RowQ
The Vault
RowQ
The Vault
CBSE Class 9 Physics · 12 questions · 29 marks
Everything around you is moving relative to something else, and this chapter gives you the vocabulary to describe that motion precisely. You will separate distance from displacement, speed from velocity, and then use three equations of motion to predict where a body will be and how fast it will be going. Graphs are the other half of the story — a distance-time or velocity-time graph often answers a question faster than any formula.
A cyclist rides once around a circular park of circumference 400 m and stops at the point where she started. What are her distance and displacement?
Answer
Distance 400 m, displacement zero — she actually travels the full 400 m path, so the distance is 400 m. But displacement is the shortest straight-line gap between her starting and finishing positions, and since both are the same point, that gap is zero.
The velocity-time graph of a moving body is a straight horizontal line parallel to the time axis. What does this tell you about the body?
Answer
The body moves with uniform velocity and zero acceleration — on a velocity-time graph the slope represents acceleration. A horizontal line has zero slope, so the acceleration is zero, while the non-zero constant height of the line shows the body keeps moving at a steady velocity.
A scooter starting from rest reaches 18 m/s in 6 s. What is its acceleration?
Answer
3 m/s² — using a = (v - u)/t with u = 0, v = 18 m/s and t = 6 s, we get a = (18 - 0)/6 = 3 m/s². The scooter's velocity increases by 3 metres per second in every second.
A satellite revolves around the Earth in a circular orbit at a constant speed. Which statement is correct?
Answer
Its speed is constant but its velocity changes, so it is accelerating — velocity is a vector, and in circular motion the direction of motion changes at every instant even though the numerical speed stays the same. A changing velocity means the satellite is accelerating, the acceleration being directed towards the centre of the orbit.
Assertion (A): The displacement of a moving object can be smaller than the distance it has travelled, but never larger. Reason (R): Displacement is measured along the straight line joining the initial and final positions, which is the shortest of all possible paths between them.
Answer
Both A and R are true and R is the correct explanation of A — since the straight line joining two points is the shortest route between them, the magnitude of displacement can at most equal the distance (when the motion is along a straight line without reversal) and is smaller for every other path. That is exactly why displacement can never exceed distance.
Distinguish between uniform motion and non-uniform motion, giving one everyday example of each.
Answer
In uniform motion a body covers equal distances in equal intervals of time, however small those intervals are, so its speed stays constant — for example, a train moving steadily at 60 km/h on a straight, clear stretch of track. In non-uniform motion the body covers unequal distances in equal intervals of time, so its speed keeps changing — for example, a bus moving through a crowded market, repeatedly slowing down and speeding up.
A car travelling at 20 m/s applies brakes and comes to a complete stop after covering 50 m. Find (i) its retardation and (ii) the time taken to stop.
Answer
Given u = 20 m/s, v = 0, s = 50 m. (i) Using v² = u² + 2as: 0 = (20)² + 2 × a × 50, so 0 = 400 + 100a, giving a = -400/100 = -4 m/s². The negative sign shows retardation, so the retardation is 4 m/s². (ii) Using v = u + at: 0 = 20 + (-4)t, so 4t = 20 and t = 5 s. The car takes 5 seconds to come to rest.
A runner completes the first 120 m of a straight track in 15 s and the next 180 m in 25 s. Calculate her average speed for the whole run.
Answer
Total distance = 120 m + 180 m = 300 m. Total time = 15 s + 25 s = 40 s. Average speed = total distance / total time = 300/40 = 7.5 m/s. Note that this is not the ordinary average of the two individual speeds (8 m/s and 7.2 m/s), because average speed must always be worked out from the totals.
A ball is thrown vertically upward from the ground with an initial speed of 25 m/s. Taking the upward direction as positive and a = -10 m/s², find the maximum height it reaches and the time it takes to get there.
