Class 9 Science Chapter 7 Motion

Key Concept: Third Equation of Motion

b) $1 \, \text{m/s}^2$

[Solution Description]

Given:

– Initial velocity ($u$) = 5 m/s

– Final velocity ($v$) = 15 m/s

– Distance ($s$) = 100 meters

Using the third equation of motion:

$2as = v^2 – u^2$

Substituting the given values:

$2a(100) = (15)^2 – (5)^2$

Simplifying:

$200a = 225 – 25$

$200a = 200$

Solving for $a$:

$a = \frac{200}{200} = 1 \text{ m/s}^2$

Therefore, the acceleration of the cyclist is $1 \, \text{m/s}^2$.

Key Concept: Uniform Motion

c) 20 meters

[Solution Description]

The object travels 5 meters every second. To find the total distance traveled in 4 seconds, multiply the distance per second by the number of seconds: $5 \, \text{m/s} \times 4 \, \text{s} = 20 \, \text{m}$.

Key Concept: Describing Motion

b) 100 meters

[Solution Description]

We use the equation $s = ut + \frac{1}{2}at^2$ to calculate the distance covered. Here, initial velocity $u = 0$, acceleration $a = $2 \, \text{m/s}^{2}$, and time $t = 10$ seconds. Substituting these values, we get:

$s = (0)(10) + \frac{1}{2}(2)(10)^2 = 0 + \frac{1}{2}(2)(100) = 100 \, \text{meters}$

Thus, the distance covered is 100 meters.

Key Concept: Velocity-Time Graph

b) By calculating the area under the graph

[Solution Description]

The distance moved by an object can be determined by calculating the area under the velocity-time graph. For uniformly accelerated motion, this area includes both rectangular and triangular regions.

Key Concept: Uniform Circular Motion

c) Straight line tangential to the circular path

[Solution Description]

When the stone is released from the circular path, it moves along the direction it was moving at that instant. This direction is tangential to the circular path at the point of release.

Key Concept: Velocity-Time Graph Interpretation

a) Acceleration

[Solution Description]

The slope of a velocity-time graph represents the rate of change of velocity, which is acceleration. Since the graph is a straight line starting from the origin, the slope is constant, indicating uniform acceleration.

Key Concept: Displacement Calculation

d) 50 km

[Solution Description]

The car’s movement forms a right-angled triangle with the legs being 30 km east and 40 km north. The displacement is the straight line distance from the starting point to the final position, which can be found using the Pythagorean theorem:

$\text{Displacement} = \sqrt{(30)^2 + (40)^2} = \sqrt{900 + 1600} = \sqrt{2500} = 50 \text{ km}$

Key Concept: Distance and Displacement

c) 85 km

[Solution Description]

To find the total distance covered by the object, we add the distance from O to A and the distance from A to B. The distance from O to A is 60 km, and the distance from A to B is 25 km. Therefore, the total distance covered is $60 \text{ km} + 25 \text{ km} = 85 \text{ km}$.

Key Concept: Position-Time Relation

c) 80 m

[Solution Description]

Given: Initial velocity $u = 10 \, \text{m/s}$, acceleration $a = 5 \, \text{m/s}^2$, time $t = 4 \, \text{s}$.

Using the equation $s = ut + \frac{1}{2}at^2$:

$s = 10 \times 4 + \frac{1}{2} \times 5 \times 16$

$s = 40 + 40$

$s = 80 \, \text{m}$.

Key Concept: Uniform Velocity and Displacement

b) 180 km

[Solution Description]

The formula for displacement when an object moves with uniform velocity is given by:

$s = v \times t$

Here, $v = 60$ km/h and $t = 3$ h.

Substituting the values, we get:

$s = 60 \times 3 = 180$ km

Therefore, the displacement of the car is 180 km.

Key Concept: Area under velocity-time graph, Displacement

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.

