Velocity, Acceleration and Rectilinear Motion

Understanding Motion

In physics, motion is the change in position of an object with respect to time. To describe motion accurately, we need fundamental concepts like displacement, velocity, and acceleration. Rectilinear motion is the simplest type of motion, where an object moves along a straight line. This is the foundation upon which more complex motion can be understood.

1. Displacement and Distance

Before we talk about velocity, we must distinguish between displacement and distance.

Distance

Distance is the total length of the path traveled by an object. It is a scalar quantity, meaning it only has magnitude and no direction. For example, if you walk 5 meters east and then 5 meters west, the total distance you traveled is 10 meters.

Displacement

Displacement is the change in position of an object. It is a vector quantity, meaning it has both magnitude and direction. It is the shortest distance between the initial and final position. In the previous example, if you walk 5 meters east and then 5 meters west, your final position is the same as your initial position. Therefore, your displacement is 0 meters.

Let's consider an object starting at position $x_1$ and moving to position $x_2$. The displacement, denoted by $\Delta x$, is given by:

$$ \Delta x = x_2 - x_1 $$

If $x_2 > x_1$, the displacement is positive, indicating movement in the positive direction. If $x_2 < x_1$, the displacement is negative, indicating movement in the negative direction.

2. Speed and Velocity

Speed and velocity are often used interchangeably in everyday language, but in physics, they have distinct meanings.

Speed

Speed is the rate at which an object covers distance. It is a scalar quantity. Average speed is calculated as the total distance traveled divided by the total time taken.

$$ \text{Average Speed} = \frac{\text{Total Distance}}{\text{Total Time}} $$

Instantaneous speed is the speed of an object at a particular moment in time.

Velocity

Velocity is the rate at which an object changes its position. It is a vector quantity, meaning it has both magnitude and direction. The magnitude of velocity is speed. Average velocity is calculated as the total displacement divided by the total time taken.

$$ \text{Average Velocity} (\vec{v}_{\text{avg}}) = \frac{\text{Total Displacement}}{\text{Total Time}} = \frac{\Delta \vec{x}}{\Delta t} $$

In rectilinear motion, we can often simplify this by considering only the magnitude and direction along the line. If the motion is along the x-axis, we can write:

$$ v_{\text{avg}} = \frac{\Delta x}{\Delta t} = \frac{x_2 - x_1}{t_2 - t_1} $$

Instantaneous velocity is the velocity of an object at a particular moment in time. It is the limit of average velocity as the time interval approaches zero. Mathematically, it is the derivative of the position with respect to time:

$$ \vec{v} = \lim_{\Delta t \to 0} \frac{\Delta \vec{x}}{\Delta t} = \frac{d\vec{x}}{dt} $$

In rectilinear motion along the x-axis:

$$ v = \frac{dx}{dt} $$

Example: A car travels 100 meters east in 10 seconds. Its displacement is 100 meters east. Its average velocity is 10 m/s east. If the car then travels 50 meters west in 5 seconds, its total displacement is 100 m east - 50 m west = 50 meters east. The total time taken is 10 s + 5 s = 15 s. The average velocity for the entire trip is 50 m / 15 s = 3.33 m/s east. The total distance traveled is 100 m + 50 m = 150 m. The average speed is 150 m / 15 s = 10 m/s.

3. Acceleration

Acceleration is the rate at which an object's velocity changes. It is also a vector quantity. In rectilinear motion, acceleration occurs when the speed or direction of motion along the straight line changes.

Average Acceleration

Average acceleration is the change in velocity divided by the time interval over which the change occurs.

$$ \vec{a}_{\text{avg}} = \frac{\Delta \vec{v}}{\Delta t} = \frac{\vec{v}_2 - \vec{v}_1}{t_2 - t_1} $$

For rectilinear motion along the x-axis:

$$ a_{\text{avg}} = \frac{\Delta v}{\Delta t} = \frac{v_2 - v_1}{t_2 - t_1} $$

If the final velocity ($v_2$) is greater than the initial velocity ($v_1$), the acceleration is positive (assuming $v_1$ and $v_2$ are in the positive direction), meaning the object is speeding up. If $v_2 < v_1$, the acceleration is negative, meaning the object is slowing down (this is often called deceleration).

