Class 9 Science Chapter 4: Describing Motion Around Us
Short Notes and NCERT Book Solutions
Part 1: Short Notes
1. Motion and Rest
An object is in motion when its position changes with time relative to a reference point. An object is at rest when its position does not change relative to that reference point.
Reference point: A point used to describe the position or motion of an object.
2. Distance and Displacement
- Distance: The total length of the path travelled.
- Displacement: The change in position from the starting point to the ending point, including direction.
- Distance is always greater than or equal to the magnitude of displacement.
- If an object returns to its starting point, its displacement is zero.
Illustration of distance and displacement. Open image source
3. Speed and Velocity
Speed tells us how much distance an object covers in a given time.
Velocity is the rate of change of displacement. It includes direction.
- SI unit of speed and velocity: metre per second (m/s).
- Speed is a scalar quantity; velocity is a vector quantity.
- Uniform motion: Equal distances are covered in equal intervals of time.
- Non-uniform motion: Distance covered or velocity changes with time.
To convert km/h into m/s, multiply by 5/18.
4. Acceleration
Acceleration is the rate of change of velocity with time. It occurs when speed changes, direction changes, or both change.
a = (v − u)/t
- u: Initial velocity
- v: Final velocity
- t: Time taken
- a: Acceleration
The SI unit of acceleration is m/s². A negative value can indicate that velocity is decreasing in the chosen positive direction.
5. Equations of Motion
These equations apply to straight-line motion with constant acceleration.
Here, s is displacement, u is initial velocity, v is final velocity, a is acceleration and t is time.
6. Position–Time Graphs
- A horizontal line means the object is at rest.
- A straight sloping line represents constant velocity.
- A curved line shows changing velocity.
- The slope of a position–time graph gives velocity.
7. Velocity–Time Graphs
- A horizontal line means constant velocity.
- The slope of the graph gives acceleration.
- The area between the graph and the time axis gives displacement, considering the sign of the area.
Illustration of motion graphs. Open image source
8. Motion in a Plane and Circular Motion
When an object moves along a circular path, it undergoes circular motion.
- In uniform circular motion, speed remains constant.
- Velocity changes because direction changes continuously.
- Therefore, uniform circular motion is accelerated motion.
- The instantaneous velocity is along the tangent to the circle.
- Distance in one complete revolution = 2πR, where R is the radius.
- Displacement after one complete revolution is zero.
Illustration of circular motion. Open image source
9. Quick Formula Revision
| Quantity | Formula | SI unit |
|---|---|---|
| Average speed | Total distance ÷ Total time | m/s |
| Average velocity | Displacement ÷ Total time | m/s |
| Acceleration | (v − u) ÷ t | m/s² |
| First equation | v = u + at | — |
| Second equation | s = ut + ½at² | m |
| Third equation | v² = u² + 2as | — |
| Circular distance | 2πR per revolution | m |
Part 2: End-of-Chapter Book Questions and Solutions
Question 1
My father went to a shop from home which is located at a distance of 250 m on a straight road. On reaching there, he discovered that he forgot to carry a cloth bag. He came home to take it, went to the shop again, bought provisions and came back home. How much was the total distance travelled by him? What was his displacement from home?
Total distance = 250 + 250 + 250 + 250 = 1000 m.
He returns home, so displacement = 0 m.
Question 2
A student runs from the ground floor to the fourth floor of a school building to collect a book and then comes down to their classroom on the second floor. If the height of each floor is 3 m, find:
(i) the total vertical distance travelled, and
(ii) their displacement from the starting point.
Ground to fourth floor = 4 × 3 = 12 m.
Fourth to second floor = 2 × 3 = 6 m.
(i) Total distance = 12 + 6 = 18 m.
(ii) Displacement = 2 × 3 = 6 m upward.
Question 3
A girl is riding her scooter and finds that its speedometer reading is constant. Is it possible for her scooter to be accelerating and if so, how?
Yes. If the scooter changes direction while turning, its velocity changes even when its speed remains constant. Therefore, it accelerates.
