33 Cards 14 Min Read

Ch 02 - Motion In A Straight Line11th Physics

Rectilinear Motion

What specifically does the term 'rectilinear motion' refer to in kinematics?

DEFINITION

Rectilinear motion is the study of motion of objects strictly along a straight line (One-Dimensional Motion).

Note: In this chapter, all objects are typically approximated as point-mass objects.

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Point Object Approximation

Under what specific boundary condition is it valid to approximate a moving object as a 'point object'?

CONCEPT

An object can be considered a point object if the size of the object is much smaller than the distance it moves in a reasonable duration of time.

Example: A train travelling between two distant stations can be considered a point object, but a spinning cricket ball hitting the ground cannot.

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Distance vs. Displacement

What is the fundamental scientific distinction between Distance and Displacement?

DISTINCTION

Distance (Path Length): The total length of the actual path traversed by an object. It is a Scalar quantity and is always $\ge 0$ .
Displacement: The shortest straight-line distance between the initial and final positions. It is a Vector quantity and can be positive, negative, or zero.

Condition: Magnitude of displacement is always less than or equal to the total distance.

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Average Speed

CONCEPT Formula

v_{avg} = \frac{\text{Total Path Length (Distance)}}{\text{Total Time Interval}}

State the formula for calculating Average Speed over a given time interval.

Variables & Units:

$v_{avg}$ = Average Speed (m/s)

Conditions/Notes:

It is a scalar quantity.
Average speed is always greater than or equal to the magnitude of average velocity.

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Average Velocity

CONCEPT Formula

\bar{v} = \frac{\Delta x}{\Delta t} = \frac{x_2 - x_1}{t_2 - t_1}

State the mathematical formula for Average Velocity in one-dimensional motion.

Variables & Units:

$\bar{v}$ = Average Velocity (m/s)
$\Delta x$ = Displacement (m)
$\Delta t$ = Time interval (s)

Conditions/Notes:

It is a vector quantity.
If an object returns to its starting point, displacement is zero, and thus average velocity is zero.

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Average Speed: Equal Distances

APPLICATION Formula

v_{avg} = \frac{2 v_1 v_2}{v_1 + v_2}

State the shortcut formula for average speed when a particle travels TWO EQUAL DISTANCES with different speeds $v_1$ and $v_2$ .

Derivation

Total Distance = $s + s = 2s$
Total Time = $t_1 + t_2 = \frac{s}{v_1} + \frac{s}{v_2}$
$v_{avg} = \frac{2s}{s(\frac{1}{v_1} + \frac{1}{v_2})} = \frac{2 v_1 v_2}{v_1 + v_2}$

Variables & Units:

$v_{avg}$ = Average speed (m/s)
$v_1, v_2$ = Individual speeds (m/s)

Conditions/Notes:

This formula represents the Harmonic Mean of the two speeds.
Strictly only applies if the distances $s_1$ and $s_2$ are exactly equal.

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Average Speed vs. Magnitude of Average Velocity

Under what strict boundary condition is the magnitude of average velocity exactly equal to the average speed?

CONCEPT

The magnitude of average velocity is equal to the average speed ONLY if the particle moves along a straight line without turning back (i.e., maintaining a constant, unidirectional motion).

Note: In all other cases (e.g., reversing direction or taking a curved path), the total path length is strictly greater than the magnitude of displacement, meaning Average Speed > Magnitude of Average Velocity.

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Average Speed: Equal Time Intervals

APPLICATION Formula

v_{avg} = \frac{v_1 + v_2}{2}

State the shortcut formula for average speed when a particle travels with different speeds $v_1$ and $v_2$ for TWO EQUAL TIME INTERVALS.

Derivation

Total Distance = $v_1 t + v_2 t$
Total Time = $t + t = 2t$
$v_{avg} = \frac{t(v_1 + v_2)}{2t} = \frac{v_1 + v_2}{2}$

Variables & Units:

$v_{avg}$ = Average speed (m/s)
$v_1, v_2$ = Individual speeds (m/s)

Conditions/Notes:

This formula represents the Arithmetic Mean of the two speeds.
Strictly only applies if $t_1 = t_2$ .

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Instantaneous Velocity

CONCEPT Formula

v = \lim_{\Delta t \to 0} \frac{\Delta x}{\Delta t} = \frac{dx}{dt}

State the differential calculus formula for Instantaneous Velocity.

Variables & Units:

v = Instantaneous Velocity (m/s)
$\frac{dx}{dt}$ = First derivative of position with respect to time.

Conditions/Notes:

This represents the velocity of an object at a specific instant of time.
Graphically, it is the slope of the tangent to the position-time (x-t) graph at that instant.

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Average Acceleration

CONCEPT Formula

\bar{a} = \frac{\Delta v}{\Delta t} = \frac{v_2 - v_1}{t_2 - t_1}

State the formula for Average Acceleration.

