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Chapter 7 Motion
In our daily lives, we observe countless objects in different states of motion or rest. Motion is fundamentally about the change in an object's position over time. While sometimes motion is directly visible, like a bird flying or a car moving, other times we infer it indirectly, such as the movement of air felt as wind, or the Earth's motion causing sunrise and sunset.
Motion can be relative; an object perceived as moving by one observer might seem stationary to another, depending on their own state of motion or choice of reference point. Motion itself can be complex, involving straight lines, curves, rotation, or vibration, often in combination. This chapter focuses on describing the simplest type of motion – motion along a straight line – using mathematical equations and graphs, and introduces the concept of uniform circular motion.
Description Of Motion
To accurately describe the position of an object, we need a **reference point**. This fixed point or object is called the **origin**. By specifying the position of an object relative to the origin, we can locate it precisely. For example, saying a school is 2 km north of the railway station uses the railway station as the reference point (origin).
Motion Along A Straight Line
The simplest form of motion is movement along a straight line. We can describe this motion using two key quantities: distance and displacement.
- **Distance:** The **total path length** covered by an object during its motion. It is a scalar quantity, meaning it is described only by its numerical value (magnitude) and does not include direction.
- **Displacement:** The **shortest straight-line distance** from the initial position to the final position of an object. It is a vector quantity, having both magnitude and direction.
Consider an object moving along a straight line from point O (the origin or reference point) to point A, then back to point B, and finally to point C, as illustrated below:
Let the distance from O to C be 25 km, O to B be 35 km, and O to A be 60 km.
- If the object moves from O to A:
- Distance covered = OA = 60 km.
- Magnitude of Displacement = OA = 60 km (direction is from O to A).
- If the object moves from O to A and then back to B:
- Distance covered = (O to A) + (A to B) = 60 km + (60 - 35) km = 60 km + 25 km = 85 km.
- Magnitude of Displacement = Shortest distance from O to B = OB = 35 km (direction is from O to B).
- If the object moves from O to A and then back to O:
- Distance covered = (O to A) + (A to O) = 60 km + 60 km = 120 km.
- Magnitude of Displacement = 0 km, because the final position (O) is the same as the initial position (O).
From these examples, we see that the distance covered is always positive and accumulates throughout the journey. The magnitude of displacement, however, depends only on the initial and final positions and can be equal to, less than, or even zero, regardless of the distance travelled. The magnitude of displacement is only equal to the distance travelled when the object moves along a straight line in a single direction.
Question 1. An object has moved through a distance. Can it have zero displacement? If yes, support your answer with an example.
Answer:
Yes, an object can have zero displacement even if it has moved through a distance. This happens when the object returns to its initial position after covering some distance. For example, if a person walks from their home to a market and then walks back to their home, the total distance covered is twice the distance between the home and the market. However, since their final position (home) is the same as their initial position (home), their total displacement is zero.
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:
Side of the square field = 10 m.
Perimeter of the square field = $4 \times \text{side} = 4 \times 10 \text{ m} = 40 \text{ m}$.
Time taken to complete one round along the boundary = 40 s.
Total time for which the farmer moves = 2 minutes 20 seconds = $(2 \times 60) \text{ s} + 20 \text{ s} = 120 \text{ s} + 20 \text{ s} = 140 \text{ s}$.
Number of rounds completed in 140 s = $\frac{\text{Total time}}{\text{Time per round}} = \frac{140 \text{ s}}{40 \text{ s}} = \frac{14}{4} = 3.5$ rounds.
After completing 3 full rounds, the farmer is back at his initial position. In the remaining 0.5 (half) round, the farmer moves along half the perimeter of the square field.
If the farmer starts from one corner (say A) and moves along the boundary A $\to$ B $\to$ C $\to$ D $\to$ A for one round, after 3.5 rounds, he will be at the corner diagonally opposite to the starting point (i.e., at corner C if starting from A).
The magnitude of displacement is the shortest distance between the initial position (A) and the final position (C). This is the length of the diagonal of the square.
