Motion In A Straight LineClass 11 Physics Notes

Introduction to Motion

Motion is simply the change in an object's position over time. It's a fundamental concept in physics and is happening all around us, from a leaf falling from a tree to planets revolving around the sun.

In this chapter, we will focus on the simplest type of motion: rectilinear motion, which is motion along a straight line. To describe this motion, we'll use key concepts like velocity and acceleration.

The Concept of a Point Object

To make studying motion easier, we often treat moving objects as point objects (or point masses). This is a useful approximation when the size of the object is much smaller than the distance it travels.

What is Kinematics?

Kinematics is the branch of physics that describes motion without getting into what causes the motion (like forces). We will study the causes of motion in a later chapter.

Instantaneous Velocity and Speed

While average velocity tells us how fast an object moved over a period, it doesn't describe its speed at any single moment. For that, we need instantaneous velocity.

Instantaneous velocity is the velocity of an object at a specific instant in time. It's what the speedometer in a car shows. We find it by looking at the average velocity over an infinitesimally small (extremely short) time interval.

Mathematically, it is defined as the rate of change of position. In calculus terms, it's the derivative of position ($x$) with respect to time ($t$).

$v = \lim_{\Delta t \rightarrow 0} \frac{\Delta x}{\Delta t} = \frac{\mathrm{d} x}{\mathrm{~d} t}$

• Graphically, the instantaneous velocity at any point on a position-time graph is the slope of the tangent to the curve at that point. A steeper slope means a higher velocity.

Instantaneous speed, or simply speed, is the magnitude (the absolute value) of the instantaneous velocity. It tells you how fast you are going, regardless of the direction.

Given

• Position equation: $x = a + bt^2$
• $a = 8.5 \text{ m}$
• $b = 2.5 \text{ m s}^{-2}$

To Find

• Instantaneous velocity at $t=0 \text{ s}$ and $t=2.0 \text{ s}$
• Average velocity between $t=2.0 \text{ s}$ and $t=4.0 \text{ s}$

Formula

• Instantaneous velocity: $v = \frac{dx}{dt}$
• Average velocity: $\bar{v} = \frac{x(t_2) - x(t_1)}{t_2 - t_1}$

Solution

First, we find the general expression for instantaneous velocity by taking the derivative of the position equation: $v = \frac{d}{dt}(a + bt^2) = 2bt$ Substituting the value of $b$: $v = 2(2.5)t = 5.0t \text{ m s}^{-1}$

Velocity at $t=0 \text{ s}$: $v = 5.0 \times 0 = 0 \text{ m s}^{-1}$

Velocity at $t=2.0 \text{ s}$: $v = 5.0 \times 2.0 = 10 \text{ m s}^{-1}$

Average velocity between $t=2.0 \text{ s}$ and $t=4.0 \text{ s}$: First, find the positions at $t=2.0 \text{ s}$ and $t=4.0 \text{ s}$. $x(2.0) = a + b(2.0)^2 = a + 4b$ $x(4.0) = a + b(4.0)^2 = a + 16b$

Now, use the average velocity formula: $\text{Average velocity} = \frac{x(4.0) - x(2.0)}{4.0 - 2.0} = \frac{(a + 16b) - (a + 4b)}{2.0}$ $\text{Average velocity} = \frac{12b}{2.0} = 6.0b$ Substituting the value of $b$: $\text{Average velocity} = 6.0 \times 2.5 = 15 \text{ m s}^{-1}$

Final Answer

• At $t=0 \text{ s}$, velocity is $0 \text{ m s}^{-1}$.
• At $t=2.0 \text{ s}$, velocity is $10 \text{ m s}^{-1}$.
• The average velocity between $2.0 \text{ s}$ and $4.0 \text{ s}$ is $15 \text{ m s}^{-1}$.

Acceleration

When an object's velocity changes, it is accelerating. Acceleration is the rate of change of velocity with respect to time.