Variables & Units:

$\bar{a}$ = Average Acceleration (m/s²)
$\Delta v$ = Change in velocity (m/s)
$\Delta t$ = Time interval (s)

Conditions/Notes:

Acceleration can result from a change in speed, a change in direction, or both.

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Instantaneous Speed vs. Velocity

What is the relationship between instantaneous speed and instantaneous velocity?

DISTINCTION

Instantaneous speed is exactly equal to the magnitude of the instantaneous velocity.

Crucial Distinction: While Average Speed can be greater than the magnitude of Average Velocity (if the path is curved or object turns back), at an infinitesimally small instant ( $dt$ ), the path is always a straight line without turning. Hence, $|dv| _start]= ds$ .

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Instantaneous Acceleration

CONCEPT Formula

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

State the differential calculus formula for Instantaneous Acceleration.

Derivation

$a = \frac{dv}{dt}$
Multiply and divide by $dx$ : $a = \frac{dv}{dx} \cdot \frac{dx}{dt}$
Since $\frac{dx}{dt} = v$ , we get $a = v \frac{dv}{dx}$

Variables & Units:

a = Instantaneous Acceleration (m/s²)
$\frac{dv}{dt}$ = First derivative of velocity w.r.t time.
$\frac{d^2x}{dt^2}$ = Second derivative of position w.r.t time.

Conditions/Notes:

Another useful chain-rule format used frequently in JEE/NEET is: $a = v \frac{dv}{dx}$

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Zero Velocity vs. Zero Acceleration

Does a particle having zero velocity at a specific instant necessarily mean it has zero acceleration at that instant? Provide an example.

CASE_STUDY

No. A particle may be momentarily at rest but still possess a non-zero acceleration.

Example: When a ball is thrown vertically upwards, at its absolute uppermost highest point, its velocity becomes strictly zero ( $v=0$ ). However, the acceleration at that exact instant continues to be the acceleration due to gravity ( $a = -9.8 \text{ m/s}^2$ ).

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1st Equation of Motion

DATA Formula

v = u + at

State the first kinematic equation relating velocity and time.

Derivation

$a = \frac{dv}{dt} \rightarrow dv = a \cdot dt$
$\int_u^v dv = a \int_0^t dt$
$v - u = at \rightarrow v = u + at$

Variables & Units:

v = Final velocity (m/s)
u ( $v_0$ ) = Initial velocity (m/s)
a = Acceleration (m/s²)
t = Time (s)

Conditions/Notes:

CRITICAL: This equation is ONLY valid if the acceleration ( $a$ ) is perfectly constant/uniform.

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Sign of Acceleration vs. Speeding Up

Does a 'negative acceleration' always imply that a particle is slowing down (retarding)? Explain the relationship between the signs of velocity and acceleration.

DISTINCTION

No. A negative acceleration simply means the acceleration vector is directed along the negative axis.

Speeding Up: Occurs when velocity and acceleration have the same sign (both positive or both negative). Example: An object falling downwards (negative velocity) under gravity (negative acceleration) is speeding up.
Slowing Down (Retardation): Occurs when velocity and acceleration have opposite signs.

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2nd Equation of Motion

DATA Formula

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

State the second kinematic equation relating position and time.

Derivation

$v = \frac{ds}{dt} \rightarrow ds = v \cdot dt$
Substitute $v = u + at$ : $ds = (u + at) dt$
$\int_0^s ds = \int_0^t (u + at) dt$
$s = ut + \frac{at^2}{2}$

Variables & Units:

s ( $x - x_0$ ) = Displacement (m)
u ( $v_0$ ) = Initial velocity (m/s)
a = Constant Acceleration (m/s²)
t = Time (s)

Conditions/Notes:

Represents the area under a uniformly accelerating v-t graph (Area of rectangle + Area of triangle).

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Slopes of Motion Graphs

What physical quantities are given by the slopes of the Position-Time (x-t) and Velocity-Time (v-t) graphs?

CONCEPT

Slope of Position-Time (x-t) Graph: Gives the Velocity ( $v = \frac{dx}{dt}$ ).
Slope of Velocity-Time (v-t) Graph: Gives the Acceleration ( $a = \frac{dv}{dt}$ ).

Note: If the x-t graph is a curve, the slope of the tangent gives instantaneous velocity. If it's a secant line, it gives average velocity.

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3rd Equation of Motion

DATA Formula

v^2 = u^2 + 2as

State the third kinematic equation relating velocity and displacement (independent of time).

