| Non-Rationalised Science NCERT Notes and Solutions (Class 6th to 10th) | ||||||||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| 6th | 7th | 8th | 9th | 10th | ||||||||||
| Non-Rationalised Science NCERT Notes and Solutions (Class 11th) | ||||||||||||||
| Physics | Chemistry | Biology | ||||||||||||
| Non-Rationalised Science NCERT Notes and Solutions (Class 12th) | ||||||||||||||
| Physics | Chemistry | Biology | ||||||||||||
Chapter 8: Motion
In our daily lives, we observe various objects that are either at rest or in motion. An object is generally perceived to be in motion if its position changes continuously with time. When an object's position does not change relative to its surroundings over time, it is considered to be at rest.
Sometimes, motion can be inferred even if not directly seen, such as deducing air movement from rustling leaves or moving dust particles. Large-scale phenomena like sunrise, sunset, and changing seasons are also consequences of motion, specifically the motion of the Earth.
The perception of motion is relative. An object might appear to be moving to one observer but stationary to another. For instance, passengers in a moving train see the outside world (trees, buildings) as moving, while they perceive fellow passengers inside the train as being at rest relative to themselves. A person standing outside observes the entire train, including passengers, as being in motion.
Motions can be complex, involving straight-line movement, circular paths, rotation, vibration, or combinations thereof. To study motion systematically, we first need methods to describe it accurately.
Describing Motion
To describe the location or position of an object, it is essential to specify a reference point. This reference point is also called the origin. The position of the object is then described relative to this origin.
For example, saying a school is 2 km north of a railway station uses the railway station as the reference point or origin. Choosing a convenient origin is crucial for describing position.
Motion Along A Straight Line
The simplest type of motion is movement along a straight path. To describe this, we use the concepts of distance and displacement.
Consider an object starting at point O (the origin) and moving along a straight line. If it moves from O to A, passing through C and B, the total length of the path covered (O to A) is the distance travelled. Distance is a scalar quantity, meaning it is described only by its numerical value (magnitude) and not direction.
If the object moves from O to A, the distance is 60 km. If it then moves back from A to C, passing through B, the additional distance covered is 35 km (A to C). The total distance travelled for the entire journey (O to A and then A to C) is $60 \text{ km} + 35 \text{ km} = 95 \text{ km}$.
Displacement, on the other hand, is the shortest distance measured from the initial position to the final position of an object. Displacement is a vector quantity, requiring both magnitude and direction for its complete description.
In the example above, if the object moves from O to A, the initial position is O and the final position is A. The shortest distance from O to A is 60 km in the positive direction (assuming rightward is positive). So, the displacement is +60 km.
If the object moves from O to A and then back to C, the initial position is O, and the final position is C. The shortest distance from O to C is 25 km. In this case, the displacement is +25 km (assuming C is at +25 km from O). The magnitude of displacement (25 km) is less than the total distance travelled (95 km).
The magnitude of displacement can be equal to the distance travelled only when the object moves along a straight line in a single direction.
If the object moves from O to A and then back to O, the initial and final positions are the same (O). In this case, the displacement is zero. However, the distance travelled is $60 \text{ km} + 60 \text{ km} = 120 \text{ km}$. Thus, an object can have zero displacement even if it has travelled a non-zero distance.
Uniform Motion And Non-uniform Motion
Motion can be classified based on how the distance covered changes over equal intervals of time.
- Uniform Motion: An object is in uniform motion if it travels equal distances in equal intervals of time along a straight line. For example, a car moving 5 metres every second in a straight line is in uniform motion. The speed remains constant.
- Non-uniform Motion: An object is in non-uniform motion if it travels unequal distances in equal intervals of time. For example, a car moving in a crowded street, speeding up and slowing down, covers different distances in successive equal time intervals. The speed is not constant.
Most motions we observe in everyday life are non-uniform.
Measuring The Rate Of Motion
Objects move at different rates. Some are fast, others are slow. To quantify how fast or slow an object is moving, we use the concept of speed.
Speed is the measure of the rate of motion. It is defined as the distance travelled by an object per unit time.
Speed is a scalar quantity, requiring only magnitude for its description.
The SI unit of speed is metre per second (m/s or m s⁻¹). Other units include centimetre per second (cm/s or cm s⁻¹) and kilometre per hour (km/h or km h⁻¹).