Instantaneous Acceleration

Instantaneous acceleration is the acceleration of an object at a particular moment in time. It is the limit of average acceleration as the time interval approaches zero. Mathematically, it is the derivative of velocity with respect to time, or the second derivative of position with respect to time:

$$ \vec{a} = \lim_{\Delta t \to 0} \frac{\Delta \vec{v}}{\Delta t} = \frac{d\vec{v}}{dt} = \frac{d^2\vec{x}}{dt^2} $$

For rectilinear motion along the x-axis:

$$ a = \frac{dv}{dt} = \frac{d^2x}{dt^2} $$

Example: A cyclist starts from rest ($v_1 = 0$ m/s) and reaches a velocity of 10 m/s in 5 seconds. Her average acceleration is:

$$ a_{\text{avg}} = \frac{10 \text{ m/s} - 0 \text{ m/s}}{5 \text{ s}} = 2 \text{ m/s}^2 $$

If the cyclist then applies the brakes and slows down to 2 m/s in 3 seconds, her acceleration during this phase is:

$$ a_{\text{avg}} = \frac{2 \text{ m/s} - 10 \text{ m/s}}{3 \text{ s}} = \frac{-8 \text{ m/s}}{3 \text{ s}} \approx -2.67 \text{ m/s}^2 $$

The negative sign indicates that the acceleration is in the opposite direction to the velocity, causing the cyclist to slow down.

4. Uniform Velocity

An object has uniform velocity if its velocity remains constant over time. This means both its speed and direction are unchanging. In rectilinear motion, this implies that the object travels equal displacements in equal intervals of time.

If an object moves with uniform velocity $\vec{v}$, its displacement $\Delta \vec{x}$ over a time interval $\Delta t$ is given by:

$$ \Delta \vec{x} = \vec{v} \Delta t $$

In this case, the acceleration is zero, as the velocity is not changing.

Example: A train moving at a constant speed of 60 km/h along a straight track has uniform velocity. If it travels for 2 hours, its displacement will be $60 \text{ km/h} \times 2 \text{ h} = 120 \text{ km}$ in the direction of the track.

5. Uniform Acceleration

An object has uniform acceleration if its acceleration remains constant over time. This means the velocity changes by the same amount in every equal time interval.

For motion with uniform acceleration, we can use a set of kinematic equations that relate displacement, initial velocity, final velocity, acceleration, and time. These equations are derived from the definitions of velocity and acceleration.

Let:

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

6. Kinematic Equations for Uniform Acceleration

These equations are fundamental for solving problems involving uniformly accelerated rectilinear motion.

Equation 1: Relating final velocity, initial velocity, acceleration, and time

From the definition of average acceleration, if acceleration is uniform ($a_{\text{avg}} = a$), then:

$$ a = \frac{v - u}{t} $$

Rearranging this gives the first kinematic equation:

$$ v = u + at $$

This equation tells us the final velocity of an object after a certain time, given its initial velocity and constant acceleration.

Equation 2: Relating displacement, initial velocity, time, and acceleration

The average velocity for uniform acceleration is the average of the initial and final velocities:

$$ v_{\text{avg}} = \frac{u + v}{2} $$

We also know that displacement is average velocity multiplied by time:

$$ s = v_{\text{avg}} \times t $$

Substituting the expression for average velocity:

$$ s = \left(\frac{u + v}{2}\right) t $$

Now, substitute $v = u + at$ from Equation 1 into this equation:

$$ s = \left(\frac{u + (u + at)}{2}\right) t $$ $$ s = \left(\frac{2u + at}{2}\right) t $$ $$ s = \left(u + \frac{1}{2}at\right) t $$

This gives us the second kinematic equation:

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

This equation allows us to calculate the displacement of an object after a certain time, given its initial velocity and constant acceleration.