Derivation

$a = v \frac{dv}{ds} \rightarrow v \cdot dv = a \cdot ds$
$\int_u^v v \cdot dv = a \int_0^s ds$
$\left[\frac{v^2}{2}\right]_u^v = a[s]_0^s$
$\frac{v^2}{2} - \frac{u^2}{2} = as \rightarrow v^2 = u^2 + 2as$

Variables & Units:

v = Final velocity (m/s)
u ( $v_0$ ) = Initial velocity (m/s)
a = Constant Acceleration (m/s²)
s ( $x - x_0$ ) = Displacement (m)

Conditions/Notes:

Highly useful when the time variable ( $t$ ) is neither given nor required.

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Areas of Motion Graphs

What physical quantities are given by the areas under the Velocity-Time (v-t) and Acceleration-Time (a-t) graphs?

CONCEPT

Area under Velocity-Time (v-t) Graph: Gives the Displacement ( $x = \int v \, dt$ ).
Area under Acceleration-Time (a-t) Graph: Gives the Change in Velocity ( $\Delta v = \int a \, dt$ ).

Boundary Condition: The x-t graph area has no significant physical meaning in basic kinematics.

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Displacement in the nth Second

APPLICATION Formula

S_{n} = u + \frac{a}{2}(2n - 1)

State the formula for calculating the displacement of an accelerating body specifically during the $n^{th}$ second of its motion.

Derivation

$S_n = S(n) - S(n-1)$
Substitute 2nd Eq: $\left(un + \frac{1}{2}an^2\right) - \left(u(n-1) + \frac{1}{2}a(n-1)^2\right)$
Simplify to yield: $u + \frac{a}{2}(2n - 1)$

Variables & Units:

$S_n$ = Displacement in the $n^{th}$ second (m)
u = Initial velocity (m/s)
a = Constant Acceleration (m/s²)
n = The specific second (e.g., 5th second)

Conditions/Notes:

Despite evaluating to meters, dimensionally, $S_n$ represents velocity because it's distance per 1 second.

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Determining Acceleration from x-t Graphs

How do you visually determine the sign of acceleration strictly from the shape or curvature of a Position-Time (x-t) graph?

CONCEPT

The sign of acceleration depends on the concavity of the curve:

Curves Upward (Concave Up): Represents Positive acceleration ( $a > 0$ ).
Curves Downward (Concave Down): Represents Negative acceleration ( $a < 0$ ).
Straight Line: Represents Zero acceleration ( $a = 0$ , uniform velocity).

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Average Velocity under Constant Acceleration

If a particle moves with strictly uniform acceleration, how can you rapidly find its average velocity without using total distance over total time?

APPLICATION

For uniform acceleration, the average velocity is simply the arithmetic mean of the initial and final velocities.

$\bar{v} = \frac{u + v}{2}$

Application: You can then quickly find displacement using: $s = \left(\frac{u + v}{2}\right) \cdot t$

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Impossible Kinematic Graphs

What are the three strict rules that determine if a kinematic graph (like position-time or speed-time) is physically impossible for 1D motion?

CONCEPT

Multiple Values: A particle cannot have two positions or two velocities at the exact same instant of time (i.e., the graph cannot have vertical lines, overlap itself, or form a closed loop).
Negative Scalars: Speed and Total Path Length (Distance) are scalar quantities that can never be negative.
Decreasing Distance: Total path length (distance) can never decrease with time; it must always be $\ge 0$ and continuously increasing or constant.

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Free Fall Equations (Dropped from Rest)

APPLICATION Formula

v = gt, \quad h = \frac{1}{2}gt^2, \quad v = \sqrt{2gh}

State the three modified kinematic equations for a body dropped from rest ( $u=0$ ) near the earth's surface.

Variables & Units:

h = Distance dropped (m)
v = Final velocity before hitting ground (m/s)
t = Time of fall (s)
g = $9.8 \text{ m/s}^2$

Conditions/Notes:

Here, downward is taken as positive for convenience since the object is strictly falling.
These equations are invalid if the object is thrown downwards with an initial velocity.

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Stopping Distance

APPLICATION Formula

d_s = \frac{-v_0^2}{2a}

State the formula for the stopping distance of a vehicle moving with initial velocity $v_0$ and uniform braking deceleration $-a$ .

Derivation

Use $v^2 = u^2 + 2as$
Set final velocity $v = 0$ .
$0 = v_0^2 + 2as \rightarrow s = \frac{-v_0^2}{2a}$

Variables & Units:

$d_s$ = Stopping distance (m)
$v_0$ = Initial velocity (m/s)
a = Deceleration (negative value) (m/s²)

Conditions/Notes:

Stopping distance is strictly proportional to the square of the initial velocity.
If you double your speed, your braking distance becomes four times longer.

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Relative Velocity in 1D

CONCEPT Formula

v_{AB} = v_A - v_B

State the formula for the relative velocity of object A with respect to object B.

