Ans: A body is said to be in uniform motion if it covers equal distances in equal intervals of time, however small the intervals may be.

Ans: The slope of a velocity–time graph represents the acceleration of the body.

Ans: The SI unit of acceleration is metre per second squared (m/s² or m s⁻²).

Ans: The area under a velocity–time graph gives the displacement of the body (or the distance travelled if the body does not change its direction).

Ans: Distance: total length of the path travelled; scalar; always positive; can be larger than displacement. Displacement: shortest distance between the initial and final positions in a specified direction; vector; can be positive, negative or zero. Example: after one complete round of a circular track, distance = circumference but displacement = zero.

Ans: Speed is the distance travelled per unit time; it is a scalar and is always positive. Velocity is the displacement per unit time, i.e. speed in a given direction; it is a vector and can be positive, negative or zero. Example: 60 km/h is a speed; 60 km/h towards the north is a velocity.

Ans: Yes. Velocity depends on both speed and direction. If a body moves at constant speed but its direction keeps changing, its velocity changes. Example: a car moving at constant speed around a circular track, or the Moon moving around the Earth, which is uniform circular motion.

Ans: Distance = 3 + 4 = 7 m. Displacement is the straight-line distance from start to end: √(3² + 4²) = √25 = 5 m, in the north-east direction.

Ans: 1 km/h = 5/18 m/s, so 72 km/h = 72 × 5/18 = 20 m/s.

Ans: u = 18 km/h = 5 m/s; v = 36 km/h = 10 m/s; t = 5 s. a = (v − u) ÷ t = (10 − 5) ÷ 5 = 1 m/s².

Ans: Radius = 100 m; circumference = 2πr = 2 × (22/7) × 100 ≈ 628 m, covered in 40 s. Time 2 min 20 s = 140 s = 3.5 rounds. Distance = 3.5 × 628 ≈ 2200 m. After 3.5 rounds he is at the point diametrically opposite to the start, so displacement = diameter = 200 m.

Ans: u = 0, a = 0.1 m/s², t = 2 min = 120 s. (a) v = u + at = 0 + 0.1 × 120 = 12 m/s. (b) s = ut + ½at² = 0 + ½ × 0.1 × 120² = ½ × 0.1 × 14400 = 720 m.

Ans: u = 0, a = g = 10 m/s², t = 5 s. Height h = ½gt² = ½ × 10 × 25 = 125 m. Final speed v = gt = 10 × 5 = 50 m/s.

Ans: A distance–time graph gives the speed, the distance travelled in a given time and whether the motion is uniform. Body at rest: a straight line parallel to the time axis. Uniform speed: a straight sloping line; slope = speed, and a steeper line means a greater speed. Non-uniform speed: a curved line, because the distance covered in equal intervals changes.

Ans: u = 90 km/h = 25 m/s; v = 54 km/h = 15 m/s; t = 10 s. a = (v − u) ÷ t = (15 − 25) ÷ 10 = −1 m/s², i.e. retardation of 1 m/s². Distance s = (v² − u²) ÷ 2a = (225 − 625) ÷ (−2) = 200 m.

Ans: Let the total distance be 2d. Time for the first half = d/40, for the second half = d/60. Total time = d/40 + d/60 = 5d/120 = d/24. Average speed = 2d ÷ (d/24) = 48 km/h. Note: it is not the simple average (50 km/h), because the times are different.

Ans: Let a body have initial velocity u at t = 0 and final velocity v at time t, with uniform acceleration a. On the v–t graph the line goes from A (u) to B (v). (1) Slope = a = (v − u) ÷ t, so v = u + at. (2) Distance s = area under the graph = rectangle (u × t) + triangle (½ × t × (v − u)). Since v − u = at, s = ut + ½at². (3) Area of trapezium: s = ½ (u + v) × t. Putting t = (v − u) ÷ a, we get s = (v² − u²) ÷ 2a, so v² = u² + 2as.

Ans: A body moving along a circular path with constant speed is in uniform circular motion. Although its speed is constant, the direction of its velocity changes at every point (velocity is always along the tangent). Since velocity is a vector, a change in direction means a change in velocity, and a change in velocity per unit time is acceleration. So uniform circular motion is accelerated. Examples: the Moon revolving around the Earth; a stone whirled in a circle with a string; an athlete running on a circular track at constant speed.

Ans: Phase 1: v = u + at = 0 + 2 × 10 = 20 m/s (maximum velocity); s₁ = ½ × 2 × 10² = 100 m. Phase 2: constant velocity of 20 m/s for 20 s; s₂ = 20 × 20 = 400 m. Phase 3: retardation a = (0 − 20) ÷ 5 = −4 m/s²; s₃ = ½ × 20 × 5 = 50 m. Total distance = 100 + 400 + 50 = 550 m.

Ans: At the highest point v = 0. Using v² = u² + 2as with a = −10: 0 = 400 − 20h, so h = 20 m. Time to rise: t = u ÷ g = 20 ÷ 10 = 2 s. The time of descent equals the time of ascent, so the total time = 4 s.