44 Kinematics Conceptual Questions & Answers

Displacement (s) and distance (d) may look similar in many motions, but they represent fundamentally different ideas. Distance (d) is the total path length traveled, while displacement (s) is the shortest straight-line distance between initial and final positions, including direction.

This means displacement depends only on starting and ending points, not the path taken.

For example, moving in a circle and returning to the start gives d > 0 but s = 0. Mathematically, displacement is a vector, while distance is scalar, which creates this key difference.

Yes, displacement (s) can be zero even when distance (d) is not zero. This happens when an object returns to its initial position after traveling some path. In such a case, the total path length (distance) is positive, but the net change in position is zero.

For example, if a person walks 5 m forward and then 5 m backward, total distance (d) = 10 m, but displacement (s) = 0.

This highlights that displacement depends only on initial and final positions, not the journey taken.

Velocity (v) is a vector because it includes both magnitude and direction, while speed only measures how fast an object moves, without direction.

Mathematically, velocity is defined as displacement per unit time, and displacement itself is a vector.

\[\boxed{v=\dfrac{s}{t}}\]

Speed, on the other hand, is based on distance, which has no direction. Therefore, speed cannot indicate whether motion is forward or backward.

\[v_{s}=\dfrac{d}{t}\]

This is why velocity can be positive or negative, but speed is always positive.

Yes, it is absolutely possible for velocity (v) = 0 while acceleration (a) is not zero. A classic example is at the highest point of vertical motion. At that instant, the object momentarily stops, so velocity becomes zero.

However, acceleration due to gravity (a = g) still acts downward. This means the object is about to change direction.

So, zero velocity does not mean no acceleration—it simply means the direction of motion is changing at that instant.

Velocity (v) and displacement (s) are both frame-dependent quantities, meaning their values change depending on the observer’s reference frame. For example, if you are sitting in a moving train, a stationary object outside appears to move backward relative to you.

Mathematically, velocities add or subtract depending on relative motion:
\[ \boxed{ \;v_{rel} = v_{obj}\, – \,v_{obs}\;} \]

However, acceleration (a) remains the same in all inertial frames. This shows that motion is not absolute but depends on the observer’s frame of reference.

Average velocity (v_avg) is defined over a finite time interval, while instantaneous velocity (v) is defined at a specific instant of time. Mathematically, average velocity is given by
\[ v_{avg} = \dfrac{\Delta s}{\Delta t} \],
where Δs is total displacement over time Δt.

Instantaneous velocity is the velocity at a particular moment, obtained by taking the limit as Δt → 0:
\[ v = \lim_{\Delta t \to 0} \dfrac{\Delta s}{\Delta t} \].
So, average velocity gives overall motion, while instantaneous velocity tells exact motion at a point.

Average velocity (v_avg) depends on displacement (s) because velocity is a vector quantity, and displacement includes direction. Distance (d) does not include direction, so it cannot describe velocity properly.

Mathematically,
\[ v_{avg} = \dfrac{s}{t} \]
where s is net displacement. If distance were used, direction would be lost, and we would only get speed, not velocity. This is why average velocity reflects net change in position, not total path traveled.

Instantaneous velocity (v) is defined as the velocity when the time interval becomes extremely small. It is the limiting value of average velocity as Δt approaches zero.

Mathematically,
\[\boxed{ v = \lim_{\Delta t \to 0} \dfrac{\Delta s}{\Delta t} = \dfrac{ds}{dt}} \]

This shows that instantaneous velocity is the rate of change of displacement with respect to time. It gives the exact velocity at any specific moment rather than over a duration.

When the slope of the position-time (s–t) graph becomes zero, the instantaneous velocity (v) also becomes zero. This is because velocity is mathematically equal to the slope of the s–t graph.

If slope = 0, then
\[ v = \dfrac{ds}{dt} = 0 \]

Physically, this means the object is momentarily at rest. However, this does not necessarily mean acceleration (a) is zero. For example, at the highest point of motion, velocity is zero but acceleration still exists.

The slope of a graph represents the rate of change of the vertical axis with respect to the horizontal axis. In an s–t graph, displacement (s) is plotted against time (t), so the slope becomes:

\[ \text{slope} = \dfrac{\Delta s}{\Delta t} \]

As the interval becomes very small, this becomes:
\[ v = \dfrac{ds}{dt} \]

Thus, the slope directly gives velocity. This is why a straight line in s–t graph indicates constant velocity, and a curved graph indicates changing velocity. Slope is the mathematical representation of how fast position changes.

Uniform motion means an object moves with constant velocity (v), i.e., it covers equal displacement (s) in equal intervals of time (t). Since velocity is not changing, there is no change in speed or direction.