Answer
At the highest point the ball momentarily stops, so v = 0, with u = 25 m/s and a = -10 m/s². Maximum height: using v² = u² + 2as, 0 = (25)² + 2(-10)s, so 0 = 625 - 20s, giving s = 625/20 = 31.25 m. Time taken: using v = u + at, 0 = 25 + (-10)t, so t = 25/10 = 2.5 s. The ball rises 31.25 m and reaches that height 2.5 s after being thrown.
Derive the second equation of motion, s = ut + ½at², using a velocity-time graph for a body moving with uniform acceleration. Then use it to find the distance covered in 8 s by a body starting at 4 m/s with an acceleration of 2 m/s².
Answer
Consider a body whose velocity increases uniformly from u to v in time t. Its velocity-time graph is a straight sloping line: it starts at height u on the velocity axis at t = 0 and reaches height v at time t. The area enclosed between this line and the time axis gives the displacement s. This area is a trapezium, which can be split into a rectangle and a triangle: Area of rectangle = base × height = t × u = ut. This is the distance the body would have covered had it continued at its initial velocity u. Area of triangle = ½ × base × height = ½ × t × (v - u). Since a = (v - u)/t, we can write (v - u) = at, so this area becomes ½ × t × at = ½at². This is the extra distance gained because of the acceleration. Adding the two parts, s = ut + ½at², which is the second equation of motion. Numerical: with u = 4 m/s, a = 2 m/s² and t = 8 s, s = (4 × 8) + ½ × 2 × (8)² = 32 + 64 = 96 m. The body covers 96 m in 8 seconds.
A metro train starts from rest at a station, accelerates uniformly at 1 m/s² for 20 s, then runs at that steady speed for 60 s, and finally decelerates uniformly and stops in 10 s. (a) Find the maximum speed reached. (b) Find the distance covered in each of the three stages. (c) Find the average speed for the entire journey.
Answer
(a) Maximum speed: during the accelerating stage u = 0, a = 1 m/s², t = 20 s. Using v = u + at, v = 0 + 1 × 20 = 20 m/s. This is the steady speed for the middle stage. (b) Stage 1 (accelerating): s₁ = ut + ½at² = 0 + ½ × 1 × (20)² = 200 m. Stage 2 (uniform speed): s₂ = speed × time = 20 × 60 = 1200 m. Stage 3 (decelerating): here u = 20 m/s, v = 0, t = 10 s, so the average velocity is (20 + 0)/2 = 10 m/s and s₃ = 10 × 10 = 100 m. (c) Total distance = 200 + 1200 + 100 = 1500 m. Total time = 20 + 60 + 10 = 90 s. Average speed = 1500/90 = 16.67 m/s (approximately). Since the train ran along a straight track without reversing, the magnitude of its average velocity is the same value.
Read the following and answer the questions that follow: A school sets up a speed-measuring activity on a straight 300 m stretch of road. Two students, Kavya and Imran, cycle the stretch. Kavya covers it in 60 s, moving steadily throughout. Imran starts from rest, speeds up uniformly for the first half of the stretch, then keeps that speed for the second half, and finishes in 50 s. (a) Calculate Kavya's speed and state whether her motion is uniform or non-uniform. (b) Sketch in words the shape of Kavya's distance-time graph and say what its slope represents. (c) Whose motion is accelerated, and during which part of the journey? (d) Calculate Imran's average speed for the full 300 m.
Answer
(a) Kavya's speed = distance / time = 300/60 = 5 m/s. Because she covers equal distances in equal intervals of time, her motion is uniform motion, and her acceleration is zero. (b) Kavya's distance-time graph is a straight line sloping upward from the origin. Its slope represents her speed; since the line is straight, the slope and hence the speed is the same at every instant, confirming uniform motion. (c) Imran's motion is accelerated during the first half of the stretch (the first 150 m), when his velocity increases uniformly from rest. In the second half he moves at a constant speed, so his acceleration there is zero. Kavya is never accelerated. (d) Imran's average speed = total distance / total time = 300/50 = 6 m/s. So although he began from rest, his overall average speed is greater than Kavya's steady 5 m/s.
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