Average and Instantaneous Acceleration

Average acceleration is the change in velocity over a time interval. $\bar{a} = \frac{v_2 - v_1}{t_2 - t_1} = \frac{\Delta v}{\Delta t}$ The SI unit for acceleration is meters per second squared ($\text{m s}^{-2}$).

Instantaneous acceleration is the acceleration at a specific instant in time. It is found by taking the limit as the time interval becomes infinitesimally small. $a = \lim_{\Delta t \rightarrow 0} \frac{\Delta v}{\Delta t} = \frac{\mathrm{d} v}{\mathrm{~d} t}$

• Graphically, the instantaneous acceleration at any point on a velocity-time graph is the slope of the tangent to the curve at that point.

Since velocity has both magnitude (speed) and direction, acceleration can occur if:

• The speed changes.
• The direction of motion changes.
• Both speed and direction change.

Acceleration can be positive, negative, or zero.

• Positive acceleration: The velocity is increasing in the positive direction. The position-time graph curves upward.
• Negative acceleration (or deceleration/retardation): The velocity is decreasing in the positive direction, or increasing in the negative direction. The position-time graph curves downward.
• Zero acceleration: The velocity is constant (uniform motion). The position-time graph is a straight line.

Area Under the Velocity-Time Graph

An important feature of a velocity-time (v-t) graph is that the area under the curve represents the displacement of the object during that time interval.

For an object moving with constant velocity $u$, the v-t graph is a horizontal line. The area under the graph between $t=0$ and $t=T$ is a rectangle with height $u$ and base $T$. $\text{Area} = u \times T$ This is exactly the formula for displacement (Displacement = velocity × time).

Kinematic Equations for Uniformly Accelerated Motion

When an object moves with constant acceleration, we can use a set of simple equations, known as kinematic equations, to relate displacement ($x$), time ($t$), initial velocity ($v_0$), final velocity ($v$), and acceleration ($a$).

Assuming the object starts at position $x=0$ at time $t=0$:

1. Velocity-Time Relation: $v = v_0 + at$
2. Position-Time Relation: $x = v_0 t + \frac{1}{2}at^2$
3. Position-Velocity Relation: $v^2 = v_0^2 + 2ax$

If the starting position at $t=0$ is $x_0$ instead of 0, the position $x$ in the equations is replaced by the displacement ($x - x_0$). The equations become:

• $v = v_0 + at$
• $x = x_0 + v_0 t + \frac{1}{2}at^2$
• $v^2 = v_0^2 + 2a(x - x_0)$

Solution

We start from the basic definitions of acceleration and velocity.

1. Deriving $v = v_0 + at$ By definition, acceleration is $a = \frac{dv}{dt}$. We can rearrange this to $dv = a \ dt$. Now, we integrate both sides. We assume the velocity is $v_0$ at $t=0$ and $v$ at time $t$. $\int_{v_0}^{v} dv = \int_{0}^{t} a \ dt$ Since $a$ is constant, we can take it out of the integral: $\int_{v_0}^{v} dv = a \int_{0}^{t} dt$ $[v]_{v_0}^{v} = a [t]_{0}^{t}$ $v - v_0 = a(t - 0)$ $v = v_0 + at$

2. Deriving $x = x_0 + v_0t + \frac{1}{2}at^2$ By definition, velocity is $v = \frac{dx}{dt}$. We rearrange this to $dx = v \ dt$. We substitute the expression for $v$ we just found: $dx = (v_0 + at) dt$. Now, we integrate both sides. We assume the position is $x_0$ at $t=0$ and $x$ at time $t$. $\int_{x_0}^{x} dx = \int_{0}^{t} (v_0 + at) dt$ $[x]_{x_0}^{x} = [v_0t + \frac{1}{2}at^2]_{0}^{t}$ $x - x_0 = (v_0t + \frac{1}{2}at^2) - (0)$ $x = x_0 + v_0t + \frac{1}{2}at^2$