If an object covers a total distance $s$ in a total time $t$, its average speed is given by:
$$ \text{Average speed} = \frac{\text{Total distance travelled}}{\text{Total time taken}} $$ $$ v = \frac{s}{t} $$For uniform motion, the speed is constant and is equal to the average speed. For non-uniform motion, the speed changes, and the average speed represents the overall rate of motion over the entire duration.
Example 8.1. An object travels 16 m in 4 s and then another 16 m in 2 s. What is the average speed of the object?
Answer:
Total distance travelled = 16 m + 16 m = 32 m
Total time taken = 4 s + 2 s = 6 s
Average speed = $\frac{\text{Total distance}}{\text{Total time}} = \frac{32 \text{ m}}{6 \text{ s}}$
Average speed = $\approx 5.33 \text{ m s}^{-1}$
The average speed of the object is approximately 5.33 m s⁻¹.
Example 8.2. The odometer of a car reads 2000 km at the start of a trip and 2400 km at the end of the trip. If the trip took 8 h, calculate the average speed of the car in km h⁻¹ and m s⁻¹.
Answer:
Distance covered by the car, $s$ = Final odometer reading - Initial odometer reading
$s = 2400 \text{ km} - 2000 \text{ km} = 400 \text{ km}$
Time taken, $t = 8 \text{ h}$
Average speed in km h⁻¹: $v_{\text{avg}} = \frac{s}{t} = \frac{400 \text{ km}}{8 \text{ h}} = 50 \text{ km h}^{-1}$
To convert to m s⁻¹, we use $1 \text{ km} = 1000 \text{ m}$ and $1 \text{ h} = 3600 \text{ s}$.
$50 \text{ km h}^{-1} = 50 \times \frac{1000 \text{ m}}{3600 \text{ s}} = \frac{50000}{3600} \text{ m s}^{-1} = \frac{500}{36} \text{ m s}^{-1} \approx 13.89 \text{ m s}^{-1}$
The average speed of the car is 50 km h⁻¹, which is approximately 13.89 m s⁻¹.
Speed With Direction
To describe the rate of motion more completely, especially in cases where direction matters, we use velocity.
Velocity is defined as the speed of an object moving in a definite direction. It is the displacement per unit time.
Velocity is a vector quantity, described by both its magnitude (speed) and direction.
The velocity of an object can change if its speed changes, if its direction of motion changes, or if both change.
For motion where velocity is changing at a uniform rate along a straight line, the average velocity is given by the arithmetic mean of the initial and final velocities:
$$ \text{Average velocity} = \frac{\text{Initial velocity} + \text{Final velocity}}{2} $$ $$ v_{\text{avg}} = \frac{u + v}{2} $$where $u$ is the initial velocity and $v$ is the final velocity.
The units of velocity are the same as speed (m/s, cm/s, km/h).
The magnitude of average velocity is equal to the average speed only when the object moves along a straight line in the same direction.
Example 8.3. Usha swims in a 90 m long pool. She covers 180 m in one minute by swimming from one end to the other and back along the same straight path. Find the average speed and average velocity of Usha.
Answer:
Length of the pool = 90 m.
Time taken = 1 minute = 60 seconds.
Distance covered = Swimming from one end to the other (90 m) and back (90 m). Total distance = 90 m + 90 m = 180 m.
Displacement = Starting point and ending point are the same (one end of the pool). Displacement = 0 m.
Average speed = $\frac{\text{Total distance}}{\text{Total time}} = \frac{180 \text{ m}}{60 \text{ s}} = 3 \text{ m s}^{-1}$.
Average velocity = $\frac{\text{Displacement}}{\text{Total time}} = \frac{0 \text{ m}}{60 \text{ s}} = 0 \text{ m s}^{-1}$.
Usha's average speed is 3 m s⁻¹, and her average velocity is 0 m s⁻¹.
Rate Of Change Of Velocity
When an object is in uniform motion along a straight line, its velocity is constant, meaning the change in velocity is zero for any time interval. However, in non-uniform motion, the velocity changes with time.
To describe how quickly the velocity of an object is changing, we use the term acceleration.
Acceleration is defined as the rate of change of velocity per unit time.
If the velocity of an object changes from an initial value $u$ to a final value $v$ in time $t$, the acceleration $a$ is given by:
$$ \text{Acceleration} = \frac{\text{Change in velocity}}{\text{Time taken}} = \frac{\text{Final velocity} - \text{Initial velocity}}{\text{Time taken}} $$ $$ a = \frac{v - u}{t} $$Acceleration is a vector quantity. The SI unit of acceleration is metre per second squared (m/s² or m s⁻²).