Variables & Units:

$v_{AB}$ = Velocity of A w.r.t B (m/s)
$v_A$ = Velocity of A w.r.t ground (m/s)
$v_B$ = Velocity of B w.r.t ground (m/s)

Conditions/Notes:

Same Direction: Speeds subtract (e.g., chasing a car).
Opposite Direction: Speeds add up mathematically because $V_B$ is negative: $v_{AB} = v_A - (-v_B) = v_A + v_B$ (e.g., two trains crossing head-on).

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Condition for Free Fall

What defines a 'Free Fall' motion, and what is the standard sign convention applied to its acceleration?

CONCEPT

An object is in free fall if it moves solely under the influence of gravity, with air resistance neglected. If we take the upward direction as the positive y-axis, gravity always acts downwards. Therefore, acceleration is strictly constant and negative:
** $a = -g = -9.8 \text{ m/s}^2$ .

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Graphical Shapes of Free Fall

What are the specific geometric shapes of the acceleration-time, velocity-time, and position-time graphs for an object dropped in free fall? (Assuming downward is the negative y-direction).

FACT

Acceleration-Time (a-t) Graph: A straight horizontal line below the time axis (at $a = -g = -9.8 \text{ m/s}^2$ ).
Velocity-Time (v-t) Graph: A straight line passing through the origin with a constant negative slope.
Position-Time (x-t) Graph: A downward-opening parabola (quadratic curve) passing through the origin.

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Galileo's Law of Odd Numbers

State Galileo's Law of Odd Numbers regarding freely falling bodies.

FACT

The distances traversed during equal intervals of time by a body falling from rest, stand to one another in the same ratio as the odd numbers beginning with unity.

Ratio:** $1: 3: 5: 7: 9: 11...$

Proof hint: Using $s = \frac{1}{2}gt^2$ , distances from start are $y_0, 4y_0, 9y_0$ . Distances in successive intervals are $y_0, (4y_0-y_0), (9y_0-4y_0) \rightarrow 1:3:5$ .

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Kinematic Equation with Initial Position

APPLICATION Formula

x = x_0 + v_0 t + \frac{1}{2}at^2

State the modified second kinematic equation for the final position $x$ of a uniformly accelerating particle, explicitly including its initial position $x_0$ at $t=0$ .

Derivation

We know displacement $s = x - x_0$
From standard equation: $s = v_0 t + \frac{1}{2}at^2$
Substitute $s$ : $x - x_0 = v_0 t + \frac{1}{2}at^2 \rightarrow x = x_0 + v_0 t + \frac{1}{2}at^2$

Variables & Units:

x = Final position coordinate (m)
x_0 = Initial position coordinate at $t=0$ (m)
v_0 = Initial velocity (m/s)
a = Constant acceleration (m/s²)
t = Time (s)

Conditions/Notes:

This general form is highly useful in relative motion problems or when the origin is not the starting point of the object.

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Reaction Time

What is 'Reaction Time' in real-world kinematics, and how does it impact total stopping distance?

DEFINITION

Definition: The time a person takes to observe a hazard, think, and take action (slam the brakes).
Impact: During the reaction time ( $t_r$ ), the vehicle continues to move at a constant initial velocity ( $v_0$ ).
Total Stopping Distance = Distance traveled during reaction time + Braking Distance = $(v_0 \times t_r) + d_s$ .

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Velocity at the Midpoint

APPLICATION Formula

v_{mid} = \sqrt{\frac{u^2 + v^2}{2}}

State the shortcut mathematical formula for the velocity of a particle at the exact geometric midpoint of its path, given it accelerates uniformly from initial velocity $u$ to final velocity $v$ .

Derivation

Let total distance be $2s$ . Midpoint is at distance $s$ .
Total path: $v^2 = u^2 + 2a(2s) \rightarrow 4as = v^2 - u^2$
Midpoint path: $v_{mid}^2 = u^2 + 2as$
Substitute $2as = \frac{v^2 - u^2}{2}$ into the midpoint equation.
$v_{mid}^2 = u^2 + \frac{v^2 - u^2}{2} = \frac{u^2 + v^2}{2}$

Variables & Units:

$v_{mid}$ = Velocity at the exact midpoint of the distance (m/s)
u = Initial velocity (m/s)
v = Final velocity (m/s)

Conditions/Notes:

Valid strictly for uniform acceleration.
Note that this is the Root Mean Square (RMS) of the initial and final velocities, NOT the arithmetic mean.

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Symmetry of Relative Velocity

How does the relative velocity of object A with respect to B mathematically relate to the relative velocity of object B with respect to A?

FACT

They are exactly equal in magnitude but opposite in direction.

Mathematically: $v_{AB} = -v_{BA}$

Explanation: If you see a car pulling ahead of you at +10 m/s, the driver of that car sees you falling behind at -10 m/s.

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