3. Deriving $v^2 = v_0^2 + 2a(x - x_0)$ We can use the chain rule to write acceleration as $a = \frac{dv}{dt} = \frac{dv}{dx} \frac{dx}{dt}$. Since $\frac{dx}{dt} = v$, we get $a = v \frac{dv}{dx}$. Rearranging gives $v \ dv = a \ dx$. Now, we integrate both sides, from initial velocity $v_0$ and position $x_0$ to final velocity $v$ and position $x$. $\int_{v_0}^{v} v \ dv = \int_{x_0}^{x} a \ dx$ Since $a$ is constant: $\int_{v_0}^{v} v \ dv = a \int_{x_0}^{x} dx$ $[\frac{v^2}{2}]_{v_0}^{v} = a [x]_{x_0}^{x}$ $\frac{v^2}{2} - \frac{v_0^2}{2} = a(x - x_0)$ $v^2 - v_0^2 = 2a(x - x_0)$ $v^2 = v_0^2 + 2a(x - x_0)$

Given

• Initial velocity, $v_0 = +20 \text{ m s}^{-1}$ (taking upward as positive)
• Initial height, $y_0 = 25.0 \text{ m}$
• Acceleration, $a = -g = -10 \text{ m s}^{-2}$ (gravity acts downward)

To Find

(a) The maximum height the ball will rise to. (b) The total time until the ball hits the ground.

Formula

$v^2 = v_0^2 + 2a(y - y_0)$ $y = y_0 + v_0t + \frac{1}{2}at^2$

Solution

(a) How high the ball will rise

At the highest point of its path, the ball's instantaneous velocity is zero. So, $v = 0$. We want to find the displacement $(y - y_0)$. Using the formula $v^2 = v_0^2 + 2a(y - y_0)$: $0^2 = (20)^2 + 2(-10)(y - y_0)$ $0 = 400 - 20(y - y_0)$ $20(y - y_0) = 400$ $(y - y_0) = \frac{400}{20} = 20 \text{ m}$

The ball will rise 20 m above the point it was thrown from. The maximum height from the ground is $25 \text{ m} + 20 \text{ m} = 45 \text{ m}$.

Answer for part (a) = The ball will rise $20 \text{ m}$ from its launch point.

(b) How long before the ball hits the ground

We want to find the total time $t$ for the entire journey, from being thrown to hitting the ground. At the final point (the ground), the position is $y = 0$. We can solve this directly using the position-time equation.

Initial position, $y_0 = 25 \text{ m}$ Final position, $y = 0 \text{ m}$ Initial velocity, $v_0 = +20 \text{ m s}^{-1}$ Acceleration, $a = -10 \text{ m s}^{-2}$

$y = y_0 + v_0t + \frac{1}{2}at^2$ $0 = 25 + 20t + \frac{1}{2}(-10)t^2$ $0 = 25 + 20t - 5t^2$ Rearranging into a standard quadratic equation ($ax^2 + bx + c = 0$): $5t^2 - 20t - 25 = 0$ Divide the entire equation by 5 to simplify: $t^2 - 4t - 5 = 0$ Factoring the quadratic equation: $(t - 5)(t + 1) = 0$ The solutions are $t = 5 \text{ s}$ or $t = -1 \text{ s}$. Since time cannot be negative, we choose the positive solution.

Answer for part (b) = The ball will hit the ground after $5 \text{ s}$.

Discussion

An object in free fall is one that is moving only under the influence of gravity. If we neglect air resistance, all objects near the Earth's surface fall with a constant downward acceleration, known as the acceleration due to gravity, $g$. The value of $g$ is approximately $9.8 \text{ m s}^{-2}$.

Let's set up a coordinate system where the upward direction is positive ($+y$).

• The acceleration is always downward, so $a = -g = -9.8 \text{ m s}^{-2}$.
• If an object is dropped from rest from position $y=0$, its initial velocity is $v_0 = 0$.