An object's motion is called accelerated motion if its velocity is changing.
- If velocity increases, acceleration is positive (in the direction of velocity).
- If velocity decreases (retardation or deceleration), acceleration is negative (opposite to the direction of velocity).
Types of Acceleration:
- Uniform Acceleration: If an object travels in a straight line and its velocity changes by equal amounts in equal intervals of time. Example: A freely falling object experiences uniform acceleration due to gravity (assuming negligible air resistance).
- Non-uniform Acceleration: If an object's velocity changes by unequal amounts in equal intervals of time. Example: A car moving on a straight road and increasing its speed unevenly.
Example 8.4. Starting from a stationary position, Rahul paddles his bicycle to attain a velocity of 6 m s⁻¹ in 30 s. Then he applies brakes such that the velocity of the bicycle comes down to 4 m s⁻¹ in the next 5 s. Calculate the acceleration of the bicycle in both the cases.
Answer:
Case 1: Starting from rest and speeding up
- Initial velocity, $u$ = 0 m s⁻¹ (stationary position)
- Final velocity, $v$ = 6 m s⁻¹
- Time taken, $t$ = 30 s
Acceleration, $a_1 = \frac{v - u}{t} = \frac{6 \text{ m s}^{-1} - 0 \text{ m s}^{-1}}{30 \text{ s}} = \frac{6}{30} \text{ m s}^{-2} = 0.2 \text{ m s}^{-2}$.
The acceleration in the first case is 0.2 m s⁻².
Case 2: Applying brakes and slowing down
- Initial velocity, $u$ = 6 m s⁻¹ (velocity before applying brakes)
- Final velocity, $v$ = 4 m s⁻¹ (velocity after 5 s of braking)
- Time taken, $t$ = 5 s
Acceleration, $a_2 = \frac{v - u}{t} = \frac{4 \text{ m s}^{-1} - 6 \text{ m s}^{-1}}{5 \text{ s}} = \frac{-2}{5} \text{ m s}^{-2} = -0.4 \text{ m s}^{-2}$.
The acceleration in the second case is -0.4 m s⁻². The negative sign indicates deceleration or retardation, meaning the acceleration is in the direction opposite to the motion.
Graphical Representation Of Motion
Graphs are useful tools for visualising and analysing motion. They show how physical quantities related to motion (like distance, velocity) change over time.
Distance–time Graphs
A distance-time graph plots the distance travelled by an object on the y-axis against time on the x-axis. These graphs can represent different types of motion, including uniform speed, non-uniform speed, or an object at rest.
Interpreting Distance-time Graphs:
- Object at Rest: If an object is at rest, its distance from the origin does not change over time. The distance-time graph is a straight line parallel to the time (x) axis.
- Uniform Speed: For uniform motion, the object covers equal distances in equal time intervals, meaning distance is directly proportional to time. The distance-time graph is a straight line with a constant slope. A steeper slope indicates higher speed.
- Non-uniform Speed (Accelerated Motion): For non-uniform motion, the object covers unequal distances in equal time intervals. The distance-time graph is a curved line. If the curve is bending upwards, it indicates increasing speed (acceleration). If it's bending downwards, it indicates decreasing speed (deceleration).
The speed of an object from a distance-time graph for uniform motion can be calculated from the slope of the line. For any two points (t₁, s₁) and (t₂, s₂) on the straight line graph, the speed ($v$) is given by the change in distance divided by the change in time:
$$ v = \frac{s_2 - s_1}{t_2 - t_1} $$Velocity-time Graphs
A velocity-time graph plots the velocity of an object on the y-axis against time on the x-axis. These graphs are particularly useful for studying accelerated motion.
Interpreting Velocity-time Graphs:
- Uniform Velocity (Zero Acceleration): If an object moves with uniform velocity, its velocity remains constant over time. The velocity-time graph is a straight line parallel to the time (x) axis.
- Uniform Acceleration: If an object moves with uniform acceleration, its velocity changes by equal amounts in equal time intervals. The velocity-time graph is a straight line with a constant slope. A positive slope indicates positive acceleration, while a negative slope indicates negative acceleration (deceleration).
- Non-uniform Acceleration: If an object's velocity changes unevenly over time, the velocity-time graph is a curved line. The shape of the curve depends on how the acceleration is changing.
Distance/Displacement from Velocity-time Graph:
A crucial aspect of velocity-time graphs is that the area enclosed by the graph and the time axis represents the magnitude of the displacement (or distance if the motion is in a straight line in one direction) of the object during that time interval.
