered = AB + BC = 300 + 100 = 400 m. Displacement AC (shortest distance) = 300 − 100 = 200 m.
2. Total time = 2 min 30 s + 1 min = 3 min 30 s = 210 s.
3. Average speed = 400/210 = 1.90 ms⁻¹. Average velocity = 200/210 = 0.952 ms⁻¹.

Distance-time graph plotted from the train data, with points A, B, C, D, E

1. (i) Total distance travelled = 50 km. Total time = 10:00 AM to 11:30 AM = 1 hour 30 minutes = 3/2 h.
2. (ii) Average speed = 50 ÷ (3/2) = 100/3 = 33.33 km h⁻¹.
3. (iii) Speed = slope of the distance-time graph. The slope is maximum between 10:00 AM and 10:30 AM, so the train was travelling at its highest speed during this interval.
4. (iv) The part of the graph between 11:00 AM and 11:15 AM has minimum slope, so the train had a minimum speed there — the train had slowed down between 40 km and 42 km.
5. (v) Speed between 10:40 AM to 11:00 AM = (40−28) km / 20 min = 12/(20/60) = 36 km/h.