1 Mark CBSE 2019
Which physical quantity is given by the slope of a distance–time graph?
Ans: The slope of a distance–time graph gives the speed of the body.
1 Mark CBSE 2020
What is the SI unit of displacement?
Ans: The SI unit of displacement is the metre (m).
1 Mark CBSE 2022
An object moves along a circle of radius r and returns to its starting point after one full revolution. What is its displacement?
Ans: The displacement is zero, because the initial and final positions are the same. The distance covered is 2πr.
2 Mark CBSE 2018
Define acceleration. When is it said to be negative?
Ans: Acceleration is the rate of change of velocity with time: a = (v − u) ÷ t; SI unit m/s². Acceleration is negative when the velocity of the body decreases with time, i.e. the acceleration is opposite to the direction of motion. Negative acceleration is called retardation or deceleration.
2 Mark CBSE 2021
The odometer of a car reads 57321.0 km when the clock shows 08:30 AM and 57336.0 km at 09:00 AM. Find the distance covered and the average speed.
Ans: Distance covered = 57336.0 − 57321.0 = 15 km. Time = 30 min = 0.5 h. Average speed = 15 ÷ 0.5 = 30 km/h (≈ 8.33 m/s).
2 Mark CBSE 2023
Under what condition is the magnitude of average velocity of an object equal to its average speed?
Ans: The magnitude of average velocity equals the average speed when the object moves along a straight line in a single direction, so that the total distance travelled equals the magnitude of its displacement.
2 Mark CBSE 2019
Write the three equations of motion for uniform acceleration and state the meaning of each symbol.
Ans: v = u + at; s = ut + ½at²; v² = u² + 2as. Here u is the initial velocity, v the final velocity, a the uniform acceleration, t the time taken and s the displacement of the body in time t.
3 Mark CBSE 2022
A bus moving at 54 km/h is stopped by applying brakes in 5 s. Find the retardation and the distance travelled before it stops.
Ans: u = 54 km/h = 15 m/s; v = 0; t = 5 s. a = (0 − 15) ÷ 5 = −3 m/s², so retardation = 3 m/s². Distance s = ut + ½at² = 15 × 5 − ½ × 3 × 25 = 75 − 37.5 = 37.5 m.
3 Mark CBSE 2020
Distinguish between uniform and non-uniform acceleration. Give one example of each.
Ans: Uniform acceleration: velocity changes by equal amounts in equal intervals of time; the v–t graph is a straight line. Example: a freely falling body. Non-uniform acceleration: velocity changes by unequal amounts in equal intervals of time; the v–t graph is a curve. Example: a car moving in heavy city traffic.
3 Mark CBSE 2018
A train starting from rest attains a velocity of 72 km/h in 5 minutes. Assuming uniform acceleration, find the acceleration and the distance travelled.
Ans: u = 0; v = 72 km/h = 20 m/s; t = 5 min = 300 s. Acceleration a = (20 − 0) ÷ 300 = 0.067 m/s² (1/15 m/s²). Distance s = ½at² = ½ × (1/15) × 300² = 3000 m = 3 km.
3 Mark CBSE 2024
A stone is thrown vertically upward with a speed of 49 m/s. Find the maximum height it reaches and the total time of flight. (g = 9.8 m/s²)
Ans: At the highest point v = 0. H = u² ÷ 2g = 49² ÷ (2 × 9.8) = 2401 ÷ 19.6 = 122.5 m. Time to reach the top = u ÷ g = 49 ÷ 9.8 = 5 s. Time of descent is also 5 s, so total time of flight = 10 s.
3 Mark CBSE 2023
The distance–time graphs of two bodies A and B are straight lines through the origin, and the line of A is steeper. Which is faster? How do you find the speed from the graph?
Ans: Body A is faster because the slope of a distance–time graph gives the speed, and a steeper line has a greater slope. Speed = (change in distance) ÷ (change in time) = Δs ÷ Δt, taken between any two points on the straight line.
5 Mark CBSE 2021
Derive the equation s = ut + ½at² for a uniformly accelerated body using a velocity–time graph.
Ans: On a v–t graph, the line starts at velocity u at t = 0 and reaches v at time t. Displacement s equals the area under the graph. The area has two parts: a rectangle of height u and width t, giving ut; and a triangle of base t and height (v − u), giving ½ × t × (v − u). Since acceleration a = (v − u) ÷ t, we have v − u = at, so the triangle area is ½ × t × at = ½at². Hence s = ut + ½at².
5 Mark CBSE 2022
A body starts from rest and moves with uniform acceleration of 4 m/s² for 5 s. It then moves with constant velocity for 10 s and is finally brought to rest in 4 s. Find the maximum velocity, the retardation and the total distance covered.
Ans: Maximum velocity v = u + at = 0 + 4 × 5 = 20 m/s. Retardation = (0 − 20) ÷ 4 = −5 m/s², i.e. 5 m/s². Distances: s₁ = ½ × 4 × 25 = 50 m; s₂ = 20 × 10 = 200 m; s₃ = ½ × 20 × 4 = 40 m. Total distance = 50 + 200 + 40 = 290 m.
5 Mark CBSE 2024
Derive v² = u² + 2as. A car moving at 20 m/s is brought to rest with a retardation of 5 m/s². Find the stopping distance.
Ans: Derivation: average velocity = (u + v) ÷ 2, so s = ((u + v) ÷ 2) × t. From v = u + at, t = (v − u) ÷ a. Therefore s = (u + v)(v − u) ÷ 2a = (v² − u²) ÷ 2a, giving v² = u² + 2as. Numerical: u = 20 m/s, v = 0, a = −5 m/s². 0 = 400 + 2(−5)s, so 10s = 400 and s = 40 m. The car stops after 40 m.
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