For uniform velocity (Fig. above, straight line parallel to x-axis), the distance covered between time $t_1$ and $t_2$ is the area of the rectangle formed by the graph, the time axis, and the vertical lines at $t_1$ and $t_2$. Distance = velocity $\times$ time interval.
For uniform acceleration (Fig. above, straight line with slope), the distance covered during a time interval is the area under the straight line segment of the graph corresponding to that interval. This area can be calculated by dividing it into simple geometric shapes (rectangles and triangles).
Equations Of Motion By Graphical Method
For objects moving with uniform acceleration along a straight line, we can derive a set of mathematical relationships between initial velocity ($u$), final velocity ($v$), acceleration ($a$), time ($t$), and distance covered ($s$). These are known as the equations of motion.
There are three main equations of motion:
- $v = u + at$ (Velocity-Time Relation)
- $s = ut + \frac{1}{2}at^2$ (Position-Time Relation)
- $2as = v^2 - u^2$ (Position-Velocity Relation)
These equations can be derived using the properties of velocity-time graphs for uniformly accelerated motion.
Consider the velocity-time graph above, representing an object starting with initial velocity $u$ at time $t=0$ and uniformly accelerating to final velocity $v$ at time $t$. The motion is represented by the straight line AB.
- Initial velocity at $t=0$ is OA = $u$.
- Final velocity at time $t$ is BC = $v$.
- Time interval is OC = $t$.
- BD represents the change in velocity in time $t$.
Equation For Velocity-time Relation
We know that acceleration ($a$) is the slope of the velocity-time graph. For the line AB, the slope is the change in velocity (BD) divided by the time interval (AD or OC).
Change in velocity, BD = BC - CD. Since CD = OA (from the graph), BD = $v - u$.
Time interval, AD = OC = $t$.
Acceleration, $a = \frac{\text{Change in velocity}}{\text{Time interval}} = \frac{BD}{AD} = \frac{v - u}{t}$
Rearranging this equation gives the first equation of motion:
$$ v - u = at $$ $$ v = u + at $$Equation For Position-time Relation
The distance ($s$) travelled by the object in time $t$ is given by the area under the velocity-time graph OABC. This area is a trapezium, but we can calculate it as the sum of the area of the rectangle OADC and the area of the triangle ABD.
Area of rectangle OADC = Length $\times$ Width = OA $\times$ OC = $u \times t = ut$.
Area of triangle ABD = $\frac{1}{2} \times$ Base $\times$ Height = $\frac{1}{2} \times$ AD $\times$ BD.
From the graph, AD = OC = $t$. From the velocity-time relation, BD = $at$.
Area of triangle ABD = $\frac{1}{2} \times t \times at = \frac{1}{2}at^2$.
Total distance, $s$ = Area of rectangle OADC + Area of triangle ABD
$s = ut + \frac{1}{2}at^2$
This is the second equation of motion.
Equation For Position–velocity Relation
The distance ($s$) travelled by the object in time $t$ can also be calculated as the area of the trapezium OABC directly.
Area of trapezium OABC = $\frac{1}{2} \times$ (Sum of parallel sides) $\times$ Height.
Parallel sides are OA ($u$) and BC ($v$). Height is OC ($t$).
$s = \frac{1}{2} \times (OA + BC) \times OC = \frac{1}{2} \times (u + v) \times t$.
From the first equation of motion ($v = u + at$), we can express $t$ as $t = \frac{v - u}{a}$.
Substitute this expression for $t$ into the trapezium area formula:
$s = \frac{1}{2} \times (u + v) \times \left(\frac{v - u}{a}\right)$
$s = \frac{(v + u)(v - u)}{2a}$
Using the algebraic identity $(v+u)(v-u) = v^2 - u^2$:
$s = \frac{v^2 - u^2}{2a}$
Rearranging gives the third equation of motion:
$$ 2as = v^2 - u^2 $$ $$ v^2 = u^2 + 2as $$Example 8.5. A train starting from rest attains a velocity of 72 km h⁻¹ in 5 minutes. Assuming that the acceleration is uniform, find (i) the acceleration and (ii) the distance travelled by the train for attaining this velocity.
