1An object has moved through a distance. Can it have zero displacement? If yes, support your answer with an example.
Answer & Reference:
Yes, an object can have zero displacement even after moving through a distance. For example, if an object travels from point O to point A and then returns back to point O, the displacement is zero because the initial and final positions are the same, but the distance covered is twice the distance between O and A. (Page 73, Section 7.1.1)
2A farmer moves along the boundary of a square field of side 10 m in 40 s. What will be the magnitude of displacement of the farmer at the end of 2 minutes 20 seconds from his initial position?
Answer & Reference:
Given: Side of square field = 10 m, time for one round = 40 s
Total time = 2 min 20 s = 140 s
Number of rounds completed = 140/40 = 3.5 rounds
After 3.5 rounds, the farmer will be at the diagonally opposite corner from the starting point.
Displacement = diagonal of square = √(10² + 10²) = √200 = 10√2 ≈ 14.14 m
(Page 73, Section 7.1.1 - Concept of displacement)
3Which of the following is true for displacement? (a) It cannot be zero. (b) Its magnitude is greater than the distance travelled by the object.
Answer & Reference:
Neither (a) nor (b) is true.
(a) Displacement can be zero when the initial and final positions are the same.
(b) The magnitude of displacement is always equal to or less than the distance travelled, never greater.
(Page 73, Section 7.1.1)
4Distinguish between speed and velocity.
Answer & Reference: Speed is the distance travelled by an object per unit time. It is a scalar quantity (has only magnitude). Velocity is the displacement of an object per unit time in a definite direction. It is a vector quantity (has both magnitude and direction).
(Page 75-76, Section 7.2.1)
5Under what condition(s) is the magnitude of average velocity of an object equal to its average speed?
Answer & Reference:
The magnitude of average velocity is equal to average speed when the object moves in a straight line without changing direction (so that distance travelled equals displacement).
(Page 76, Example 7.3 - Usha swimming example shows when they are not equal)
6What does the odometer of an automobile measure?
Answer & Reference:
The odometer of an automobile measures the total distance travelled by the vehicle.
(Page 74, Activity 7.4)
7What does the path of an object look like when it is in uniform motion?
Answer & Reference:
When an object is in uniform motion, it moves in a straight line at constant speed. Therefore, its path is a straight line.
(Page 74, Section 7.1.2 - Uniform motion definition)
8During an experiment, a signal from a spaceship reached the ground station in five minutes. What was the distance of the spaceship from the ground station? The signal travels at the speed of light, that is, 3 × 10⁸ m s⁻¹.
9When will you say a body is in (i) uniform acceleration? (ii) non-uniform acceleration?
Answer & Reference:
(i) A body is in uniform acceleration when its velocity changes by equal amounts in equal intervals of time.
(ii) A body is in non-uniform acceleration when its velocity changes by unequal amounts in equal intervals of time.
(Page 77, Section 7.3 - Acceleration definition)
10A bus decreases its speed from 80 km h⁻¹ to 60 km h⁻¹ in 5 s. Find the acceleration of the bus.
Answer & Reference:
Given: u = 80 km/h = 80 × (5/18) = 22.22 m/s
v = 60 km/h = 60 × (5/18) = 16.67 m/s
t = 5 s
Acceleration a = (v - u)/t = (16.67 - 22.22)/5 = -5.55/5 = -1.11 m/s²
Negative sign indicates deceleration.
(Page 77, Equation 7.3)
11A train starting from a railway station and moving with uniform acceleration attains a speed 40 km h⁻¹ in 10 minutes. Find its acceleration.
Answer & Reference:
Given: u = 0, v = 40 km/h = 40 × (5/18) = 11.11 m/s
t = 10 minutes = 600 s
Acceleration a = (v - u)/t = (11.11 - 0)/600 = 0.0185 m/s²
(Page 77, Equation 7.3)
12What is the nature of the distance-time graphs for uniform and non-uniform motion of an object?