Acceleration (a) is defined as the rate of change of velocity:

\[ a = \dfrac{dv}{dt} \]

In uniform motion, velocity remains constant, so \( \dfrac{dv}{dt} = 0 \). Therefore, acceleration becomes zero. This shows that uniform motion is a zero-acceleration motion.

Acceleration (a) represents how velocity (v) changes with time. If acceleration is zero, it means there is no change in velocity at any instant.

Mathematically,
\[ a = \dfrac{dv}{dt} = 0 \Rightarrow dv = 0 \]

\[ dv = 0 \Rightarrow v=\, constant\]

This implies velocity does not change with time, so it remains constant. Hence, both magnitude and direction of velocity stay the same, resulting in uniform motion.

Acceleration is defined as rate of change of velocity. For constant acceleration, we directly integrate:

\[
a = \dfrac{dv}{dt}
\]

\[
a \, dt = dv
\]

\[
\int_{0}^{t} a \, dt = \int_{u}^{v} dv
\]

\[
at = v – u
\]

\[
\boxed{v = u + at}
\]

This shows velocity increases linearly with time under constant acceleration.

Since velocity is rate of change of displacement, we integrate velocity to get displacement:

\[
v = \dfrac{ds}{dt} = u + at
\]

\[
ds = (u + at)\, dt
\]

\[
\int_{0}^{s} ds = \int_{0}^{t} (u + at)\, dt
\]

\[
s = \int_{0}^{t} u\,dt + \int_{0}^{t} at\,dt
\]

\[
\boxed{s = ut + \dfrac{1}{2}at^2}
\]

This shows displacement grows non-linearly due to acceleration.

We eliminate time by combining two equations of motion,

\[
v = u + at \Rightarrow t = \dfrac{v \,-\, u}{a}
\]

\[
s = ut + \dfrac{1}{2}at^2
\]

\[
s = u\left(\dfrac{v – u}{a}\right) + \dfrac{1}{2}a\left(\dfrac{v – u}{a}\right)^2
\]

\[
s = \dfrac{u(v – u)}{a} + \dfrac{(v – u)^2}{2a}
\]

\[
s = \dfrac{2u(v – u) + (v – u)^2}{2a}
\]

\[
s = \dfrac{v^2 – u^2}{2a}
\]

\[
\boxed{v^2 = u^2 + 2as}
\]

This equation directly connects velocity and displacement without time.

Displacement is equal to the area under the velocity-time graph. For uniform acceleration, the graph is a straight line.

\[
s = \text{Area under v–t graph}
\]

\[
s = \text{Rectangle} + \text{Triangle}
\]

\[
s = ut + \dfrac{1}{2}(v – u)t
\]

\[
v – u = at
\]

\[
\boxed{s = ut + \dfrac{1}{2}at^2}
\]

This gives a geometric interpretation of displacement.

Velocity (v) is defined as displacement per unit time
\[ v = \dfrac{s}{t} \]

Rearranging this equation algebraically, we get
\[ s = vt \]

This derivation assumes velocity is constant (uniform motion). So, displacement is directly proportional to time. This means if time increases, displacement increases linearly.

In uniform motion, displacement (s) is given by:
\[ s = vt \]

Since velocity (v) is constant, displacement is directly proportional to time:
\[ s \propto t \]

If time doubles (t → 2t), then
\[ s’ = v(2t) = 2vt = 2s \]

So displacement also doubles. This shows a linear relationship between displacement and time in uniform motion.

The three standard equations of motion are:

\[
v = u + at
\]

\[
s = ut + \dfrac{1}{2}at^2
\]

\[
v^2 = u^2 + 2as
\]

These equations are valid only when acceleration (a) is constant. This means the rate of change of velocity remains uniform throughout the motion. If acceleration varies with time, these equations cannot be directly used.

These equations connect displacement (s), velocity (v), time (t), and acceleration (a), helping us analyze motion without needing all variables at once.

Acceleration (a) is defined as the rate of change of velocity:

\[
a = \dfrac{v – u}{t}
\]

Rearranging the equation:

\[
at = v – u
\]

Adding u on both sides:

\[
v = u + at
\]

This shows that final velocity (v) increases linearly with time (t) when acceleration is constant. It directly comes from the definition of acceleration.

Constant acceleration (a) means velocity changes by equal amounts in equal time intervals. From the equation:

\[
v = u + at
\]

we see that velocity (v) depends linearly on time (t). This is because a and u are constants, so v increases in a straight-line manner with time.

Graphically, this produces a straight line in the velocity-time (v–t) graph. Hence, constant acceleration always results in linear velocity change.