The kinematic equations for this specific case become:

• Velocity: $v = 0 + (-g)t = -9.8t \text{ m s}^{-1}$
• Position: $y = 0 + (0)t + \frac{1}{2}(-g)t^2 = -4.9t^2 \text{ m}$
• Velocity-Position: $v^2 = 0^2 + 2(-g)y = -19.6y \text{ m}^2\text{s}^{-2}$

These equations show:

• The velocity becomes more negative (i.e., its downward speed increases) linearly with time.
• The distance fallen increases with the square of time.

Proof

Let an object fall from rest ($v_0 = 0$) under constant acceleration $g$. We will look at the distance it covers in successive equal time intervals, which we'll call $\tau$. The position $y$ at any time $t$ is given by $y = \frac{1}{2}gt^2$ (taking downward as positive for simplicity).

Let's calculate the position at times $t = 0, \tau, 2\tau, 3\tau, ...$

• At $t=0$: $y(0) = 0$
• At $t=\tau$: $y(\tau) = \frac{1}{2}g\tau^2$
• At $t=2\tau$: $y(2\tau) = \frac{1}{2}g(2\tau)^2 = 4 \times (\frac{1}{2}g\tau^2)$
• At $t=3\tau$: $y(3\tau) = \frac{1}{2}g(3\tau)^2 = 9 \times (\frac{1}{2}g\tau^2)$

Let's define a basic unit of distance $y_0 = \frac{1}{2}g\tau^2$. The positions are then $0, y_0, 4y_0, 9y_0, ...$

Now, let's find the distance traveled during each interval:

• First interval (0 to $\tau$): $d_1 = y(\tau) - y(0) = y_0 - 0 = y_0$
• Second interval ($\tau$ to $2\tau$): $d_2 = y(2\tau) - y(\tau) = 4y_0 - y_0 = 3y_0$
• Third interval ($2\tau$ to $3\tau$): $d_3 = y(3\tau) - y(2\tau) = 9y_0 - 4y_0 = 5y_0$
• Fourth interval ($3\tau$ to $4\tau$): $d_4 = y(4\tau) - y(3\tau) = 16y_0 - 9y_0 = 7y_0$

The ratio of the distances traversed in successive intervals is: $d_1 : d_2 : d_3 : d_4 : ... = y_0 : 3y_0 : 5y_0 : 7y_0 : ...$ $= 1 : 3 : 5 : 7 : ...$ This proves Galileo's law.

Given

• Initial velocity = $v_0$
• Final velocity, $v = 0$ (since the vehicle stops)
• Acceleration = $a$ (which will be a negative value, representing deceleration)

To Find

An expression for the stopping distance, $d_s$.

Formula

$v^2 = v_0^2 + 2ax$

Solution

Let the stopping distance be $d_s$. So, $x = d_s$. Substitute the known values into the kinematic equation: $0^2 = v_0^2 + 2ad_s$ Now, solve for $d_s$: $-v_0^2 = 2ad_s$ $d_s = \frac{-v_0^2}{2a}$

Final Answer The stopping distance is $d_s = \frac{-v_0^2}{2a}$. This shows that the stopping distance is proportional to the square of the initial velocity ($d_s \propto v_0^2$). If you double your speed, you need four times the distance to stop.

Given

• Distance traveled by ruler, $d = 21.0 \text{ cm} = 0.21 \text{ m}$
• The ruler is dropped, so its initial velocity $v_0 = 0$.
• The ruler is in free fall, so its acceleration is $a = g = 9.8 \text{ m s}^{-2}$ (taking downward as positive).

To Find

The reaction time, $t_r$.

Formula

$x = v_0t + \frac{1}{2}at^2$

Solution

Here, $x = d$ and $t = t_r$. $d = (0)t_r + \frac{1}{2}gt_r^2$ $d = \frac{1}{2}gt_r^2$ Now, we rearrange the formula to solve for $t_r$: $t_r^2 = \frac{2d}{g}$ $t_r = \sqrt{\frac{2d}{g}}$ Substitute the given values: $t_r = \sqrt{\frac{2 \times 0.21}{9.8}} \text{ s}$ $t_r \approx 0.2 \text{ s}$

Final Answer The estimated reaction time is approximately $0.2$ seconds.