[Solution Description]

The assertion states that the area under a velocity-time graph always gives the displacement of the object. This is true because the area under the velocity-time graph represents the integral of velocity with respect to time, which is displacement.

The reason states that displacement is the product of velocity and time when the velocity is constant. This is also true because if the velocity is constant, the area under the velocity-time graph forms a rectangle, and the area of a rectangle is given by the product of its height (velocity) and width (time).

However, the reason does not fully explain the assertion. The assertion holds true even when the velocity is not constant, as the area under the curve still gives the displacement. Therefore, while both the assertion and reason are true, the reason is not the correct explanation of the assertion.

Key Concept: Displacement and Distance

a) 0 m

[Solution Description]

Displacement is defined as the shortest distance between the initial and final positions of an object. Since the object returns to its starting point, the displacement is zero.

Key Concept: Position-Time Relation, Velocity-Time Relation

b) $20 \, \text{m/s}, 75 \, \text{m}$

[Solution Description]

Given $u = 10 \, \text{m/s}$, $a = 2 \, \text{m/s}^2$, and $t = 5 \, \text{s}$, we first find the final velocity using the velocity-time relation:

$v = u + at$

Substituting the known values:

$v = 10 + 2 \times 5 = 10 + 10 = 20 \, \text{m/s}$

Next, we use the position-time relation to find the distance:

$s = ut + \frac{1}{2}at^2$

Substituting the known values:

$s = 10 \times 5 + \frac{1}{2} \times 2 \times (5)^2$

Simplifying:

$s = 50 + \frac{1}{2} \times 2 \times 25 = 50 + 25 = 75 \, \text{m}$

So, the final velocity is $20 \, \text{m/s}$ and the distance covered is $75 \, \text{m}$.

Key Concept: Negative Acceleration

b) -5 m/s$^2$

[Solution Description]

The initial velocity $u$ is 15 m/s, the final velocity $v$ is 5 m/s, and the time taken $t$ is 2 seconds. Using the formula for acceleration:

$a = \frac{v – u}{t}$

Substituting the given values:

$a = \frac{5 – 15}{2} = \frac{-10}{2} = -5 \, \text{m/s}^2$

Therefore, the acceleration of the cyclist is -5 m/s$^2$, indicating negative acceleration.

Key Concept: Types of Motion

b) The motion is controlled and serves human needs.

[Solution Description]

Hydro-electric power is generated by harnessing the controlled motion of water, typically in a river or reservoir. This controlled motion is used to drive turbines, converting kinetic energy into electrical energy. Therefore, the motion involved is controlled and serves a beneficial purpose.

Key Concept: Second Equation of Motion

c) 96 meters

[Solution Description]

Using the second equation of motion: $s = ut + \frac{1}{2} at^2$. Here, initial velocity $u = 10$ m/s, acceleration $a = 2$ m/s$^2$, and time $t = 6$ s. Substituting these values:

$s = 10 \times 6 + \frac{1}{2} \times 2 \times (6)^2$

Simplifying:

$s = 60 + \frac{1}{2} \times 2 \times 36$

$s = 60 + \frac{1}{2} \times 72 = 60 + 36 = 96 \, \text{meters}$

The distance covered by the particle is 96 meters.

Key Concept: Distance and Displacement

b) 80 km

[Solution Description] The total distance covered by the object is the sum of the distances traveled in each part of the journey. The object travels from O to A (40 km) and then back to O (40 km). So, the total distance is $40 + 40 = 80$ km.

Key Concept: Uniform Circular Motion

b) $\pi$ m/s

[Solution Description]

The speed $v$ can be calculated using the formula $v = \frac{2\pi r}{t}$. Given $r = 5$ meters and $t = 10$ seconds, we substitute these values into the formula:

$v = \frac{2\pi \times 5}{10} = \frac{10\pi}{10} = \pi \text{ m/s}$

Key Concept: Distance-Time Graph: Uniform Motion

d) 50 km/h

[Solution Description]

The slope of a distance-time graph represents speed. To find the slope, we use the formula: $v = \frac{\Delta s}{\Delta t}$. Here, $\Delta s = 150$ km and $\Delta t = 3$ hours. Thus, $v = \frac{150}{3} = 50$ km/h. Therefore, the slope is 50 km/h.