Negative acceleration means acceleration acts in the opposite direction to velocity. Using:

\[
v = u + at
\]

if a is negative, then the term at reduces the value of velocity over time. This is called retardation or deceleration.

If the object keeps slowing down, velocity may become zero and then change direction. So, negative acceleration causes velocity to decrease linearly with time.

Using the equation:

\[
s = ut + \dfrac{1}{2}at^2
\]

If u = 0, then:

\[
s = \dfrac{1}{2}at^2
\]

This shows that displacement (s) is proportional to the square of time:

\[
\boxed{s \propto t^2}
\]

So, if time doubles, displacement becomes four times. This means motion is non-linear and accelerates over time.

From the equation:

\[
s = ut + \dfrac{1}{2}at^2
\]

If acceleration becomes 2a, then:

\[
s’ = ut + \dfrac{1}{2}(2a)t^2 = ut + at^2
\]

Comparing with original displacement:

\[
\boxed{s’- s = (ut + at^2) – (ut + \dfrac{1}{2}at^2) = \frac{1}{2}at}
\]

The acceleration-dependent part doubles, but the ut term remains same. So displacement does not fully double unless u = 0.

Thus, displacement increases more rapidly with higher acceleration, but the exact change depends on initial velocity.

The slope of a velocity-time (v–t) graph represents acceleration (a). This is because slope measures how velocity changes with time.

Mathematically, slope is:
\[
\text{slope} = \dfrac{\Delta v}{\Delta t}
\]

From definition of acceleration:
\[
a = \dfrac{\Delta v}{\Delta t}
\]

So, slope of v–t graph directly gives acceleration. If slope is constant, acceleration is constant. If slope increases, acceleration increases.

Displacement (s) can be calculated from the area under the velocity-time (v–t) graph.

Mathematically,
\[
s = \int v \, dt
\]

For some cases:
1. Rectangle area → constant velocity → \( s = vt \)
2. Triangle area → uniformly accelerated motion → \( s = \dfrac{1}{2}vt \)

So, the total area between the graph and time axis gives displacement. This is because velocity is rate of change of displacement.

The area under the acceleration-time (a–t) graph represents the change in velocity.

Mathematically:
\[
\text{Area} = \int a \, dt = \Delta v
\]

So, if we know initial velocity (u), then:
\[
v = u + \int a \, dt
\]

This means acceleration tells how velocity changes, and its area gives total change in velocity over time.

In a velocity-time (v–t) graph, slope is defined as:
\[
\text{slope} = \dfrac{\Delta v}{\Delta t}
\]

From the definition of acceleration:
\[
a = \dfrac{dv}{dt}
\]

So slope is simply the derivative of velocity with respect to time. That is why slope of v–t graph equals acceleration.

This shows that graphs are direct visual representations of mathematical definitions.

When the v–t graph is curved, it means acceleration (a) is not constant. Since slope of the graph gives acceleration, a changing slope implies changing acceleration.

Mathematically,
\[
a = \dfrac{dv}{dt}
\]

If dv/dt changes with time, acceleration varies. This type of motion is called non-uniform acceleration.

Acceleration due to gravity (g) is the acceleration with which objects fall toward the Earth due to gravitational force. Its average value near Earth’s surface is approximately 9.8 m/s² downward.

It is considered constant because the height (h) of objects in most problems is very small compared to Earth’s radius. From Newton’s law:

\[
g = \dfrac{GM}{R^2}
\]

Since R (distance from Earth’s center) does not change significantly for small heights, g remains nearly constant. Thus, uniform acceleration assumption works near Earth’s surface.

At maximum height, the object momentarily stops before coming down, so velocity (v) becomes zero. However, gravity is still acting downward.

Acceleration depends on force, not velocity. Since gravitational force is always present.

\[
a = g
\]

So even when velocity is zero, acceleration is not zero. This means the object is about to change direction. Zero velocity does not imply zero acceleration.

Time of flight (T) for a vertically projected object is given by:

\[
v = u – gt,\quad v=0 \Rightarrow t_{\text{up}} = \dfrac{u}{g}
\]

\[
T = 2t_{\text{up}}
\]

\[
T = \dfrac{2u}{g}
\]

This shows:
\[
\boxed{T \propto \dfrac{1}{g}}
\]

So, if gravity (g) increases, time of flight decreases. Physically, stronger gravity pulls the object down faster, reducing total time in air. Thus, higher g → shorter flight time.

For an object falling from height (h), final velocity (v) is given by:
\[
v^2 = 2gh
\]

\[
v = \sqrt{2gh}
\]

This shows that velocity increases with square root of height,
\[
v \propto \sqrt{h}
\]

So, if height increases 4 times, velocity doubles. This relationship comes from energy or kinematics and shows non-linear dependence on height.

