Physics: Class 11 (Book 1) — Chapter 2: Motion in a Straight Line

• Kinematics : The branch of mechanics that describes the motion of objects without considering the forces or causes that produce the motion.
• Rectilinear Motion : Motion along a straight line path. It represents the simplest form of one-dimensional motion.
• Point Object Approximation : An approximation where an object's size is neglected, treating it as a mathematical point. This is valid when the distance moved by the object is much larger than its linear dimensions.

The instantaneous velocity $v$ at any instant $t$ is defined as the limit of the average velocity as the time interval $\Delta t$ becomes infinitesimally small:

In calculus, $\frac{dx}{dt}$ is the differential coefficient of position $x$ with respect to time $t$, representing the instantaneous rate of change of position.

• Instantaneous Speed : The magnitude of instantaneous velocity at any given instant. Unlike average speed, which can exceed the magnitude of average velocity, instantaneous speed is always exactly equal to the magnitude of instantaneous velocity.

• Average Acceleration : The change in velocity divided by the time interval:$$\bar{a} = \frac{v_2 - v_1}{t_2 - t_1} = \frac{\Delta v}{\Delta t}$$
• Instantaneous Acceleration : The limit of average acceleration as the time interval $\Delta t$ approaches zero:$$a = \lim_{\Delta t \to 0} \frac{\Delta v}{\Delta t} = \frac{dv}{dt}$$

• Slope of $v-t$ Graph : The slope of the tangent to the velocity-time curve at any instant gives the instantaneous acceleration.
• Area under $v-t$ Graph : The area under the velocity-time curve over a given time interval represents the displacement of the object during that interval.

1. First Equation (Velocity-Time Relation): $$v = v_0 + at$$
2. Second Equation (Position-Time Relation): $$x = x_0 + v_0 t + \frac{1}{2} at^2$$
3. Third Equation (Velocity-Position Relation): $$v^2 = v_0^2 + 2a(x - x_0)$$

Using the definitions of acceleration and velocity, we can derive these equations rigorously:

1. Start with $a = \frac{dv}{dt} \implies dv = a \, dt$. Integrating both sides from $t = 0$ (where velocity is $v_0$) to $t$ (where velocity is $v$):$$\int_{v_0}^{v} dv = \int_{0}^{t} a \, dt \implies v - v_0 = at \implies v = v_0 + at$$
2. Next, use $v = \frac{dx}{dt} \implies dx = v \, dt = (v_0 + at) \, dt$. Integrating both sides from $t = 0$ (where position is $x_0$) to $t$ (where position is $x$):$$\int_{x_0}^{x} dx = \int_{0}^{t} (v_0 + at) \, dt \implies x - x_0 = v_0 t + \frac{1}{2} at^2$$
3. Finally, write acceleration as $a = \frac{dv}{dt} = \frac{dv}{dx} \frac{dx}{dt} = v \frac{dv}{dx} \implies v \, dv = a \, dx$. Integrating both sides:$$\int_{v_0}^{v} v \, dv = \int_{x_0}^{x} a \, dx \implies \frac{v^2 - v_0^2}{2} = a(x - x_0) \implies v^2 = v_0^2 + 2a(x - x_0)$$

An object released near the Earth's surface experiences a constant downward acceleration due to gravity, denoted by $g \approx 9.8\text{ m/s}^2$ (neglecting air resistance). Taking the upward direction as positive, $a = -g$.

• Velocity: $v = -gt$
• Displacement: $y = -\frac{1}{2} gt^2$
• Velocity-displacement: $v^2 = -2gy$

The distances traversed by a freely falling body starting from rest during equal successive intervals of time stand to one another in the same ratio as the odd numbers beginning with unity: $1 : 3 : 5 : 7 : \dots$

The distance a vehicle travels before coming to rest after brakes are applied is called the stopping distance ($d_s$). It depends on the initial velocity $v_0$ and the deceleration $-a$:

Thus, stopping distance is directly proportional to the square of the initial velocity ($d_s \propto v_0^2$).

Reaction time ($t_r$) is the time a person takes to observe, think, and act. It can be measured using a simple ruler-drop experiment where a ruler is caught after being dropped from rest:

where $d$ is the distance the ruler falls before being caught.