JEE Main - Frame of reference and motion in a straight line E-Book Study Material

Frame of Reference and Motion in a Straight Line

1. Frame of Reference

A frame of reference is essential to describe the motion of an object. It's a coordinate system along with a clock, used to measure the position, velocity, and acceleration of an object. Imagine you're on a train. To describe your movement, you need a point of reference. Are you moving relative to the seat you're sitting on, or relative to the ground outside the train? This choice of reference point defines your frame of reference.

There are two main types of frames of reference:

1.1 Inertial Frame of Reference

An inertial frame of reference is one where Newton's first law of motion (the law of inertia) holds true. In simpler terms, an object at rest stays at rest, and an object in motion continues in motion with constant velocity, unless acted upon by a net external force. These frames are either at rest or moving with a constant velocity relative to another inertial frame. The Earth is often approximated as an inertial frame, though technically it's not, due to its rotation and orbital motion.

Key Characteristics:

Newton's first law is valid.
No fictitious forces are present.
Acceleration is zero if no net force acts on the object.

1.2 Non-Inertial Frame of Reference

A non-inertial frame of reference is one that is accelerating with respect to an inertial frame. In these frames, Newton's first law does not hold true unless we introduce fictitious forces (also called pseudo-forces). Examples include a rotating merry-go-round, an accelerating car, or a lift moving with non-constant velocity.

Key Characteristics:

Newton's first law is not valid without fictitious forces.
Fictitious forces (like centrifugal force, Coriolis force) are observed.
These frames are accelerating.

Example: Consider a person standing inside a car. If the car is moving at a constant speed on a straight road (an inertial frame, approximately), the person feels no special force. However, if the car suddenly accelerates forward, the person feels pushed back into the seat. This "push" is due to the non-inertial nature of the accelerating car.

2. Motion in a Straight Line

Motion in a straight line, also known as rectilinear motion, is the simplest type of motion. It's a one-dimensional motion where an object moves along a straight path. To describe this motion, we only need one coordinate (e.g., 'x') along with time.

We use several key concepts to analyze this motion:

2.1 Position, Distance, and Displacement

Position: The location of an object in space relative to a chosen origin in a frame of reference. In one dimension, it's often represented by a single coordinate, say 'x'.

Distance: The total length of the path traveled by an object. It's a scalar quantity and is always non-negative.

Displacement: The change in position of an object. It's a vector quantity, meaning it has both magnitude and direction. It is the straight-line distance between the initial and final positions.

Formula: Displacement ($\Delta x$) = Final Position ($x_f$) - Initial Position ($x_i$)

Example: Suppose a car moves 5 meters to the east and then 5 meters to the west. The total distance traveled is 5 m + 5 m = 10 meters. If the car starts at position $x=0$, moves to $x=5$, and then returns to $x=0$, the displacement is $\Delta x = 0 - 0 = 0$ meters.

Memory Trick: Distance is the *total path covered*, like the odometer reading. Displacement is the *net change in location*, like the difference between your starting point and ending point.

2.2 Speed and Velocity

Speed: The rate at which an object covers distance. It is a scalar quantity. Average Speed = Total Distance / Total Time

Velocity: The rate at which an object changes its position. It is a vector quantity, indicating both the speed and the direction of motion. Average Velocity ($\vec{v}_{avg}$) = Displacement ($\Delta \vec{x}$) / Time Interval ($\Delta t$)

Instantaneous Velocity: The velocity of an object at a specific moment in time. It is the limit of average velocity as the time interval approaches zero. $\vec{v} = \lim_{\Delta t \to 0} \frac{\Delta \vec{x}}{\Delta t} = \frac{d\vec{x}}{dt}$

Example: If a car travels 100 km in 2 hours, its average speed is 100 km / 2 h = 50 km/h. If it travels 100 km north in 2 hours, its average velocity is 50 km/h north.

Key Point: Speed is the magnitude of velocity. An object can have a constant speed but changing velocity if its direction of motion changes (e.g., moving in a circle). For motion in a straight line without changing direction, speed and the magnitude of velocity are the same.

2.3 Acceleration

Acceleration: The rate at which an object's velocity changes. It is a vector quantity. Average Acceleration ($\vec{a}_{avg}$) = Change in Velocity ($\Delta \vec{v}$) / Time Interval ($\Delta t$)

Instantaneous Acceleration: The acceleration of an object at a specific moment in time. $\vec{a} = \lim_{\Delta t \to 0} \frac{\Delta \vec{v}}{\Delta t} = \frac{d\vec{v}}{dt}$

If the velocity is constant, the acceleration is zero. Acceleration occurs when there is a change in speed, or a change in direction, or both.

Example: When a car speeds up, it has positive acceleration. When it slows down (brakes), it has negative acceleration (often called deceleration). If the car turns a corner at constant speed, it is still accelerating because its velocity (which includes direction) is changing.

3. Equations of Motion for Uniform Acceleration

When an object moves in a straight line with constant acceleration, we can use a set of kinematic equations to relate its displacement, initial velocity, final velocity, acceleration, and time. These equations are fundamental for solving problems involving rectilinear motion with constant acceleration.

Let:

$u$ = initial velocity
$v$ = final velocity
$a$ = constant acceleration
$t$ = time interval
$s$ = displacement

The three main equations of motion are:

$v = u + at$
This equation relates final velocity, initial velocity, acceleration, and time. It comes directly from the definition of acceleration.
$s = ut + \frac{1}{2}at^2$
This equation relates displacement, initial velocity, acceleration, and time. It tells us how far an object travels under constant acceleration.
$v^2 = u^2 + 2as$
This equation relates final velocity, initial velocity, acceleration, and displacement. It's useful when time is not given or not needed.

Sometimes, a fourth equation is also useful:

$s = \frac{(u+v)}{2}t$
This equation relates displacement, average velocity (which is $\frac{u+v}{2}$ for constant acceleration), and time.

Shortcut for derivation: These equations can be derived assuming average velocity for constant acceleration is $\frac{u+v}{2}$ and acceleration is $\frac{v-u}{t}$. 1. $a = \frac{v-u}{t} \implies v = u + at$ 2. $s = (\text{average velocity}) \times t = \frac{u+v}{2} \times t$. Substitute $v$ from eq 1: $s = \frac{u+(u+at)}{2} \times t = \frac{2u+at}{2} \times t = ut + \frac{1}{2}at^2$. 3. From eq 2, $t = \frac{2s}{u+v}$. Substitute this into eq 1: $v = u + a \left(\frac{2s}{u+v}\right)$. Rearranging gives $v^2 = u^2 + 2as$.

4. Motion Under Gravity

A very common application of motion in a straight line with constant acceleration is the motion of objects under the influence of gravity near the Earth's surface. In the absence of air resistance, all objects fall with the same constant acceleration, denoted by $g$. The value of $g$ is approximately $9.8 \, m/s^2$. For simplicity in many problems, it's often approximated as $10 \, m/s^2$.

We can adapt the kinematic equations for motion under gravity. We need to choose a coordinate system and stick to it.

4.1 Upward Motion

If an object is thrown upwards, its initial velocity ($u$) is positive (if upward is taken as positive). The acceleration due to gravity ($g$) acts downwards, so it's negative ($a = -g$).

The equations become:

$v = u - gt$
$s = ut - \frac{1}{2}gt^2$
$v^2 = u^2 - 2gs$

At the highest point of its trajectory, the object's instantaneous velocity ($v$) is zero.

Example: A ball is thrown upwards with an initial velocity of 20 m/s. How high does it go, and how long does it take to reach the highest point? (Take $g = 10 \, m/s^2$) Here, $u = 20 \, m/s$, $v = 0 \, m/s$ (at highest point), $a = -10 \, m/s^2$. Using $v^2 = u^2 + 2as$: $0^2 = (20)^2 + 2(-10)s$ $0 = 400 - 20s$ $20s = 400 \implies s = 20$ meters. Using $v = u + at$: $0 = 20 + (-10)t$ $10t = 20 \implies t = 2$ seconds.

4.2 Downward Motion

If an object is dropped from rest, its initial velocity ($u$) is zero. The acceleration due to gravity ($g$) acts downwards. If we choose downward as the positive direction, then $a = +g$.

The equations become:

$v = gt$
$s = \frac{1}{2}gt^2$
$v^2 = 2gs$

If the object is thrown downwards with an initial velocity $u$, then $a = +g$ (if downward is positive).