Diagonal of a square = $\sqrt{\text{side}^2 + \text{side}^2} = \sqrt{10^2 + 10^2} = \sqrt{100 + 100} = \sqrt{200} = 10\sqrt{2} \text{ m}$.
The magnitude of displacement of the farmer at the end of 2 minutes 20 seconds is $10\sqrt{2}$ m (approximately 14.14 m).
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:
Neither (a) nor (b) is true for displacement. Displacement can be zero (as seen in Question 1), and its magnitude is always less than or equal to the distance travelled (as seen in the example with Fig 7.1).
Therefore, the correct answer is: **Neither (a) nor (b) is true.**
Uniform Motion And Non-Uniform Motion
The type of motion can be classified based on whether the object covers equal distances in equal time intervals.
- **Uniform Motion:** An object is in uniform motion if it covers **equal distances in equal intervals of time** along a straight line. In uniform motion, the object's speed and direction remain constant. For example, a car moving on a straight highway at a steady speed of 60 km/h.
- **Non-Uniform Motion:** An object is in non-uniform motion if it covers **unequal distances in equal intervals of time**. This can happen if the object's speed changes or its direction changes (or both). Examples include a car moving in city traffic, a bouncing ball, or a person jogging in a park.
The time intervals considered should be small for this definition to be precise.
Measuring The Rate Of Motion
The rate at which an object moves tells us how fast or slow it is going. This rate can be measured using speed or velocity.
Speed
**Speed** is the measure of the **distance travelled by an object per unit time**. It is a scalar quantity, described only by its magnitude.
$$ \text{Speed} = \frac{\text{Distance travelled}}{\text{Time taken}} $$
The standard international (SI) unit of speed is **metres per second (m/s or m s⁻¹)**. Other common units include centimetres per second (cm/s) and kilometres per hour (km/h).
For objects in non-uniform motion (where speed changes over time), we often describe the rate of motion using **average speed**.
$$ \text{Average Speed} = \frac{\text{Total distance travelled}}{\text{Total time taken}} $$
If an object travels a total distance $s$ in total time $t$, its average speed $v_{avg}$ is given by:
$$ v_{avg} = \frac{s}{t} $$
Example 7.1. 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 = $\frac{\text{Total distance travelled}}{\text{Total time taken}} = \frac{32 \text{ m}}{6 \text{ s}} = 5.33 \text{ m/s}$
The average speed of the object is approximately 5.33 m/s.
Example 7.2. 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, $s = \text{Final reading} - \text{Initial reading} = 2400 \text{ km} - 2000 \text{ km} = 400 \text{ km}$.
Time elapsed, $t = 8 \text{ h}$.
Average speed in km h⁻¹:
$$ v_{avg} = \frac{s}{t} = \frac{400 \text{ km}}{8 \text{ h}} = 50 \text{ km/h} $$
To convert km/h to m/s, we use the conversion factors: 1 km = 1000 m and 1 h = 3600 s.
$$ 50 \text{ km/h} = 50 \times \frac{1000 \text{ m}}{3600 \text{ s}} = 50 \times \frac{10}{36} \text{ m/s} = 50 \times \frac{5}{18} \text{ m/s} = \frac{250}{18} \text{ m/s} \approx 13.89 \text{ m/s} $$
The average speed of the car is 50 km/h or approximately 13.89 m/s.
Speed With Direction (Velocity)
To provide a more complete description of motion, we need to include the direction of motion along with the speed. This quantity is called **velocity**.
Velocity is the speed of an object moving in a definite direction. It is a vector quantity, having both magnitude (speed) and direction.
$$ \text{Velocity} = \frac{\text{Displacement}}{\text{Time taken}} $$
The velocity of an object can be uniform (constant speed and direction) or variable. Velocity can change if the object's speed changes, its direction of motion changes, or both.
For motion where the velocity is changing at a uniform rate (uniform acceleration), the **average velocity** can be calculated as the arithmetic mean of the initial and final velocities:
$$ \text{Average Velocity}, v_{av} = \frac{\text{Initial velocity} (u) + \text{Final velocity} (v)}{2} $$
The SI unit of velocity is the same as speed: **metres per second (m/s or m s⁻¹)**.