Answer & Reference:
For uniform motion, the distance-time graph is a straight line with constant slope.
For non-uniform motion, the distance-time graph is a curved line (non-linear).
(Page 78-79, Section 7.4.1 - Distance-time graphs)
13What can you say about the motion of an object whose distance-time graph is a straight line parallel to the time axis?
Answer & Reference:
If the distance-time graph is a straight line parallel to the time axis, it means the distance is not changing with time. Therefore, the object is at rest.
(Page 81, Question 2 - Implied from graph interpretation)
14What can you say about the motion of an object if its speed-time graph is a straight line parallel to the time axis?
Answer & Reference:
If the speed-time graph is a straight line parallel to the time axis, it means the speed is constant with time. Therefore, the object is moving with uniform speed (zero acceleration).
(Page 81, Question 3 - Velocity-time graph concept)
15What is the quantity which is measured by the area occupied below the velocity-time graph?
Answer & Reference:
The area under the velocity-time graph gives the displacement (or distance, if direction is not considered) of the object in a given time interval.
(Page 80, Section 7.4.2 - Velocity-time graphs)
16A bus starting from rest moves with a uniform acceleration of 0.1 m s⁻² for 2 minutes. Find (a) the speed acquired, (b) the distance travelled.
Answer & Reference:
Given: 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)(120)² = 0.05 × 14400 = 720 m
(Page 82, Equations 7.5 and 7.6)
17A train is travelling at a speed of 90 km h⁻¹. Brakes are applied so as to produce a uniform acceleration of –0.5 m s⁻². Find how far the train will go before it is brought to rest.
Answer & Reference:
Given: u = 90 km/h = 25 m/s, v = 0, a = -0.5 m/s²
Using v² = u² + 2as
0 = (25)² + 2(-0.5)s
0 = 625 - s
s = 625 m
(Page 82, Equation 7.7)
18A trolley, while going down an inclined plane, has an acceleration of 2 cm s⁻². What will be its velocity 3 s after the start?
Answer & Reference:
Given: u = 0, a = 2 cm/s² = 0.02 m/s², t = 3 s
v = u + at = 0 + (0.02)(3) = 0.06 m/s
(Page 82, Equation 7.5)
19A racing car has a uniform acceleration of 4 m s⁻². What distance will it cover in 10 s after start?
Answer & Reference:
Given: u = 0, a = 4 m/s², t = 10 s
s = ut + ½at² = 0 + ½(4)(10)² = 2 × 100 = 200 m
(Page 82, Equation 7.6)
20A stone is thrown in a vertically upward direction with a velocity of 5 m s⁻¹. If the acceleration of the stone during its motion is 10 m s⁻² in the downward direction, what will be the height attained by the stone and how much time will it take to reach there?
Answer & Reference:
Given: u = 5 m/s, v = 0 (at highest point), a = -10 m/s²
Using v² = u² + 2as
0 = (5)² + 2(-10)s
0 = 25 - 20s
s = 25/20 = 1.25 m
Using v = u + at
0 = 5 + (-10)t
t = 5/10 = 0.5 s
(Page 83, Equations 7.5 and 7.7)
21An athlete completes one round of a circular track of diameter 200 m in 40 s. What will be the distance covered and the displacement at the end of 2 minutes 20 s?
Answer & Reference:
Given: Diameter = 200 m, so radius r = 100 m
Time for one round = 40 s
Total time = 2 min 20 s = 140 s
Number of rounds = 140/40 = 3.5 rounds
Distance covered = 3.5 × circumference = 3.5 × 2πr = 3.5 × 2 × 3.14 × 100 = 2198 m
After 3.5 rounds, athlete is at diametrically opposite point
Displacement = diameter = 200 m
(Page 85, Exercise 1 - Circular motion concept from Section 7.6)
22Joseph jogs from one end A to the other end B of a straight 300 m road in 2 minutes 30 seconds and then turns around and jogs 100 m back to point C in another 1 minute. What are Joseph’s average speeds and velocities in jogging (a) from A to B and (b) from A to C?