Key Concept: Displacement and Distance

b) 20 km

[Solution Description]

The object starts at point O, moves to point A (30 km), and then returns to point B (10 km away from A). Displacement is the shortest distance from the initial position (O) to the final position (B). Since B is 20 km away from O (30 km – 10 km), the displacement is 20 km.

Key Concept: Average Velocity Calculation

c) 40 m/s

[Solution Description]

The average velocity is given by:

$v_{av} = \frac{u + v}{2}$

Here, initial velocity $u$ is 60 m/s and final velocity $v$ is 20 m/s.

Substituting the values:

$v_{av} = \frac{60 + 20}{2} = \frac{80}{2} = 40 \, \text{m/s}$

Therefore, the average velocity of the train is 40 m/s.

Key Concept: Velocity-Time Graph

c) 200 km

[Solution Description]

The distance covered by an object moving with uniform velocity can be calculated using the formula:

$s = v \times t$

Here, the velocity $v$ is 50 km/h and the time $t$ is 4 hours. Substituting these values into the formula:

$s = 50 \text{ km/h} \times 4 \text{ h} = 200 \text{ km}$

Thus, the car will cover a distance of 200 km in 4 hours.

Key Concept: Average Velocity

b) 20 m/s

[Solution Description]

The average velocity $v_{av}$ can be calculated using the formula:

$v_{av} = \frac{u + v}{2}$

Here, the initial velocity $u = 10$ m/s, and the final velocity $v = 30$ m/s. Plugging in these values:

$v_{av} = \frac{10 + 30}{2} = \frac{40}{2} = 20 \text{ m/s}$

Therefore, the average velocity of the car is $20$ m/s.

Key Concept: Motion and Direction

c) Tangential to the circular path

[Solution Description]

The stone moves along a straight line tangential to the circular path at the point of release. This happens because once released, the stone continues to move in the direction it was moving at that instant.

Key Concept: Calculating Distance from a Velocity-Time Graph

c) 250 km

[Solution Description]

The distance traveled by the train can be calculated using the area under the velocity-time graph. Since the velocity increases uniformly, the graph forms a trapezoid. The area of a trapezoid is given by $\frac{1}{2} \times (\text{sum of parallel sides}) \times \text{height}$. Here, the parallel sides are 20 km/h and 80 km/h, and the height is 5 hours.

Step 1: Identify the parallel sides and height.

Parallel sides = 20 km/h and 80 km/h, Height = 5 hours

Step 2: Calculate the area of the trapezoid.

Area = $\frac{1}{2} \times (20 + 80) \times 5$ = $\frac{1}{2} \times 100 \times 5$ = 250 km

Therefore, the distance traveled by the train is 250 km.

Key Concept: Relative Motion

b) 20 km/h

[Solution Description]

To find the relative speed of car B with respect to car A, we subtract the speed of car A from the speed of car B. Since both cars are moving in the same direction, the relative speed is $80 \text{ km/h} – 60 \text{ km/h} = 20 \text{ km/h}$.

Key Concept: Uniform Circular Motion

b) $10\pi$ seconds

[Solution Description]

The formula for the speed in uniform circular motion is given by:

$v = \frac{2\pi r}{t}$

Where $v$ is the speed, $r$ is the radius, and $t$ is the time taken to complete one full round.

Given:

$v = 10 \, \text{m/s}, \quad r = 50 \, \text{m}$

Rearrange the formula to solve for $t$:

$t = \frac{2\pi r}{v}$

Substitute the values:

$t = \frac{2 \times \pi \times 50}{10} = \frac{100\pi}{10} = 10\pi \, \text{seconds}$

Therefore, the time taken is $10\pi$ seconds.