According to Newton’s second law:
\[
F = ma
\]

Gravitational force is:
\[
F = \dfrac{GMm}{R^2} = ma
\]

\[
a = \dfrac{GMm}{R^2m}
\]

\[
a = \dfrac{GM}{R^2} = g
\]

This shows acceleration does not depend on mass of the object. So all objects fall with same acceleration if air resistance is ignored. This is a fundamental and surprising result of physics.

Displacement (s) in uniformly accelerated motion is given by:
\[
s = ut + \dfrac{1}{2}at^2
\]

If initial velocity becomes 2u:
\[
s’ = (2u)t + \dfrac{1}{2}at^2 = 2ut + \dfrac{1}{2}at^2
\]

Comparing with original s, only the ut term doubles. So displacement increases, but does not exactly double unless acceleration term is negligible. Thus, s depends linearly on u.

Time of flight is given by:
\[
T = \dfrac{2u}{g}
\]

This shows:
\[
T \propto \dfrac{1}{g}
\]

So, if gravity (g) increases, time of flight decreases. Physically, stronger gravity pulls the object down faster, reducing total time in air. Hence, T is inversely proportional to g.

Velocity is given by:
\[
v = u + at
\]

If acceleration becomes 3a:
\[
v’ = u + 3at
\]

This means the rate of increase of velocity becomes three times faster. The slope of v–t graph increases, but initial velocity remains same. So, velocity grows more rapidly with time.

Using:
\[
s = ut + \dfrac{1}{2}at^2
\]

If time becomes 2t:
\[
s’ = u(2t) + \dfrac{1}{2}a(2t)^2 = 2ut + 2at^2
\]

Original:
\[
s = ut + \dfrac{1}{2}at^2
\]

So displacement increases more than double due to the \( t^2 \) term. Hence, s has a quadratic dependence on time.

The relation between velocity and displacement is:
\[
v^2 = u^2 + 2as
\]

This shows:
\[
v = \sqrt{u^2 + 2as}
\]

So velocity depends on square root of displacement. This means velocity increases non-linearly with displacement. This equation removes time and directly connects v and s.

Acceleration vector is the time derivative of velocity:
\[
\vec{a} = \dfrac{d\vec{v}}{dt}
\]

So,
\[
\vec{a} = \dfrac{d}{dt}(3t^2)\hat{i} + \dfrac{d}{dt}(4t)\hat{j}
\]

\[
\boxed{\;\vec{a} = (6t)\hat{i} + (4)\hat{j}\;}
\]

Thus, acceleration is obtained by differentiating each component of velocity separately.

Velocity is obtained by integrating acceleration:
\[
\vec{v} = \int \vec{a}\, dt
\]

\[
\vec{v} = \int 2\,dt \hat{i} + \int 3t\,dt \hat{j}
\]

\[
\boxed{\;\vec{v} = (2t)\hat{i} + \left(\dfrac{3t^2}{2}\right)\hat{j}\;}
\]

Since initial velocity is zero, no constant term appears. So, velocity is the integral of acceleration.

Displacement is obtained by integrating velocity:
\[
\vec{r} = \int \vec{v}\, dt
\]

\[
\vec{r} = \int 2t\,dt \hat{i} + \int 5t^2\,dt \hat{j}
\]

\[
\boxed{\;\vec{r} = (t^2)\hat{i} + \left(\dfrac{5t^3}{3}\right)\hat{j}\;}
\]

Assuming initial position is zero. Thus, position vector is the integral of velocity vector.

Velocity is the derivative of position:
\[
\vec{v} = \dfrac{d\vec{r}}{dt}
\]

\[
\vec{v} = (3t^2)\hat{i} + (4t)\hat{j}
\]

Acceleration is derivative of velocity:
\[
\vec{a} = \dfrac{d\vec{v}}{dt}
\]

\[
\boxed{\;\vec{a} = (6t)\hat{i} + (4)\hat{j}\;}
\]

So, differentiation gives velocity and acceleration step-by-step.

First find velocity by integrating acceleration:
\[
\vec{v} = \int \vec{a}\, dt = \int 4t\,dt \hat{i} + \int 6\,dt \hat{j}
\]

\[
\vec{v} = (2t^2)\hat{i} + (6t)\hat{j}
\]

Now integrate velocity to get displacement:
\[
\vec{r} = \int \vec{v}\, dt
\]

\[
\vec{r} = \int 2t^2\,dt \hat{i} + \int 6t\,dt \hat{j}
\]

\[
\boxed{\;\vec{r} = \left(\dfrac{2t^3}{3}\right)\hat{i} + (3t^2)\hat{j}\;}
\]

Thus, double integration of acceleration gives displacement.