Answer & Reference:
(a) From A to B:
Distance = 300 m, Displacement = 300 m, Time = 150 s
Average speed = 300/150 = 2 m/s
Average velocity = 300/150 = 2 m/s
(b) From A to C:
Distance = 300 + 100 = 400 m, Displacement = 300 - 100 = 200 m, Time = 150 + 60 = 210 s
Average speed = 400/210 ≈ 1.90 m/s
Average velocity = 200/210 ≈ 0.95 m/s
(Page 85, Exercise 2 - Based on concepts from Section 7.2)
23Abdul, while driving to school, computes the average speed for his trip to be 20 km h⁻¹. On his return trip along the same route, there is less traffic and the average speed is 30 km h⁻¹. What is the average speed for Abdul’s trip?
Answer & Reference:
Let distance to school = d km
Time to school = d/20 hours
Time back = d/30 hours
Total distance = 2d km
Total time = d/20 + d/30 = (3d + 2d)/60 = 5d/60 = d/12 hours
Average speed = Total distance/Total time = 2d/(d/12) = 24 km/h
(Page 85, Exercise 3 - Average speed concept from Section 7.2)
24A motorboat starting from rest on a lake accelerates in a straight line at a constant rate of 3.0 m s⁻² for 8.0 s. How far does the boat travel during this time?
Answer & Reference:
Given: u = 0, a = 3.0 m/s², t = 8.0 s
s = ut + ½at² = 0 + ½(3.0)(8.0)² = 1.5 × 64 = 96 m
(Page 85, Exercise 4 - Equation 7.6 from Section 7.5)
25A driver of a car travelling at 52 km h⁻¹ applies the brakes and accelerates uniformly in the opposite direction. The car stops in 5 s. Another driver going at 3 km h⁻¹ in another car applies his brakes slowly and stops in 10 s. On the same graph paper, plot the speed versus time graphs for the two cars. Which of the two cars travelled farther after the brakes were applied?
Answer & Reference:
First car: u = 52 km/h = 14.44 m/s, v = 0, t = 5 s
Distance = area under graph = ½ × 5 × 14.44 = 36.1 m
Second car: u = 3 km/h = 0.83 m/s, v = 0, t = 10 s
Distance = area under graph = ½ × 10 × 0.83 = 4.15 m
The first car travelled farther after brakes were applied.
(Page 85-86, Exercise 5 - Velocity-time graph concept from Section 7.4.2)
26Fig 7.10 shows the distance-time graph of three objects A, B and C. Study the graph and answer the following questions: (a) Which of the three is travelling the fastest? (b) Are all three ever at the same point on the road? (c) How far has C travelled when B passes A? (d) How far has B travelled by the time it passes C?
Answer & Reference:
Since the actual graph (Fig 7.10) is not provided in the PDF, we'll explain based on general principles:
(a) The object with the steepest slope on the distance-time graph is travelling fastest (B).
(b) If the three lines don't intersect at a common point, they are never at the same position.
(c) & (d) These require reading specific values from the graph.
(Page 86, Exercise 6 - Distance-time graph interpretation from Section 7.4.1)
27A ball is gently dropped from a height of 20 m. If its velocity increases uniformly at the rate of 10 m s⁻², with what velocity will it strike the ground? After what time will it strike the ground?
Answer & Reference:
Given: u = 0, s = 20 m, a = 10 m/s²
Using v² = u² + 2as
v² = 0 + 2(10)(20) = 400
v = √400 = 20 m/s
Using v = u + at
20 = 0 + 10t
t = 2 s
(Page 86, Exercise 7 - Equations of motion from Section 7.5)
28The speed-time graph for a car is shown in Fig. 7.11. (a) Find how far does the car travel in the first 4 seconds. Shade the area on the graph that represents the distance travelled by the car during the period. (b) Which part of the graph represents uniform motion of the car?
