6 Most SAQ’s of Motion in a Straight Line Chapter in Inter 1st Year Physics (TS/AP)

4 Marks

SAQ-1 : A car travels the first third of a distance with a speed of 10Kmph, the second third at 20Kmph and the last third at 60Kmph. What is its mean speed over the entire distance ?

Introduction

In this explanation, we will solve a problem related to calculating the mean speed of a car that travels different distances at various speeds. We will use the formula of mean speed, distance, and time $$V_{\text{avg}} = \frac{D}{T}$$ to find the average speed of the car over the entire journey.

Problem

A car travels a total distance ‘d’ in three parts, each with a different speed. The speeds of the car during these parts are v₁​=10km/h, v₂​=20km/h, and v₃​=60km/h. We need to calculate the mean speed of the car over the entire distance.

Step-by-Step Solution

1. Divide the Total Distance: Let’s assume the total distance is ‘d’. So, each part of the distance will be $$\frac{d}{3}$$
2. Calculate Time for Each Part: Time (t) is given by the formula $$t = \frac{\text{distance}}{\text{speed}}$$
   • For the first part (v_1​=10km/h): $$t_1 = \frac{d/3}{10}$$
   • For the second part (v₂​=20km/h): $$t_2 = \frac{d/3}{20}$$​
   • For the third part (v_3​=60km/h): $$t_3 = \frac{d/3}{60}$$
3. Calculate the Mean Speed: Mean speed (V_avg​) is given by $$V_{\text{avg}} = \frac{\text{Total distance}}{\text{Total time taken}}$$
   • Total distance is ‘d’, and total time is the sum of t_1​,t_2​, and t₃​.
4. Determine Mean Speed:
   $$V_{\text{avg}} = \frac{d}{t_1 + t_2 + t_3}$$
   $$V_{\text{avg}} = \frac{d}{\frac{d}{30} + \frac{d}{60} + \frac{d}{180}}$$
   Simplify the equation to find V_avg​.

Summary

The mean speed of the car over the entire distance is not 180 km/h. This result is incorrect as the calculation of the mean speed involves dividing the total distance by the sum of the individual times taken for each section. Speed is a scalar quantity that indicates how fast an object is moving. It’s important to note that velocity is a different concept, being a vector quantity that includes both speed and direction.

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SAQ-2 : A ball is dropped from the roof of a tall building and simultaneously another ball is thrown horizontally with some velocity from the same roof. Which ball lands first? Explain your  answer.

Introduction

In this scenario, gravity and projectile motion are the key factors. A ball is dropped vertically from a tall building’s roof, falling solely under the influence of gravity. Simultaneously, another ball is thrown horizontally with an initial velocity ‘u’ from the same height. We aim to determine which ball lands first: the vertically free-falling ball or the horizontally thrown ball.

Analysis

1. Vertical Motion of the Dropped Ball: The motion is described by the equation $$h = \frac{1}{2}gt^2$$, where ‘h’ is the height of the building, ‘g’ is the acceleration due to gravity, and ‘t’ is the time taken to reach the ground. Solving for ‘t’, we find $$t = \sqrt{\frac{2h}{g}}$$
2. Horizontal Motion of the Thrown Ball: The horizontal motion is unaffected by gravity, as there is no vertical acceleration involved. The horizontal distance (X) covered by the ball is given by X=ut, where ‘u’ is the initial horizontal velocity and ‘t’ is the time to reach the ground.

Summary

Both balls, irrespective of their motion (vertical or horizontal), are subject to the same gravitational pull. Since the horizontal velocity of the thrown ball does not influence its vertical acceleration, both balls will experience the same fall time under gravity. Therefore, both the vertically dropped ball and the horizontally thrown ball will land simultaneously. This conclusion is a fundamental principle of physics, demonstrating that the horizontal component of motion does not affect the vertical component in projectile motion.

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SAQ-3 : Derive the equation of motion x=v_0 t+1/2 at^2 using appropriate graph.

Introduction

In this explanation, we will derive the equation of motion $$x = v_0t + \frac{1}{2}at^2$$ using a velocity-time graph. We utilize the concept of the area under the graph to calculate the distance covered by a moving object.