Equation 3: Relating final velocity, initial velocity, acceleration, and displacement

We can derive a third equation that does not involve time. From Equation 1, we can express time as:

$$ t = \frac{v - u}{a} $$

Substitute this expression for $t$ into Equation 2:

$$ s = u\left(\frac{v - u}{a}\right) + \frac{1}{2}a\left(\frac{v - u}{a}\right)^2 $$ $$ s = \frac{uv - u^2}{a} + \frac{1}{2}a\frac{(v - u)^2}{a^2} $$ $$ s = \frac{uv - u^2}{a} + \frac{(v - u)^2}{2a} $$

Multiply both sides by $2a$ to clear the denominators:

$$ 2as = 2(uv - u^2) + (v - u)^2 $$ $$ 2as = 2uv - 2u^2 + (v^2 - 2uv + u^2) $$ $$ 2as = 2uv - 2u^2 + v^2 - 2uv + u^2 $$ $$ 2as = v^2 - u^2 $$

Rearranging this gives the third kinematic equation:

$$ v^2 = u^2 + 2as $$

This equation is useful when the time interval is unknown or not required.

Equation 4: Relating displacement, average velocity, and time (alternative form)

This equation is essentially a restatement of the definition of displacement using average velocity when acceleration is uniform.

$$ s = \left(\frac{u + v}{2}\right) t $$

This equation is particularly useful when the initial and final velocities are known, and we need to find the displacement over a given time.

7. Motion Under Gravity

A common example of uniformly accelerated motion is the motion of objects under the influence of gravity near the Earth's surface. When air resistance is neglected, objects fall with a constant acceleration, known as the acceleration due to gravity, denoted by $g$.

The value of $g$ is approximately $9.8 \, \text{m/s}^2$ on Earth. Its direction is always downwards, towards the center of the Earth.

When applying the kinematic equations to motion under gravity:

• We usually set up a coordinate system. For instance, we can define the upward direction as positive.
• If an object is thrown upwards, its initial velocity ($u$) is positive, and the acceleration due to gravity ($a$) is negative ($-g$), because it acts downwards. The velocity decreases as the object rises, becomes zero at the highest point, and then becomes negative as it falls.
• If an object is dropped from rest, its initial velocity ($u$) is zero, and the acceleration ($a$) is negative ($-g$). Its velocity becomes increasingly negative as it falls.
• If an object is thrown downwards, its initial velocity ($u$) is negative, and the acceleration ($a$) is also negative ($-g$).

Example: A ball is dropped from a height of 50 meters. How long does it take to reach the ground, and what is its velocity upon impact? (Assume $g = 9.8 \, \text{m/s}^2$ and neglect air resistance).

We choose the downward direction as positive.

• Initial velocity, $u = 0 \, \text{m/s}$ (dropped from rest)
• Displacement, $s = 50 \, \text{m}$ (downwards)
• Acceleration, $a = g = 9.8 \, \text{m/s}^2$ (downwards)

We need to find time ($t$) and final velocity ($v$).

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

$$ 50 = (0)t + \frac{1}{2}(9.8)t^2 $$ $$ 50 = 4.9t^2 $$ $$ t^2 = \frac{50}{4.9} \approx 10.204 $$ $$ t = \sqrt{10.204} \approx 3.19 \text{ seconds} $$

Now, using the equation $v = u + at$:

$$ v = 0 + (9.8)(3.19) $$ $$ v \approx 31.26 \text{ m/s} $$

So, it takes approximately 3.19 seconds for the ball to reach the ground, and its impact velocity is approximately 31.26 m/s downwards.

8. Graphical Representation of Motion

Graphs are powerful tools to visualize and analyze motion. For rectilinear motion, we commonly use position-time ($x-t$), velocity-time ($v-t$), and acceleration-time ($a-t$) graphs.

Position-Time (x-t) Graph

• Slope: The slope of the position-time graph represents the velocity of the object.
• Uniform Velocity: A straight line with a constant slope indicates uniform velocity. The steeper the slope, the higher the velocity. A horizontal line means the object is at rest (zero velocity).
• Non-uniform Velocity: A curved line indicates non-uniform velocity (changing velocity).

Velocity-Time (v-t) Graph

• Slope: The slope of the velocity-time graph represents the acceleration of the object.
• Uniform Velocity: A horizontal line indicates uniform velocity (zero acceleration).
• Uniform Acceleration: A straight line with a constant slope indicates uniform acceleration. A positive slope means positive acceleration (speeding up if velocity is positive), and a negative slope means negative acceleration (slowing down if velocity is positive).
• Non-uniform Acceleration: A curved line indicates non-uniform acceleration.
• Area Under the Curve: The area under the velocity-time graph represents the displacement of the object.