Key Concept: Graphical Representation of Motion

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

The assertion states that the velocity-time graph for uniformly accelerated motion is a straight line with a positive slope. This is true because in uniformly accelerated motion, the velocity increases at a constant rate, resulting in a straight-line graph with a positive slope.

The reason states that in uniformly accelerated motion, the velocity changes by equal amounts in equal intervals of time. This is also true as it defines uniformly accelerated motion. Moreover, the reason correctly explains the assertion because the constant change in velocity leads to the straight-line nature of the velocity-time graph.

Therefore, both the assertion and reason are true, and the reason correctly explains the assertion.

Key Concept: Time Period of Revolution

b) 43200 seconds

[Solution Description]

The time period $T$ of the satellite’s revolution is given as 12 hours. To convert this into seconds:

$T = 12 \times 3600 = 43200 \text{ seconds}$

Therefore, the time period of the satellite’s revolution is 43200 seconds.

Key Concept: Displacement Concept

d) 0 meters

[Solution Description]

The displacement of an object is the shortest distance between its initial and final positions, considering direction. In this case, the object starts at point O and returns to point O. Since the initial and final positions are the same, the displacement is zero.

Displacement = Final Position – Initial Position = O – O = 0

Key Concept: Distance Covered with Uniform Acceleration

b) 12.5 m

[Solution Description]

Given: initial velocity ($u$) = 0 m/s, acceleration ($a$) = 1 m/s$^2$, time ($t$) = 5 s. Using the equation $s = ut + \frac{1}{2}at^2$, we can calculate the distance:

$s = 0 \times 5 + \frac{1}{2} \times 1 \times (5)^2 = \frac{1}{2} \times 25 = 12.5 \text{ m}$

Key Concept: Displacement and Distance

b) Displacement: 0 m, Distance: 20 m

[Solution Description]

The person walks 10 meters east and then 10 meters west. Displacement is the shortest distance from the initial to the final position. Here, the person ends up at the starting point, so the displacement is zero. The total distance covered is $10\,\text{m} + 10\,\text{m} = 20\,\text{m}$.

Key Concept: Perception of Motion

d) The relative motion makes it seem so

[Solution Description]

The trees appear to move backward due to the relative motion between the bus and the trees. As the bus moves forward, the trees seem to move in the opposite direction relative to the passengers.

Key Concept: Average velocity, Initial velocity, Final velocity

b) $20 \, \text{m/s}^{-1}$

[Solution Description]

The average velocity when the velocity changes uniformly is given by the arithmetic mean of the initial and final velocities. Using the formula:

$v_{av} = \frac{u + v}{2} = \frac{10 \text{ m s}^{-1} + 30 \text{ m s}^{-1}}{2} = \frac{40 \text{ m s}^{-1}}{2} = 20 \text{ m s}^{-1}$

Key Concept: Uniform Circular Motion

b) $5\pi \text{ s}$

[Solution Description]

To find the time taken to complete one full circle, we use the formula for the circumference of a circle and the definition of speed. The circumference $C$ of a circle is given by $C = 2\pi r$, where $r$ is the radius. The speed $v$ is related to the time $t$ by $v = \frac{C}{t}$. Solving for $t$, we get:

$t = \frac{C}{v} = \frac{2\pi r}{v}$

Substituting the given values, $r = 50 \text{ m}$ and $v = 20 \text{ m/s}$, we have:

$t = \frac{2\pi \times 50}{20} = \frac{100\pi}{20} = 5\pi \text{ s}$

Therefore, the time taken by the car to complete one full circle is $5\pi \text{ s}$.

Key Concept: Angular Velocity

b) $10\pi$

[Solution Description]

To find the total angle turned by the wheel, we use the formula:

$\theta = \omega \times t$

However, since the number of rotations is given, we first calculate the time taken to complete these rotations. Let’s denote

1 rotation = $2\pi$ radians,

Therefore, total angle $\theta = \text{no. of rotations} \times 2\pi = 5 \times 2\pi = 10\pi$ radians.

Key Concept: Damped Oscillations, Time Period

b) $\frac{2\pi}{\sqrt{\omega_0^2 – \beta^2}}$

[Solution Description]

For a damped oscillator, the angular frequency $\omega_d$ is given by:

$\omega_d = \sqrt{\omega_0^2 – \beta^2}$

The time period $T$ is related to the angular frequency by:

$T = \frac{2\pi}{\omega_d}$

Substituting $\omega_d$:

$T = \frac{2\pi}{\sqrt{\omega_0^2 – \beta^2}}$

Hence, the time period of the damped oscillator is $\frac{2\pi}{\sqrt{\omega_0^2 – \beta^2}}$.

Key Concept: Average Speed, Average Velocity, Displacement

c) 3 m/s

[Solution Description]

To find the average velocity, we need to calculate the displacement and divide it by the total time taken. The cyclist rides east for $40 \, \text{m}$ and then west for $10 \, \text{m}$. The net displacement is:

$\text{Displacement} = 40 \, \text{m} – 10 \, \text{m} = 30 \, \text{m}$

The total time taken is $10 \, \text{s}$. Therefore, the average velocity is:

$\text{Average velocity} = \frac{30 \, \text{m}}{10 \, \text{s}} = 3 \, \text{m/s}$

Thus, the magnitude of the average velocity is $3 \, \text{m/s}$.

Key Concept: Uniform Motion

d) Uniform motion

[Solution Description]

Uniform motion is defined as the motion of an object where it covers equal distances in equal intervals of time. In this case, the object covers 10 meters every second for 5 seconds, which satisfies the condition for uniform motion.

Key Concept: Distance-Time Graph

c) A straight line

[Solution Description]

When an object moves with uniform speed, it covers equal distances in equal intervals of time. This means that the distance travelled by the object is directly proportional to the time taken. Therefore, the distance-time graph for an object moving with uniform speed is a straight line.

Key Concept: Types of Motion

c) A planet orbiting the sun

[Solution Description]

Circular motion refers to the movement of an object along a circular path. An example of this is the motion of a planet around the sun, which follows a circular or elliptical path.

Key Concept: First Equation of Motion

c) 25 m/s

[Solution Description]

Using the first equation of motion: $v = u + at$. Here, the initial velocity $u = 10$ m/s, acceleration $a = 3$ m/s$^2$, and time $t = 5$ s. Plugging in these values, we get:

$v = 10 + (3 \times 5) = 10 + 15 = 25 \text{ m/s}$

The final velocity is 25 m/s.

Key Concept: Position-Velocity Relation

b) 2 m/s$^2$

[Solution Description]

Given: Initial velocity ($u$) = 10 m/s, Final velocity ($v$) = 30 m/s, Distance ($s$) = 200 m. Using the equation of motion $2as = v^2 – u^2$, we substitute the values: $2a(200) = (30)^2 – (10)^2 \Rightarrow 400a = 900 – 100 \Rightarrow 400a = 800 \Rightarrow a = 2 \, \text{m/s}^2$. Therefore, the acceleration is 2 m/s$^2$.

Key Concept: Reference Point

c) Both library and hospital

[Solution Description]

In this case, the position of the school is described with respect to two different locations: the library and the hospital. The reference points are the library and the hospital. Therefore, the correct answer is that the school is using both the library and the hospital as reference points.

Key Concept: Uniform Circular Motion, Tangential Velocity

c) Tangentially to the circular path

[Solution Description]

When an object moves in uniform circular motion, its velocity at any point is tangential to the circular path. Therefore, when the hammer is released, it will move in the direction of the tangent to the circular path at the point of release.