Step-by-Step Derivation

1. Velocity-Time Graph: A velocity-time graph represents the variation of velocity of a moving object with respect to time. Its shape varies depending on the type of motion.
2. Area Under the Velocity-Time Graph: The distance covered by a moving object is equal to the area under its velocity-time graph. For our derivation, we consider a graph with initial velocity v_0​ and final velocity v.
3. Defining the Graph: Let the velocity-time graph be represented as a v-t graph. In the case of accelerated motion, the graph is a straight line inclined to the time axis, indicating a constant acceleration. For motion with zero acceleration, the graph is a horizontal line parallel to the time axis.
4. Calculating Distance (s):
   • Area of Triangle ABE: The area of the triangular region ABE represents the change in velocity over time. It is calculated as: $$\text{Area of ABE} = \frac{1}{2} \times \text{Base} \times \text{Height} = \frac{1}{2} \times t \times (v – v_0)$$
   • Area of Rectangle ECD: The area of the rectangular region ECD represents the initial velocity over time. It is calculated as: $$\text{Area of ECD} = \text{Length} \times \text{Breadth} = t \times v_0$$
5. Total Distance Covered (s): The total distance s is the sum of the areas of ABE and ECD. $$s = \text{Area of ABCD} = \text{Area of ABE} + \text{Area of ECD} = \frac{1}{2} \times t \times (v – v_0) + t \times v_0$$
6. Applying the First Equation of Motion: According to the first equation of motion, v = v₀​+at, where a is the acceleration.
7. Solving for Distance (s): Using the first equation of motion, we substitute v − v₀​ = at into the equation for s, yielding $$s = v_0t + \frac{1}{2}at^2$$

Summary

We have successfully derived the equation of motion $$x = v_0t + \frac{1}{2}at^2$$ using the velocity-time graph and the concept of the area under the graph. The variable a represents the acceleration of the body, and the area under any v-t graph equates to the distance traveled by the object. This derivation demonstrates a fundamental relationship between an object’s velocity, acceleration, and the distance it covers over time.

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SAQ-4 : Show that the maximum height reached by a projectile launched at an angle of 45^o is one quarter of its range.

Introduction

We aim to demonstrate that the maximum height reached by a projectile launched at an angle of 45 degrees is one quarter of its range. This will be achieved using the standard equations for range and maximum height in projectile motion.

Given Equations:

1. Range (R): $$R = \frac{u^2 \sin(2\theta)}{g}$$
2. Maximum Height (H): $$H = \frac{u^2 \sin^2(\theta)}{2g}$$

When θ = 45 Degrees:

• Range (Rmax): $$R_{\text{max}} = \frac{u^2 \sin(90^\circ)}{g}$$
  Since $$\sin(90^\circ) = 1$$, $$R_{\text{max}} = \frac{u^2}{g}$$
• Maximum Height (Hmax): $$H_{\text{max}} = \frac{u^2 \sin^2(45^\circ)}{2g}$$
  Since $$\sin(45^\circ) = \frac{1}{\sqrt{2}}, H_{\text{max}} = \frac{u^2 \left(\frac{1}{\sqrt{2}}\right)^2}{2g}$$
  Simplifying, $$H_{\text{max}} = \frac{u^2}{4g}$$

Comparing Hmax and Rmax at θ = 45 Degrees:

• Hmax: $$H_{\text{max}} = \frac{u^2}{4g}$$
• Rmax: $$R_{\text{max}} = \frac{u^2}{g}$$

Ratio of Hmax to Rmax:

• $$\frac{H_{\text{max}}}{R_{\text{max}}} = \frac{\frac{u^2}{4g}}{\frac{u^2}{g}}$$
• Simplifying, $$\frac{H_{\text{max}}}{R_{\text{max}}} = \frac{1}{4}$$

Summary

When a projectile is launched at an angle of 45 degrees, the maximum height (Hmax) it reaches is indeed one quarter of its range (Rmax). This conclusion is derived from the fundamental equations of projectile motion and does not depend on approximations. The ratio $$\frac{H_{\text{max}}}{R_{\text{max}}}$$​​ is precisely 1/4​, demonstrating the elegance and symmetry in the physics of projectile motion.

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SAQ-5 : How is average velocity its different from instantaneous velocity?

Introduction

Velocity is a key concept in physics that describes the speed and direction of a moving object. There are two significant types of velocity: average velocity and instantaneous velocity. Grasping the distinctions between these is crucial for a comprehensive understanding of motion.

Average Velocity

1. Definition: Average velocity is defined as the ratio of the total displacement of an object to the time taken for that displacement.
2. Formula: $$V_{\text{avg}} = \frac{X_{\text{final}} – X_{\text{initial}}}{T_{\text{final}} – T_{\text{initial}}}$$
3. Meaning: Average velocity represents an effective uniform velocity over the entire duration of the journey, irrespective of actual speed variations during the journey.

Instantaneous Velocity

1. Definition: Instantaneous velocity refers to the velocity of an object at a specific moment in time, indicating its speed and direction at that exact instance.
2. Formula: $$V_{\text{instant}} = \frac{dX}{dT}​$$
3. Meaning: Instantaneous velocity provides a precise measurement of velocity at a particular instant, considering the actual motion of the object.

Relationship between Average and Instantaneous Velocity

1. When an object’s velocity is constant, or its acceleration is zero, the average velocity and instantaneous velocity are the same.
2. Under these conditions, all instantaneous velocities during the motion are equal to each other and to the average velocity.
3. However, in many real-world situations, an object’s velocity changes over time, leading to different values for instantaneous velocity compared to average velocity.

Summary

Velocity, indicating how fast an object moves and in which direction, has two primary forms: average velocity and instantaneous velocity. Average velocity provides an effective measure of uniform motion over a time interval, while instantaneous velocity gives the exact speed and direction at a particular moment. The two types of velocity coincide when an object moves at a constant speed or has no acceleration. Otherwise, they differ, reflecting the varying speeds during motion. Understanding these differences is fundamental to comprehending motion and its mathematical representation in physics.

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SAQ-6 : Give an example where the velocity of an object is Zero but its acceleration is not zero.

Introduction

In physics, velocity and acceleration are pivotal concepts that describe an object’s motion. Velocity indicates the speed and direction of an object’s movement, while acceleration measures the change in velocity over time. Contrary to common belief, there are instances where an object can have a velocity of zero while still experiencing non-zero acceleration. This concept is crucial in understanding the dynamics of motion.

Definition of Acceleration and Velocity

1. Acceleration: The rate of change in velocity of an object over time.
2. Velocity: The speed and direction of an object’s movement over time.

Example 1: Pendulum Motion

1. In the motion of a pendulum, at the peak of its swing, its velocity momentarily becomes zero as it changes direction.
2. Despite having zero velocity at this point, the pendulum experiences non-zero acceleration due to kinetic energy and the restoring force acting on it.

Example 2: Constant Horizontal Velocity with Downward Acceleration

1. Imagine a ball moving horizontally with a constant velocity but also experiencing a constant downward acceleration due to gravity (denoted as ‘-g’).
2. The ball’s motion can be represented as $$x = \frac{1}{2} (-g) t^2$$, where ‘t’ is time.
3. In this scenario, the velocity (horizontal) and acceleration (vertical) are in different directions.

Example 3: Centripetal Acceleration in Planetary Motion

1. In planetary motion, such as the Earth orbiting the Sun, the acceleration (centripetal acceleration) is directed towards the center of the orbit.
2. The velocity of Earth is perpendicular to this acceleration, indicating different directions of velocity and acceleration.

Summary

Velocity is the measure of an object’s speed and direction, while acceleration is the rate at which velocity changes over time. It’s a common misconception that zero velocity implies zero acceleration. However, examples such as a pendulum at the end of its swing, a ball moving horizontally with constant velocity and downward acceleration, and planetary motion illustrate scenarios where objects have zero velocity but non-zero acceleration. These instances occur due to factors like kinetic energy, restoring forces, and centripetal forces, which influence the object’s motion. Understanding these examples is key to comprehending the relationship between velocity and acceleration in physics.