Answer & Reference:
Since the actual graph (Fig 7.11) is not provided in the PDF, we'll explain based on general principles:
(a) Distance = area under speed-time graph for first 4 seconds. This would be calculated based on the shape of the graph.
(b) The part of the graph where speed is constant (horizontal line) represents uniform motion.
(Page 86, Exercise 8 - Speed-time graph concept from Section 7.4.2)
29State which of the following situations are possible and give an example for each of these: (a) an object with a constant acceleration but with zero velocity (b) an object moving with an acceleration but with uniform speed. (c) an object moving in a certain direction with an acceleration in the perpendicular direction.
Answer & Reference:
(a) Possible. Example: When a ball is thrown upward, at its highest point, velocity is zero but acceleration due to gravity is constant (10 m/s² downward).
(b) Possible. Example: Uniform circular motion - speed is constant but direction changes, so there is centripetal acceleration.
(c) Possible. Example: In uniform circular motion, velocity is tangential while acceleration (centripetal) is perpendicular to it, toward the center.
(Page 86, Exercise 9 - Acceleration concepts from Sections 7.3 and 7.6)
30An artificial satellite is moving in a circular orbit of radius 42250 km. Calculate its speed if it takes 24 hours to revolve around the earth.
Answer & Reference:
Given: r = 42250 km = 4.225 × 10⁷ m, t = 24 h = 86400 s
Circumference = 2πr = 2 × 3.14 × 4.225 × 10⁷ = 2.654 × 10⁸ m
Speed v = distance/time = 2.654 × 10⁸ / 86400 ≈ 3072 m/s
(Page 86, Exercise 10 - Uniform circular motion from Section 7.6, Equation 7.8)
T1Table 7.1: Motion of objects A and B
Answer & Reference:
Time
Distance travelled by object A (m)
Distance travelled by object B (m)
9:30 am
10
12
9:45 am
20
19
10:00 am
30
23
10:15 am
40
35
10:30 am
50
37
10:45 am
60
41
11:00 am
70
44
Object A shows uniform motion (covers 10 m every 15 minutes).
Object B shows non-uniform motion (covers unequal distances in equal time intervals).
(Page 74, Table 7.1 and Activity 7.5)
T2Table 7.2: Distance travelled by a car at regular time intervals
Answer & Reference:
Time (s)
Distance (m)
0
0
2
1
4
4
6
9
8
16
10
25
12
36
This table shows non-uniform motion as the car covers increasing distances in equal time intervals.
(Page 78, Table 7.2)
T3Table 7.3: Velocity of a car at regular instants of time
Answer & Reference:
Time (s)
Velocity (m/s)
Velocity (km/h)
0
0
0
5
2.5
9
10
5.0
18
15
7.5
27
20
10.0
36
25
12.5
45
30
15.0
54
This table shows uniformly accelerated motion as velocity increases by 2.5 m/s every 5 seconds.
(Page 79, Table 7.3)
T4Table 7.4: Distances of stations B and C from A and times of arrival and departure of the train
Answer & Reference:
Station
Distance from A (km)
Time of arrival (hours)
Time of departure (hours)
A
0
08:00
08:15
B
120
11:15
11:30
C
180
13:00
13:15
This table is used for plotting distance-time graph in Activity 7.9.
(Page 80, Table 7.4)
T5Table 7.5: Distance covered by Feroz and Sania at different times on their bicycles
Answer & Reference:
Time
Distance travelled by Feroz (km)
Distance travelled by Sania (km)
8:00 am
0
0
8:05 am
1.0
0.8
8:10 am
1.9
1.6
8:15 am
2.8
2.3
8:20 am
3.6
3.0
8:25 am
-
3.6
This table is used for plotting distance-time graph in Activity 7.10.
(Page 81, Table 7.5)