Question 4
A car starts from rest and its velocity reaches 24 m s–1 in 6 s. Find the average acceleration and the distance travelled in these 6 s.
u = 0 m/s, v = 24 m/s, t = 6 s.
a = (v − u)/t = 24/6 = 4 m/s².
s = (u + v)/2 × t = (0 + 24)/2 × 6 = 72 m.
Question 5
A motorbike moving with initial velocity 28 m s–1 and constant acceleration stops after travelling 98 m. Find the acceleration of the motorbike and the time taken to come to a stop.
u = 28 m/s, v = 0 m/s, s = 98 m.
Using v² = u² + 2as:
0 = 28² + 2 × a × 98.
a = −784/196 = −4 m/s².
Using v = u + at: 0 = 28 − 4t, so t = 7 s.
Question 6
Fig. 4.27 shows a position-time graph of two objects A and B that are moving along the parallel tracks in the same direction. Do objects A and B ever have equal velocity? Justify your answer.
No. The velocity is the slope of a position–time graph. In Fig. 4.27, the two straight lines have different slopes, so the objects have different constant velocities. Even if the lines cross, that only means the objects are at the same position at that instant.
Question 7
A graph in Fig. 4.28 shows the change in position with time for two objects A and B moving in a straight line from 0 to 10 seconds. Choose the correct option(s).
(i) The average velocity of both over the 10 s time interval is equal since they have the same initial and final positions.
(ii) The average speeds of both over the 10 s time interval are equal since both cover equal distance in equal time.
(iii) The average speed of A over the 10 s time interval is lower than that of B since it covers a shorter distance than B in 10 seconds.
(iv) The average speed of A over the 10 s time interval is greater than that of B since B’s speed is lower than A’s in some segments.
Correct options: (i) and (iv).
(i) Both objects have the same initial and final positions and the same time interval, so their average velocities are equal.
(iv) Object A covers a greater total distance than B in the same time, so A has the greater average speed.
Options (ii) and (iii) are incorrect.
Question 8
A truck driver driving at the speed of 54 km h–1 notices a road sign with a speed limit of 40 km h–1 (Fig. 4.29) for trucks. He slows down to 36 km h–1 in 36 s. What was the distance travelled by him during this time? Assume the acceleration to be constant while slowing down.
u = 54 × 5/18 = 15 m/s.
v = 36 × 5/18 = 10 m/s.
Average velocity = (15 + 10)/2 = 12.5 m/s.
Distance = 12.5 × 36 = 450 m.
Question 9
A car starts from rest and accelerates uniformly to 20 m s–1 in 5 seconds. It then travels at 20 m s–1 for 10 seconds and finally applies the brake (with uniform acceleration) to stop in 6 seconds. Find the total distance travelled.
First 5 s: distance = (0 + 20)/2 × 5 = 50 m.
Next 10 s: distance = 20 × 10 = 200 m.
Last 6 s: distance = (20 + 0)/2 × 6 = 60 m.
Total distance = 50 + 200 + 60 = 310 m.
Question 10
A bus is travelling at 36 km h–1 when the driver sees an obstacle 30 m ahead. The driver takes 0.5 seconds to react before pressing the brake. Once the brake is applied, the velocity of the bus reduces with constant acceleration of 2.5 m s–2. Will the bus be able to stop before reaching the obstacle?
Initial speed = 36 × 5/18 = 10 m/s.
Reaction distance = 10 × 0.5 = 5 m.
Braking distance: 0 = 10² − 2 × 2.5 × s, so s = 20 m.
Total stopping distance = 5 + 20 = 25 m.
Yes. The bus stops 5 m before the obstacle.
Question 11
A student said, “The Earth moves around the Sun”. In this context, discuss whether an object kept on the Earth can be considered to be at rest.
An object can be considered at rest relative to the Earth’s surface because its position does not change relative to the ground. But it is moving relative to the Sun along with the Earth. Rest and motion depend on the reference point.
Question 12
The velocity-time graph from 0 s to 120 s for a cyclist is shown in Fig. 4.30. Shade the areas (in different colours) representing the displacement of the cyclist
(i) while cyclist is moving with constant velocity.