Example 7.3. 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.
Total distance covered in 1 minute (60 s) = 180 m.
Since she swims from one end to the other and back along the same path, her starting point and ending point are the same.
Total displacement in 1 minute = 0 m.
Average speed = $\frac{\text{Total distance travelled}}{\text{Total time taken}} = \frac{180 \text{ m}}{60 \text{ s}} = 3 \text{ m/s}$.
Average velocity = $\frac{\text{Total displacement}}{\text{Total time taken}} = \frac{0 \text{ m}}{60 \text{ s}} = 0 \text{ m/s}$.
Usha's average speed is 3 m/s, and her average velocity is 0 m/s.
Question 1. Distinguish between speed and velocity.
Answer:
The key differences between speed and velocity are:
| Feature | Speed | Velocity |
|---|---|---|
| Definition | Distance covered per unit time. | Displacement per unit time (speed in a definite direction). |
| Type of quantity | Scalar (magnitude only). | Vector (magnitude and direction). |
| Value | Always non-negative (>= 0). | Can be positive, negative, or zero (direction matters). |
| Change | Changes when speed changes. | Changes if speed changes, direction changes, or both. |
Question 2. Under what condition(s) is the magnitude of average velocity of an object equal to its average speed?
Answer:
The magnitude of average velocity is equal to the average speed when the object moves along a **straight line in the same direction**. In this case, the magnitude of displacement is equal to the total distance travelled, and therefore average velocity magnitude equals average speed.
Question 3. What does the odometer of an automobile measure?
Answer:
The odometer of an automobile measures the **distance travelled** by the vehicle.
Question 4. What does the path of an object look like when it is in uniform motion?
Answer:
When an object is in uniform motion, its path is a **straight line**. (By definition, uniform motion refers to motion along a straight line with constant speed and direction).
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 × 10⁸ m s⁻¹.
Answer:
Speed of the signal (speed of light), $v = 3 \times 10^8 \text{ m/s}$.
Time taken by the signal to reach the ground station, $t = 5 \text{ minutes}$.
First, convert the time into seconds: $t = 5 \times 60 \text{ s} = 300 \text{ s}$.
Distance = Speed $\times$ Time
$s = v \times t = (3 \times 10^8 \text{ m/s}) \times (300 \text{ s})$
$s = (3 \times 10^8) \times (3 \times 10^2) \text{ m}$
$s = 9 \times 10^{(8+2)} \text{ m} = 9 \times 10^{10} \text{ m}$.
The distance of the spaceship from the ground station was $9 \times 10^{10}$ metres.
Rate Of Change Of Velocity (Acceleration)
In non-uniform motion, the velocity of an object changes over time. The rate at which this velocity changes is called **acceleration**.
Acceleration is the change in velocity per unit time.
If the initial velocity of an object is $u$ at time $t=0$ and its final velocity becomes $v$ after time $t$, the acceleration $a$ is given by:
$$ \text{Acceleration}, a = \frac{\text{Change in velocity}}{\text{Time taken}} = \frac{\text{Final velocity} - \text{Initial velocity}}{\text{Time taken}} $$
$$ a = \frac{v - u}{t} $$
The SI unit of acceleration is **metres per second squared (m/s² or m s⁻²)**.
- If the velocity of the object increases, the acceleration is positive (in the direction of velocity). This is called **acceleration**.
- If the velocity decreases, the acceleration is negative (opposite to the direction of velocity). This is called **deceleration** or **retardation**.
Like velocity, acceleration can also be uniform or non-uniform.
- **Uniform Acceleration:** An object has uniform acceleration if its velocity changes by **equal amounts in equal intervals of time** along a straight line. The motion of a freely falling body under gravity (ignoring air resistance) is an example of uniformly accelerated motion.