Answer:
Given:
- Initial velocity, $u = 0$ m s⁻¹ (starting from rest)
- Final velocity, $v = 72 \text{ km h}^{-1}$
- Time, $t = 5 \text{ minutes}$
First, convert units to SI units (m and s):
- $v = 72 \text{ km h}^{-1} = 72 \times \frac{1000 \text{ m}}{3600 \text{ s}} = 72 \times \frac{5}{18} \text{ m s}^{-1} = 4 \times 5 \text{ m s}^{-1} = 20 \text{ m s}^{-1}$
- $t = 5 \text{ minutes} = 5 \times 60 \text{ s} = 300 \text{ s}$
(i) To find acceleration ($a$), use the first equation of motion: $v = u + at$.
$20 \text{ m s}^{-1} = 0 \text{ m s}^{-1} + a \times 300 \text{ s}$
$a = \frac{20 \text{ m s}^{-1}}{300 \text{ s}} = \frac{2}{30} \text{ m s}^{-2} = \frac{1}{15} \text{ m s}^{-2}$
The acceleration is $\frac{1}{15}$ m s⁻² (approximately 0.067 m s⁻²).
(ii) To find the distance ($s$), use the third equation of motion: $v^2 = u^2 + 2as$.
$(20 \text{ m s}^{-1})^2 = (0 \text{ m s}^{-1})^2 + 2 \times \left(\frac{1}{15} \text{ m s}^{-2}\right) \times s$
$400 \text{ m}^2\text{s}^{-2} = 0 + \frac{2}{15} \text{ m s}^{-2} \times s$
$s = \frac{400 \text{ m}^2\text{s}^{-2}}{\frac{2}{15} \text{ m s}^{-2}} = 400 \times \frac{15}{2} \text{ m} = 200 \times 15 \text{ m} = 3000 \text{ m}$
Distance $s = 3000 \text{ m} = 3 \text{ km}$.
The train's acceleration is $\frac{1}{15}$ m s⁻² and it travels 3 km to attain the final velocity.
Example 8.6. A car accelerates uniformly from 18 km h⁻¹ to 36 km h⁻¹ in 5 s. Calculate (i) the acceleration and (ii) the distance covered by the car in that time.
Answer:
Given:
- Initial velocity, $u = 18 \text{ km h}^{-1}$
- Final velocity, $v = 36 \text{ km h}^{-1}$
- Time, $t = 5 \text{ s}$
Convert velocities to SI units (m s⁻¹):
- $u = 18 \text{ km h}^{-1} = 18 \times \frac{1000 \text{ m}}{3600 \text{ s}} = 18 \times \frac{5}{18} \text{ m s}^{-1} = 5 \text{ m s}^{-1}$
- $v = 36 \text{ km h}^{-1} = 36 \times \frac{1000 \text{ m}}{3600 \text{ s}} = 36 \times \frac{5}{18} \text{ m s}^{-1} = 2 \times 5 \text{ m s}^{-1} = 10 \text{ m s}^{-1}$
(i) To find acceleration ($a$), use $v = u + at$.
$10 \text{ m s}^{-1} = 5 \text{ m s}^{-1} + a \times 5 \text{ s}$
$10 - 5 = 5a$
$5 = 5a$
$a = \frac{5}{5} \text{ m s}^{-2} = 1 \text{ m s}^{-2}$
The acceleration of the car is 1 m s⁻².
(ii) To find the distance ($s$), use $s = ut + \frac{1}{2}at^2$.
$s = (5 \text{ m s}^{-1}) \times (5 \text{ s}) + \frac{1}{2} \times (1 \text{ m s}^{-2}) \times (5 \text{ s})^2$
$s = 25 \text{ m} + \frac{1}{2} \times 1 \text{ m s}^{-2} \times 25 \text{ s}^2$
$s = 25 \text{ m} + 12.5 \text{ m} = 37.5 \text{ m}$
The car covers a distance of 37.5 m during this time.
Example 8.7. The brakes applied to a car produce an acceleration of 6 m s⁻² in the opposite direction to the motion. If the car takes 2 s to stop after the application of brakes, calculate the distance it travels during this time.
Answer:
Given:
- Acceleration, $a = -6 \text{ m s}^{-2}$ (opposite direction to motion, so negative)
- Time taken to stop, $t = 2 \text{ s}$
- Final velocity, $v = 0 \text{ m s}^{-1}$ (car stops)
First, find the initial velocity ($u$) before brakes were applied, using $v = u + at$.
$0 \text{ m s}^{-1} = u + (-6 \text{ m s}^{-2}) \times (2 \text{ s})$
$0 = u - 12 \text{ m s}^{-1}$
$u = 12 \text{ m s}^{-1}$
The initial velocity of the car was 12 m s⁻¹.