(ii) when the velocity of cyclist is decreasing.
Also, calculate the displacement and average acceleration in the 120 s time interval.
(i) Shade the rectangular area from 20 s to 100 s for constant velocity.
(ii) Shade the trapezium from 100 s to 120 s for decreasing velocity.
From 0–20 s, velocity increases from 0 to 3 m/s:
Displacement = ½ × 20 × 3 = 30 m.
From 20–100 s, velocity is 3 m/s:
Displacement = 80 × 3 = 240 m.
From 100–120 s, velocity decreases from 3 m/s to 2 m/s:
Displacement = ½ × (3 + 2) × 20 = 50 m.
Total displacement = 30 + 240 + 50 = 320 m.
Average acceleration = (2 − 0)/120 = 0.0167 m/s².
Question 13
A girl is preparing for her first marathon by running on a straight road. She uses a smartwatch to calculate her running speed at different intervals. The graph (Fig. 4.31) depicts her velocity versus time. Estimate the distance she ran based on the graph.
Distance is the area under the velocity–time graph. The velocity is approximately between 6.5 km/h and 7.5 km/h over 6 hours.
Average velocity ≈ 7 km/h.
Distance ≈ average velocity × time = 7 × 6 = 42 km.
Question 14
On entering a state highway, a car continues to move with a constant velocity of 6 m s–1 for 2 minutes and then accelerates with a constant acceleration 1 m s–2 for 6 seconds. Find the displacement of the car on the state highway in the 2 min 6 s time interval by drawing a velocity-time graph for its motion.
2 minutes = 120 s.
First part: s₁ = 6 × 120 = 720 m.
Second part: s₂ = ut + ½at² = 6 × 6 + ½ × 1 × 6² = 54 m.
Total displacement = 720 + 54 = 774 m.
Graph: horizontal at 6 m/s from 0 to 120 s, then a straight rising line to 12 m/s at 126 s.
Question 15
Two cars A and B start moving with a constant acceleration from rest, in a straight line. Car A attains a velocity of 5 m s–1 in 5 s. Car B attains a velocity of 3 m s–1 in 10 s. Plot the velocity-time graphs for both the cars in the same graph. Using the graph, calculate the displacement in the two time intervals mentioned (Hint: Calculate the acceleration in both cases. Then calculate their velocities at five instants of time to plot the graph).
Car A acceleration = 5/5 = 1 m/s².
Car B acceleration = 3/10 = 0.3 m/s².
Plot a straight line for A from (0,0) to (5,5) and for B from (0,0) to (10,3).
| Time (s) | Car A velocity (m/s) | Time (s) | Car B velocity (m/s) |
|---|---|---|---|
| 0 | 0 | 0 | 0 |
| 1 | 1 | 2 | 0.6 |
| 2 | 2 | 4 | 1.2 |
| 3 | 3 | 6 | 1.8 |
| 4 | 4 | 8 | 2.4 |
| 5 | 5 | 10 | 3 |
Car A displacement = area of triangle = ½ × 5 × 5 = 12.5 m.
Car B displacement = area of triangle = ½ × 10 × 3 = 15 m.
Question 16
Rohan studies science from 6 PM to 7:30 PM at home. Consider the tip of the minute’s hand of the wall clock. During the given time interval, what is its:
(i) distance travelled,
(ii) displacement,
(iii) speed, and
(iv) velocity.
The length of the minute’s hand is 7 cm (Fig. 4.32).
Time = 90 minutes = 5400 s. Radius = 7 cm. The minute hand completes 1.5 revolutions.
(i) Distance: 1.5 × 2πR = 1.5 × 2 × π × 7 = 21π ≈ 66 cm.
(ii) Displacement: The tip ends opposite its starting point. Displacement = diameter = 2 × 7 = 14 cm downward.
(iii) Average speed: 66/5400 ≈ 0.0122 cm/s.
(iv) Average velocity: 14/5400 ≈ 0.00259 cm/s downward.