Acceleration-Time (a-t) Graph

• Uniform Acceleration: A horizontal line indicates uniform acceleration.
• Non-uniform Acceleration: A varying line indicates non-uniform acceleration.
• Area Under the Curve: The area under the acceleration-time graph represents the change in velocity ($\Delta v$).

9. Relative Velocity

Velocity is always measured with respect to a frame of reference. When we talk about the velocity of an object, we implicitly assume a reference frame (e.g., the ground). Relative velocity is the velocity of an object as observed from a particular frame of reference.

Consider three objects A, B, and C. Let $\vec{v}_{AB}$ be the velocity of A relative to B, $\vec{v}_{BC}$ be the velocity of B relative to C, and $\vec{v}_{AC}$ be the velocity of A relative to C.

The rule for adding velocities (Galilean transformation for speeds much less than the speed of light) is:

$$ \vec{v}_{AC} = \vec{v}_{AB} + \vec{v}_{BC} $$

This means the velocity of A as seen by C is the velocity of A as seen by B, plus the velocity of B as seen by C.

Example: A person is walking on a moving train. The train is moving east at 30 m/s relative to the ground. The person is walking west on the train at 2 m/s relative to the train. What is the velocity of the person relative to the ground?

Let:

• Ground be frame C.
• Train be frame B.
• Person be object A.

We are given:

• $\vec{v}_{BC}$ (velocity of train relative to ground) = 30 m/s East.
• $\vec{v}_{AB}$ (velocity of person relative to train) = 2 m/s West.

We want to find $\vec{v}_{AC}$ (velocity of person relative to ground).

Let's define East as positive and West as negative.

• $\vec{v}_{BC} = +30 \, \text{m/s}$
• $\vec{v}_{AB} = -2 \, \text{m/s}$

Using the relative velocity equation:

$$ \vec{v}_{AC} = \vec{v}_{AB} + \vec{v}_{BC} $$ $$ \vec{v}_{AC} = -2 \, \text{m/s} + 30 \, \text{m/s} $$ $$ \vec{v}_{AC} = +28 \, \text{m/s} $$

The velocity of the person relative to the ground is 28 m/s East.

10. Non-Uniform Acceleration

When acceleration is not constant, the kinematic equations derived for uniform acceleration cannot be directly applied. In such cases, we must use calculus.

• If acceleration $a(t)$ is a function of time, then velocity is found by integrating acceleration with respect to time:
$$ v(t) = \int a(t) dt + C_1 $$ The constant $C_1$ is determined by the initial velocity ($v(0)$).
• Similarly, position is found by integrating velocity with respect to time:
$$ x(t) = \int v(t) dt + C_2 $$ The constant $C_2$ is determined by the initial position ($x(0)$).

Example: An object starts from rest and its acceleration is given by $a(t) = 2t \, \text{m/s}^2$, where $t$ is in seconds. Find its velocity and position at $t = 3$ seconds.

Initial conditions: $u = v(0) = 0 \, \text{m/s}$, $x(0) = 0 \, \text{m}$.

Velocity:

$$ v(t) = \int a(t) dt = \int 2t \, dt = t^2 + C_1 $$

Using $v(0) = 0$:

$$ 0 = (0)^2 + C_1 \implies C_1 = 0 $$

So, $v(t) = t^2$.

At $t = 3$ s, $v(3) = (3)^2 = 9 \, \text{m/s}$.

Position:

$$ x(t) = \int v(t) dt = \int t^2 \, dt = \frac{t^3}{3} + C_2 $$

Using $x(0) = 0$:

$$ 0 = \frac{(0)^3}{3} + C_2 \implies C_2 = 0 $$

So, $x(t) = \frac{t^3}{3}$.

At $t = 3$ s, $x(3) = \frac{(3)^3}{3} = \frac{27}{3} = 9 \, \text{m}$.

Thus, at $t = 3$ seconds, the object's velocity is 9 m/s and its position is 9 meters from the starting point.