Now, find the distance ($s$) travelled during braking using $s = ut + \frac{1}{2}at^2$.
$s = (12 \text{ m s}^{-1}) \times (2 \text{ s}) + \frac{1}{2} \times (-6 \text{ m s}^{-2}) \times (2 \text{ s})^2$
$s = 24 \text{ m} + \frac{1}{2} \times (-6 \text{ m s}^{-2}) \times (4 \text{ s}^2)$
$s = 24 \text{ m} - 12 \text{ m} = 12 \text{ m}$
Alternatively, using $v^2 = u^2 + 2as$:
$(0 \text{ m s}^{-1})^2 = (12 \text{ m s}^{-1})^2 + 2 \times (-6 \text{ m s}^{-2}) \times s$
$0 = 144 \text{ m}^2\text{s}^{-2} - 12 \text{ m s}^{-2} \times s$
$12s = 144$
$s = \frac{144}{12} \text{ m} = 12 \text{ m}$
The car travels 12 m during the time the brakes are applied until it stops. This is why maintaining a safe distance between vehicles is important.
Uniform Circular Motion
We have learned that acceleration occurs when velocity changes, either in magnitude or direction or both. A special case arises when an object moves with constant speed but its direction continuously changes.
Consider an object moving along a closed path. If the path is a rectangle, to complete one round while maintaining constant speed along the sides, the object must change direction at each of the four corners. If the path is a hexagon, it changes direction at six corners. As the number of sides of a regular closed polygon increases, the path gradually approaches the shape of a circle. Moving along a circular path involves continuously changing direction.
Uniform Circular Motion is the motion of an object along a circular path at a constant speed.
Even though the speed is constant, the velocity is continuously changing because the direction of motion is changing at every point on the circular path. A change in velocity means there is acceleration. Therefore, uniform circular motion is an example of accelerated motion.
If an object moves in a circular path of radius $r$ and completes one revolution in time $t$, its speed ($v$) is given by the distance covered (circumference) divided by the time taken:
$$ v = \frac{\text{Circumference}}{\text{Time period}} = \frac{2\pi r}{t} $$When an object in uniform circular motion is released, it moves in a straight line tangential to the circular path at the point of release. This confirms that the direction of velocity was tangential to the circle at that instant.
Examples of Uniform Circular Motion:
- Motion of the Moon around the Earth.
- Motion of a satellite in a circular orbit around the Earth.
- A cyclist riding at a constant speed on a circular track.
- The tip of a second hand of a watch moving in a circle.
Intext Questions
Page No. 100
Question 1. An object has moved through a distance. Can it have zero displacement? If yes, support your answer with an example.
Answer:
Question 2. A 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:
Question 3. Which 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:
Page No. 102
Question 1. Distinguish between speed and velocity.
Answer:
Question 2. Under what condition(s) is the magnitude of average velocity of an object equal to its average speed?
Answer:
Question 3. What does the odometer of an automobile measure?
Answer:
Question 4. What does the path of an object look like when it is in uniform motion?
Answer:
Question 5. During 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, $3 \times 10^8 m s^{-1}$.
Answer:
Page No. 103
Question 1. When will you say a body is in (i) uniform acceleration? (ii) non-uniform acceleration?
Answer:
Question 2. A bus decreases its speed from 80 km $h^{-1}$ to 60 km $h^{-1}$ in 5 s. Find the acceleration of the bus.
Answer:
Question 3. A 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:
Page No. 107
Question 1. What is the nature of the distance-time graphs for uniform and non-uniform motion of an object?
Answer:
Question 2. What can you say about the motion of an object whose distance-time graph is a straight line parallel to the time axis?
Answer:
Question 3. What can you say about the motion of an object if its speed-time graph is a straight line parallel to the time axis?
Answer:
Question 4. What is the quantity which is measured by the area occupied below the velocity-time graph?
Answer:
Page No. 109 - 110
Question 1. A 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:
Question 2. A 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:
Question 3. A 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:
Question 4. A racing car has a uniform acceleration of 4 $m s^{-2}$. What distance will it cover in 10 s after start?
Answer:
Question 5. A 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:
Exercises
Question 1. An 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:
Question 2. Joseph 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:
Question 3. Abdul, 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:
Question 4. A 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:
Question 5. A 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:
Question 6. Fig 8.11 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:
Question 7. A 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:
Question 8. The speed-time graph for a car is shown is Fig. 8.12.
(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:
Question 9. State 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:
Question 10. An 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: