Rectilinear Motion - Acceleration in Linear Motion

Acceleration describes how quickly the velocity of an object changes with time. It is a vector quantity, meaning it has both magnitude and direction. The dimensions of acceleration are [L¹ M⁰ T⁻²]. Understanding acceleration is essential for analyzing motion, especially when objects speed up, slow down, or change direction. This concept is fundamental in physics and helps explain everyday phenomena like falling objects and moving vehicles.

Acceleration is defined as the rate of change of velocity with time.

Average acceleration is calculated when an object has velocities \[\vec v_1\] and \[\vec v_2\] at times t₁ and t₂:

\[\vec{a}=\frac{\vec{v_2}-\vec{v_1}}{t_2-t_1}\]

where:

• \[\vec a\] = average acceleration
• \[\vec v_1\] = velocity at time t₁
• \[\vec v_2\] = velocity at time t₂

Instantaneous acceleration is the limiting value of average acceleration when the time interval approaches zero:

\[\vec{a}=\lim_{\Delta t\to0}\frac{\Delta\vec{v}}{\Delta t}=\frac{d\vec{v}}{dt}\]

where:

• \[\vec a\] = instantaneous acceleration
• \[d\vec{v}\] = infinitesimal change in velocity
• dt = infinitesimal change in time

The instantaneous acceleration at a given time equals the slope of the tangent to the velocity versus time curve at that time.

• Vector quantity (has both magnitude and direction)
• Dimensions: [L¹ M⁰ T⁻²]
• Can be positive (velocity increasing) or negative (velocity decreasing)
• Can be uniform (constant) or nonuniform (changing)
• Measured as a change in velocity per unit time
• For uniform acceleration, the velocity-time graph is linear
• For nonuniform acceleration, the velocity-time graph is nonlinear

Case (a): Zero acceleration (Constant Velocity)

• Horizontal line on v-t graph
• Velocity remains constant
• Shaded area under graph = displacement = v₀(t₂ − t₁)

Object moving with constant velocity.

Case (b): Positive Uniform Acceleration

• Straight line sloping upward
• The magnitude of velocity increases uniformly with time
• Area under the curve = displacement

Object moving with velocity (v) along +ve x-axis with uniform acceleration along the same direction.

Case (c): Negative Uniform Acceleration

• Straight line sloping downward
• Acceleration opposite to the velocity direction
• Velocity decreases uniformly with time
• Area under the curve = displacement

Object moving with velocity (v) with negative uniform acceleration.

Case (d): Nonuniform Acceleration

• Curved line on v-t graph
• Acceleration changes with time
• Average acceleration shown by a straight line
• Instantaneous acceleration = slope of tangent at that point

Object moving with nonuniform acceleration.

Displacement from the Velocity-Time Graph

The area under the velocity-time curve represents displacement:

Area = \[\int_{t_1}^{t_2}vdt=\int_{t_1}^{t_2}\frac{dx}{dt}dt=\int_{t_1}^{t_2}dx=x(t_2)-x(t_1)\]

This equals the displacement of the object from t₁ to t₂.

Consider an object at position x = 0 at time t = 0, with initial velocity u and final velocity v at time t.

Derivation of the equation of motion for motion with uniform acceleration.

First Equation:

Acceleration = slope of the velocity-time line

a = \[\frac{v-u}{t-0}=\frac{v-u}{t}\]

v = u + at

Second Equation:

Displacement = area under v-t graph = area of triangle + area of rectangle

s = \[\frac {1}{2}\](v − u)t + ut

Using the first equation:

s = ut + \[\frac {1}{2}\]at²

Third Equation:

Using average velocity:

s = \[v_{av}\cdot t=\frac{v+u}{2}\cdot t=\frac{(v+u)(v-u)}{2a}\]

s = \[\frac{v^2-u^2}{2a}\]

v² − u² = 2as

• Helps analyze the motion of objects that speed up or slow down
• Essential for understanding velocity-time graphs
• Area under the v-t curve gives displacement directly
• The slope of the v-t curve gives the acceleration value
• Enables calculation of motion parameters using three equations
• Applies to real-world situations like free fall
• Foundation for understanding forces and dynamics
• Critical for solving practical motion problems

Free Fall

• The most common example of uniform rectilinear motion with uniform acceleration.
• The body falls under Earth's gravity with zero initial velocity.
• Air resistance is negligible for small displacements.
• Acceleration due to gravity acts along the vertical direction.
• Assumed constant over distances small compared to Earth's radius.
• Velocity and acceleration are both along the vertical direction.

Problem

A stone is thrown vertically upward from the ground with a velocity of 15 m/s. Simultaneously, a ball is dropped from rest from a height of 30 m above. Using g = 10 m/s², find (i) how long until they meet, and (ii) the height above ground where they meet.

Solution step‑by‑step

1. Step 1: Set up equations
   Both objects move for the same time t₀. Using s = ut + \[\frac {1}{2}\] gt²:
   For stone (moving upward):
   s_stone = 15t₀ − \[\frac {1}{2}\](10)t₀² = 15t₀ − 5t₀²
   For ball (falling downward):
   s_ball = \[\frac {1}{2}\](10)t₀² = 5t₀²
2. Step 2: Apply the condition
   When they meet, the total distance = 30 m:
   s_stone + s_ball = 30
   15t₀ − 5t₀² + 5t₀² = 30
   15t₀ = 30
   t₀ = 2 seconds
3. Step 3: Find meeting height
   Distance traveled by the stone in 2 seconds:
   s_stone = 15(2) − 5(2)² = 30 − 20 = 10 m

Answer:

• Time of meeting = 2 seconds
• Height from ground = 10 meters

Vehicle Motion

• A car accelerating from rest at a traffic signal shows positive acceleration
• Applying the brakes causes negative acceleration (deceleration)
• Racing cars demonstrate high positive acceleration during starts

Sports Applications

• A cricket ball thrown vertically upward experiences negative acceleration due to gravity
• An athlete sprinting from the starting blocks shows positive acceleration
• A football kicked upward slows down due to gravitational acceleration

Elevator Motion

• Starting upward from the ground floor shows positive acceleration
• Stopping at a floor demonstrates negative acceleration
• Passengers feel changes in acceleration as weight changes

Amusement Park Rides

• Roller coasters experience varying acceleration through loops and drops
• Free-fall rides demonstrate near-constant gravitational acceleration