$v = u + gt$
$s = ut + \frac{1}{2}gt^2$
$v^2 = u^2 + 2gs$

Example: A stone is dropped from a height of 45 meters. How long does it take to reach the ground? (Take $g = 10 \, m/s^2$) Here, $u = 0 \, m/s$, $s = 45 \, m$, $a = +10 \, m/s^2$ (taking downward as positive). Using $s = ut + \frac{1}{2}at^2$: $45 = (0)t + \frac{1}{2}(10)t^2$ $45 = 5t^2$ $t^2 = 9 \implies t = 3$ seconds.

Important Note on Signs: Always define your positive direction (upwards or downwards) and be consistent. If upward is positive, acceleration due to gravity is $-g$. If downward is positive, acceleration due to gravity is $+g$.

5. Relative Velocity

Relative velocity is the velocity of an object as observed from a particular frame of reference. When dealing with motion in a straight line, it simplifies considerably.

Let $v_{AB}$ be the velocity of object A relative to object B. If both objects are moving along the same straight line:

If they move in the same direction: $v_{AB} = v_A - v_B$
If they move in opposite directions: $v_{AB} = v_A - (-v_B) = v_A + v_B$ (if velocities are measured with respect to a common stationary frame, and $v_A, v_B$ are their respective velocities)

More formally, if $\vec{v}_A$ is the velocity of A with respect to the ground, and $\vec{v}_B$ is the velocity of B with respect to the ground, then the velocity of A with respect to B is: $\vec{v}_{AB} = \vec{v}_A - \vec{v}_B$

Example: Two cars, A and B, are moving along a straight road. Car A moves east at 60 km/h, and Car B moves east at 40 km/h. What is the velocity of A relative to B? Let East be the positive direction. $v_A = +60 \, km/h$ $v_B = +40 \, km/h$ $v_{AB} = v_A - v_B = 60 - 40 = +20 \, km/h$. This means Car A is moving east at 20 km/h relative to Car B.

Example 2: Car A moves east at 60 km/h. Car B moves west at 40 km/h. What is the velocity of A relative to B? $v_A = +60 \, km/h$ $v_B = -40 \, km/h$ $v_{AB} = v_A - v_B = 60 - (-40) = 60 + 40 = +100 \, km/h$. This means Car A is moving east at 100 km/h relative to Car B.

Relative Velocity Shortcut: To find the velocity of A relative to B, imagine yourself in B. Your frame is non-inertial if B is accelerating, but if B is moving at constant velocity, your frame is inertial. You 'see' A moving with a velocity that is A's actual velocity minus your own velocity.

6. Graphical Representation of Motion

Graphs are powerful tools to visualize and analyze motion. For motion in a straight line, we commonly use position-time, velocity-time, and acceleration-time graphs.

6.1 Position-Time Graph ($x$ vs $t$)

Slope: Represents velocity. A steeper slope means higher velocity.
Straight Line with constant slope: Constant velocity (uniform motion).
Horizontal line: Object is at rest (zero velocity).
Curved line: Changing velocity (acceleration).
Positive slope: Velocity is positive (moving in the positive direction).
Negative slope: Velocity is negative (moving in the negative direction).

6.2 Velocity-Time Graph ($v$ vs $t$)

Slope: Represents acceleration. A steeper slope means higher acceleration.
Straight line with constant slope: Constant acceleration.
Horizontal line: Constant velocity (zero acceleration).
Area under the curve: Represents displacement.
Positive $v$: Motion in the positive direction.
Negative $v$: Motion in the negative direction.
Crosses the time axis: The object changes direction.

Example: A velocity-time graph shows a line starting at $v=0$, increasing linearly to $v=10 \, m/s$ at $t=5 \, s$, and then staying constant at $v=10 \, m/s$ until $t=10 \, s$. From $t=0$ to $t=5 \, s$: The slope is constant and positive, meaning constant positive acceleration. The area under the curve is $\frac{1}{2} \times 5 \times 10 = 25 \, m$, which is the displacement. From $t=5 \, s$ to $t=10 \, s$: The slope is zero, meaning zero acceleration (constant velocity). The area under the curve is $10 \times (10-5) = 50 \, m$, which is the displacement during this interval.

6.3 Acceleration-Time Graph ($a$ vs $t$)

Area under the curve: Represents the change in velocity ($\Delta v$).
Horizontal line: Constant acceleration.
Zero line: Zero acceleration (constant velocity).

Graphical Analysis Tip: Remember the relationships: Slope of x-t graph is v. Slope of v-t graph is a. Area under v-t graph is displacement ($\Delta x$). Area under a-t graph is change in velocity ($\Delta v$).