- **Non-Uniform Acceleration:** An object has non-uniform acceleration if its velocity changes at a **non-uniform rate** (i.e., by unequal amounts in equal intervals of time). For example, a car accelerating on a crowded road where the driver has to change speed frequently.
Example 7.4. 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 accelerating
Initial velocity, $u = 0$ m/s (stationary position)
Final velocity, $v = 6$ m/s
Time taken, $t = 30$ s
Acceleration, $a = \frac{v - u}{t} = \frac{6 \text{ m/s} - 0 \text{ m/s}}{30 \text{ s}} = \frac{6 \text{ m/s}}{30 \text{ s}} = 0.2 \text{ m/s}^2$.
The acceleration in the first case is 0.2 m/s².
Case 2: Applying brakes and decelerating
Initial velocity, $u = 6$ m/s (This is the velocity just before applying brakes)
Final velocity, $v = 4$ m/s
Time taken, $t = 5$ s
Acceleration, $a = \frac{v - u}{t} = \frac{4 \text{ m/s} - 6 \text{ m/s}}{5 \text{ s}} = \frac{-2 \text{ m/s}}{5 \text{ s}} = -0.4 \text{ m/s}^2$.
The acceleration in the second case is -0.4 m/s² (indicating deceleration or retardation).
Question 1. When will you say a body is in (i) uniform acceleration? (ii) non-uniform acceleration?
Answer:
(i) A body is in **uniform acceleration** if its velocity changes by equal amounts in equal intervals of time along a straight line. For example, the motion of a stone falling freely under gravity (assuming negligible air resistance).
(ii) A body is in **non-uniform acceleration** if its velocity changes by unequal amounts in equal intervals of time. For example, the motion of a car in traffic, where the speed increases and decreases irregularly.
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:
Initial speed, $u = 80$ km/h. Convert to m/s: $u = 80 \times \frac{1000 \text{ m}}{3600 \text{ s}} = 80 \times \frac{5}{18} \text{ m/s} = \frac{400}{18} \text{ m/s} = \frac{200}{9} \text{ m/s} \approx 22.22 \text{ m/s}$.
Final speed, $v = 60$ km/h. Convert to m/s: $v = 60 \times \frac{1000 \text{ m}}{3600 \text{ s}} = 60 \times \frac{5}{18} \text{ m/s} = \frac{300}{18} \text{ m/s} = \frac{50}{3} \text{ m/s} \approx 16.67 \text{ m/s}$.
Time taken, $t = 5$ s.
Assuming motion is along a straight line, initial velocity $u = \frac{200}{9}$ m/s and final velocity $v = \frac{50}{3}$ m/s.
Acceleration, $a = \frac{v - u}{t} = \frac{\frac{50}{3} \text{ m/s} - \frac{200}{9} \text{ m/s}}{5 \text{ s}}$
$a = \frac{\frac{150}{9} \text{ m/s} - \frac{200}{9} \text{ m/s}}{5 \text{ s}} = \frac{-\frac{50}{9} \text{ m/s}}{5 \text{ s}} = -\frac{50}{9 \times 5} \text{ m/s}^2 = -\frac{10}{9} \text{ m/s}^2$.
$a \approx -1.11 \text{ m/s}^2$.
The acceleration of the bus is $-\frac{10}{9}$ m/s² (or approximately -1.11 m/s²), which indicates deceleration.
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:
Initial velocity, $u = 0$ km/h (starting from rest).
Final velocity, $v = 40$ km/h. Convert to m/s: $v = 40 \times \frac{5}{18} \text{ m/s} = \frac{200}{18} \text{ m/s} = \frac{100}{9} \text{ m/s}$.
Time taken, $t = 10$ minutes. Convert to seconds: $t = 10 \times 60 \text{ s} = 600 \text{ s}$.
Acceleration, $a = \frac{v - u}{t} = \frac{\frac{100}{9} \text{ m/s} - 0 \text{ m/s}}{600 \text{ s}}$
$a = \frac{100}{9 \times 600} \text{ m/s}^2 = \frac{1}{9 \times 6} \text{ m/s}^2 = \frac{1}{54} \text{ m/s}^2$.
The acceleration of the train is $\frac{1}{54}$ m/s² (approximately 0.0185 m/s²).
Graphical Representation Of Motion
Graphs are powerful tools for visually representing the relationship between physical quantities and understanding the nature of motion.
Distance–Time Graphs
A **distance-time graph** plots the distance travelled by an object on the y-axis against the time taken on the x-axis. A suitable scale is chosen for both axes.
- **Uniform Speed:** For an object moving at uniform speed, the distance covered is directly proportional to the time taken. The distance-time graph is a **straight line**. The slope of this straight line represents the **speed** of the object.
If we take two points (t₁, s₁) and (t₂, s₂) on the graph, the speed $v$ is calculated as:
$$ v = \frac{s_2 - s_1}{t_2 - t_1} = \text{Slope of the graph} $$
- **Object at Rest:** If an object is at rest, its position (distance from the origin) does not change with time. The distance-time graph for an object at rest is a **straight line parallel to the time (x) axis**.
- **Non-Uniform Speed:** For an object moving with non-uniform speed, the distance covered in equal time intervals is unequal. The distance-time graph for non-uniform motion is a **curved line**. The shape of the curve indicates how the speed is changing (e.g., getting steeper if speeding up, flattening if slowing down).
Velocity–Time Graphs
A **velocity-time graph** plots the velocity of an object on the y-axis against the time taken on the x-axis.
- **Uniform Velocity:** For an object moving at uniform velocity, its velocity remains constant over time. The velocity-time graph is a **straight line parallel to the time (x) axis**.
The **area enclosed by the velocity-time graph and the time axis** represents the **distance** (or magnitude of displacement) travelled by the object during that time interval. For uniform velocity (a rectangle shape on the graph), the distance $s$ from time $t_1$ to $t_2$ is:
$$ s = \text{Velocity} \times \text{Time Interval} = v \times (t_2 - t_1) $$
This corresponds to the area of the rectangle formed by the graph line, the time axis, and the perpendiculars at $t_1$ and $t_2$.
- **Uniform Acceleration:** For an object moving with uniform acceleration, its velocity changes by equal amounts in equal time intervals. The velocity-time graph is a **straight line with a slope**. The **slope of the velocity-time graph** represents the **acceleration** of the object.
Again, the **area under the velocity-time graph** represents the distance travelled. For uniform acceleration, this area might be a triangle, rectangle, or a trapezium, depending on the initial velocity and the time interval.
- **Non-Uniform Acceleration:** For non-uniform acceleration, the velocity-time graph is a **curved line**. The slope of the curve at any point gives the instantaneous acceleration. Interpreting distance from the area under such a curve often requires more advanced mathematical techniques.
Question 1. What is the nature of the distance-time graphs for uniform and non-uniform motion of an object?
Answer:
For uniform motion, the distance-time graph is a **straight line**. For non-uniform motion, the distance-time graph is a **curved line**.
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:
If the distance-time graph is a straight line parallel to the time axis, it means the distance of the object from the origin is not changing with time. Therefore, the object is **at rest**.
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:
If the speed-time graph is a straight line parallel to the time axis, it means the speed of the object is constant over time. If the motion is along a straight line in one direction, this represents **uniform motion** (constant velocity). If the motion involves changing direction while maintaining constant speed (like uniform circular motion), it represents motion with **constant speed but changing velocity** (i.e., accelerated motion).
Question 4. What is the quantity which is measured by the area occupied below the velocity-time graph?
Answer:
The area occupied below the velocity-time graph measures the **distance travelled** or the magnitude of the displacement of the object during the corresponding time interval.
Equations Of Motion
For an object moving along a straight line with **uniform acceleration**, we can use a set of equations to relate its initial velocity, final velocity, acceleration, time taken, and the distance covered. These are known as the **equations of motion**.
There are three main equations of motion:
- **Velocity-Time Relation:** $$ v = u + at $$ (This equation relates the final velocity ($v$), initial velocity ($u$), acceleration ($a$), and time taken ($t$).)
- **Position-Time Relation:** $$ s = ut + \frac{1}{2}at^2 $$ (This equation relates the distance covered ($s$), initial velocity ($u$), acceleration ($a$), and time taken ($t$).)
- **Position-Velocity Relation:** $$ v^2 - u^2 = 2as \quad \text{or} \quad 2as = v^2 - u^2 $$ (This equation relates the final velocity ($v$), initial velocity ($u$), acceleration ($a$), and distance covered ($s$). It can be derived from the first two equations by eliminating time $t$.)
Where:
- $u$ = initial velocity
- $v$ = final velocity
- $a$ = uniform acceleration
- $t$ = time taken
- $s$ = distance travelled
These equations are valid only when the acceleration is constant and the motion is along a straight line.
Example 7.5. 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 (starting from rest)
Final velocity, $v = 72$ km/h. Convert to m/s: $v = 72 \times \frac{1000 \text{ m}}{3600 \text{ s}} = 72 \times \frac{5}{18} \text{ m/s} = 4 \times 5 \text{ m/s} = 20 \text{ m/s}$.
Time taken, $t = 5$ minutes. Convert to seconds: $t = 5 \times 60 \text{ s} = 300 \text{ s}$.
(i) To find acceleration ($a$), use the first equation of motion, $v = u + at$:
$20 \text{ m/s} = 0 \text{ m/s} + a \times 300 \text{ s}$
$a = \frac{20 \text{ m/s}}{300 \text{ s}} = \frac{1}{15} \text{ m/s}^2$.
The acceleration of the train is $\frac{1}{15}$ m/s².
(ii) To find the distance travelled ($s$), we can use the third equation of motion, $v^2 - u^2 = 2as$:
$(20 \text{ m/s})^2 - (0 \text{ m/s})^2 = 2 \times \left(\frac{1}{15} \text{ m/s}^2\right) \times s$
$400 \text{ m}^2/\text{s}^2 = \frac{2}{15} \text{ m/s}^2 \times s$
$s = \frac{400 \text{ m}^2/\text{s}^2}{\frac{2}{15} \text{ m/s}^2} = 400 \times \frac{15}{2} \text{ m} = 200 \times 15 \text{ m} = 3000 \text{ m}$.
The distance travelled is 3000 m, which is equal to 3 km.
Example 7.6. 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, $u = 18$ km/h. Convert to m/s: $u = 18 \times \frac{5}{18} \text{ m/s} = 5 \text{ m/s}$.
Final velocity, $v = 36$ km/h. Convert to m/s: $v = 36 \times \frac{5}{18} \text{ m/s} = 2 \times 5 \text{ m/s} = 10 \text{ m/s}$.
Time taken, $t = 5$ s.
(i) To find acceleration ($a$), use the first equation of motion, $v = u + at$:
$10 \text{ m/s} = 5 \text{ m/s} + a \times 5 \text{ s}$
$10 \text{ m/s} - 5 \text{ m/s} = a \times 5 \text{ s}$
$5 \text{ m/s} = a \times 5 \text{ s}$
$a = \frac{5 \text{ m/s}}{5 \text{ s}} = 1 \text{ m/s}^2$.
The acceleration of the car is 1 m/s².
(ii) To find the distance covered ($s$), use the second equation of motion, $s = ut + \frac{1}{2}at^2$:
$s = (5 \text{ m/s}) \times (5 \text{ s}) + \frac{1}{2} \times (1 \text{ m/s}^2) \times (5 \text{ s})^2$
$s = 25 \text{ m} + \frac{1}{2} \times 1 \text{ m/s}^2 \times 25 \text{ s}^2$
$s = 25 \text{ m} + 12.5 \text{ m} = 37.5 \text{ m}$.
The distance covered by the car in that time is 37.5 m.
Example 7.7. 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, $a = -6$ m/s² (opposite direction to motion means negative acceleration)
Time taken to stop, $t = 2$ s
Final velocity, $v = 0$ m/s (car stops)
We need to find the initial velocity ($u$) first using the first equation of motion, $v = u + at$:
$0 \text{ m/s} = u + (-6 \text{ m/s}^2) \times (2 \text{ s})$
$0 = u - 12 \text{ m/s}$
$u = 12 \text{ m/s}$.
Now, to find the distance ($s$) travelled during this time, use the second equation of motion, $s = ut + \frac{1}{2}at^2$:
$s = (12 \text{ m/s}) \times (2 \text{ s}) + \frac{1}{2} \times (-6 \text{ m/s}^2) \times (2 \text{ s})^2$
$s = 24 \text{ m} + \frac{1}{2} \times (-6 \text{ m/s}^2) \times (4 \text{ s}^2)$
$s = 24 \text{ m} + (-3 \text{ m/s}^2) \times (4 \text{ s}^2)$
$s = 24 \text{ m} - 12 \text{ m} = 12 \text{ m}$.
The car travels a distance of 12 metres before it stops after applying the brakes. This example highlights the importance of maintaining a safe distance between vehicles.
Uniform Circular Motion
We know that acceleration occurs when velocity changes. Velocity can change by altering its magnitude (speed) or its direction, or both. What happens if an object moves at a constant speed but its direction is continuously changing?
Consider an object moving along a closed polygonal path with many sides. As the number of sides increases, the path becomes closer and closer to a circle. When the number of sides becomes infinite, the path is a circle, and the direction of motion changes continuously at every point.
**Uniform circular motion** is defined as the motion of an object along a circular path at a **constant speed**. Although the speed is constant, the velocity is continuously changing because the direction of motion changes at every point on the circle. Therefore, uniform circular motion is considered an **accelerated motion**.
The acceleration in uniform circular motion is always directed towards the center of the circle (centripetal acceleration), causing the change in direction of the velocity vector.
If an object moves in a circle of radius $r$ and takes time $t$ to complete one full revolution (the circumference, $2\pi r$), its speed $v$ is given by:
$$ v = \frac{\text{Circumference}}{\text{Time period}} = \frac{2\pi r}{t} $$
Examples of uniform circular motion:
- Motion of the Moon around the Earth.
- Motion of artificial satellites in circular orbits around the Earth.
- A cyclist riding on a circular track at a constant speed.
- The tip of the hands of a clock.
When an object moving in a circular path is released (like a stone whirled on a string), it moves along a straight line tangential to the circle at the point of release. This is because at that instant, the velocity is directed tangentially, and without the force pulling it towards the center, it continues motion in that direction according to inertia.
Intext Questions
Page No. 74
Question 1. An object has moved through a distance. Can it have zero displacement? If yes, support your answer with an example.
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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?
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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.
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Page No. 75
Question 1. Distinguish between speed and velocity.
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Question 2. Under what condition(s) is the magnitude of average velocity of an object equal to its average speed?
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Question 3. What does the odometer of an automobile measure?
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Question 4. What does the path of an object look like when it is in uniform motion?
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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, $3 \times 10^8\ m\ s^{-1}$.
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Page No. 77
Question 1. When will you say a body is in (i) uniform acceleration? (ii) non-uniform acceleration?
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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.
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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.
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Page No. 81
Question 1. What is the nature of the distance-time graphs for uniform and non-uniform motion of an object?
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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?
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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?
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Question 4. What is the quantity which is measured by the area occupied below the velocity-time graph?
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Page No. 82 - 83
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.
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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.
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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?
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Question 4. A racing car has a uniform acceleration of $4\ m\ s^{-2}$. What distance will it cover in 10 s after start?
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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?
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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 $20\ s$?
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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?
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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?
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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?
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Question 5. A driver of a car travelling at $52\ km\ h^{-1}$ applies the brakes Shade the area on the graph that represents the distance travelled by the car during the period.
Which part of the graph represents uniform motion of the car?
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Question 6. Fig 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?
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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?
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Question 8. The speed-time graph for a car is shown is 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?
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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.
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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.
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