Motion — NCERT Solutions, AI Tutor & Practice

Distance is the total length of the path travelled by an object, regardless of direction. It is a scalar quantity (has only magnitude).

Displacement is the shortest straight-line distance between the initial and final positions of an object, measured in a specified direction. It is a vector quantity (has both magnitude and direction).

Key differences:

| Distance | Displacement | |----------|--------------| | Scalar quantity | Vector quantity | | Always positive | Can be positive, negative or zero | | Depends on path taken | Depends only on end points | | $\geq$ magnitude of displacement | $\leq$ distance |

For example, if a person walks 4 m East and then 3 m North, the distance = $4 + 3 = 7\ \text{m}$, but the displacement = $\sqrt{4^2 + 3^2} = 5\ \text{m}$ towards North-East.

Given:

• First part: distance $= 16\ \text{m}$, time $= 4\ \text{s}$
• Second part: distance $= 16\ \text{m}$, time $= 2\ \text{s}$

Formula: $\text{Average speed} = \frac{\text{Total distance}}{\text{Total time}}$

Substitution: $\text{Total distance} = 16 + 16 = 32\ \text{m}$ $\text{Total time} = 4 + 2 = 6\ \text{s}$ $\text{Average speed} = \frac{32}{6} = 5.33\ \text{m/s}$

Answer: The average speed of the object is approximately $5.33\ \text{m/s}$.

Uniform motion: An object is said to be in uniform motion if it covers equal distances in equal intervals of time, however small these intervals may be.

Example: A car moving at a steady $60\ \text{km/h}$ on a straight highway.

Non-uniform motion: An object is said to be in non-uniform motion if it covers unequal distances in equal intervals of time.

Example: A bus moving through crowded city traffic, speeding up and slowing down.

Note: In uniform motion the speed remains constant, so acceleration is zero. In non-uniform motion the speed changes, so the object is accelerated.

Given:

• Initial velocity $u = 0\ \text{m/s}$ (starts from rest)
• Final velocity $v = 20\ \text{m/s}$
• Time $t = 10\ \text{s}$

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

Substitution: $a = \frac{20 - 0}{10} = \frac{20}{10} = 2\ \text{m/s}^2$

Answer: The acceleration of the bus is $2\ \text{m/s}^2$.

Given:

• Initial velocity $u = 72\ \text{km/h}$
• Final velocity $v = 0\ \text{m/s}$
• Time $t = 20\ \text{s}$

Convert to m/s: $u = 72 \times \frac{5}{18} = 20\ \text{m/s}$

Acceleration: $a = \frac{v - u}{t} = \frac{0 - 20}{20} = -1\ \text{m/s}^2$

The negative sign indicates deceleration (retardation).

Distance travelled (using $v^2 = u^2 + 2as$): $0 = (20)^2 + 2(-1)s$ $0 = 400 - 2s$ $s = \frac{400}{2} = 200\ \text{m}$

Answer: Acceleration $= -1\ \text{m/s}^2$ and distance travelled $= 200\ \text{m}$.

For an object moving with uniform acceleration $a$, having initial velocity $u$, final velocity $v$, displacement $s$ after time $t$:

First equation (velocity–time): $v = u + at$ (relates velocity, acceleration and time — no displacement)

Second equation (position–time): $s = ut + \frac{1}{2}at^2$ (relates displacement, time and acceleration — no final velocity)

Third equation (velocity–position): $v^2 = u^2 + 2as$ (relates velocities, acceleration and displacement — no time)

Where:

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

Given:

• Initial velocity $u = 49\ \text{m/s}$ (upward)
• At maximum height, final velocity $v = 0\ \text{m/s}$
• Acceleration $a = -g = -9.8\ \text{m/s}^2$ (gravity acts downward, opposing motion)

Maximum height (using $v^2 = u^2 + 2as$): $0 = (49)^2 + 2(-9.8)s$ $0 = 2401 - 19.6\,s$ $s = \frac{2401}{19.6} = 122.5\ \text{m}$

Time to reach maximum height (using $v = u + at$): $0 = 49 + (-9.8)t$ $t = \frac{49}{9.8} = 5\ \text{s}$

Answer: Maximum height $= 122.5\ \text{m}$ and time taken $= 5\ \text{s}$.

Slope of a distance–time graph: The slope equals $\dfrac{\text{change in distance}}{\text{change in time}}$, which is the speed of the object.

$\text{Slope} = \frac{s_2 - s_1}{t_2 - t_1} = \text{speed}$

Graph for uniform motion: For uniform motion, the speed is constant, so the slope is constant. This gives a straight line (with constant slope) on the distance–time graph.

• A steeper straight line means a higher speed.
• A line parallel to the time axis (horizontal) means the object is at rest (zero speed).
• A curved line indicates non-uniform motion (changing speed).

The area under a velocity–time graph gives the displacement (distance travelled) of the object.

This is because: $\text{velocity} \times \text{time} = \frac{\text{distance}}{\text{time}} \times \text{time} = \text{distance}$

For a uniformly accelerated body starting from initial velocity $u$ and reaching velocity $v$ in time $t$, the velocity–time graph is a straight (sloping) line. The area under it is a trapezium:

$\text{Distance} = \text{Area of trapezium} = \frac{1}{2}(u + v)\,t$

This can be split into:

• Area of rectangle $= u \times t$
• Area of triangle $= \frac{1}{2} \times t \times (v - u) = \frac{1}{2}at^2$

Adding these gives $s = ut + \frac{1}{2}at^2$, which is the second equation of motion.

Uniform circular motion: When an object moves along a circular path with constant speed, its motion is called uniform circular motion.

Example: A stone tied to a thread and whirled in a circle, or the tip of a clock's second hand.

Is it accelerated? — Yes.

Justification: Velocity is a vector, having both magnitude and direction. In uniform circular motion, the magnitude of velocity (speed) stays constant, but its direction changes continuously as the object moves around the circle. Since velocity changes (due to changing direction), there is acceleration. Therefore, uniform circular motion is an example of accelerated motion.

Speed in circular motion: $v = \frac{2\pi r}{t}$ where $r$ is the radius and $t$ is the time for one revolution.

Given:

• Initial velocity $u = 18\ \text{km/h}$
• Final velocity $v = 36\ \text{km/h}$
• Time $t = 5\ \text{s}$

Convert to m/s: $u = 18 \times \frac{5}{18} = 5\ \text{m/s}$ $v = 36 \times \frac{5}{18} = 10\ \text{m/s}$

(a) Acceleration: $a = \frac{v - u}{t} = \frac{10 - 5}{5} = 1\ \text{m/s}^2$

(b) Distance covered (using $s = ut + \frac{1}{2}at^2$): $s = (5)(5) + \frac{1}{2}(1)(5)^2$ $s = 25 + \frac{1}{2}(25) = 25 + 12.5 = 37.5\ \text{m}$

Answer: (a) Acceleration $= 1\ \text{m/s}^2$; (b) Distance $= 37.5\ \text{m}$.

Speed: The rate at which an object covers distance. It is a scalar quantity (magnitude only). $\text{Speed} = \frac{\text{distance}}{\text{time}}$

Velocity: The rate at which an object changes its position, i.e., displacement per unit time in a specified direction. It is a vector quantity (magnitude and direction). $\text{Velocity} = \frac{\text{displacement}}{\text{time}}$

| Speed | Velocity | |-------|----------| | Scalar | Vector | | No direction | Has direction | | Always positive | Can be positive, negative or zero |

Can speed be constant while velocity changes? — Yes.

In uniform circular motion, the speed (magnitude) stays constant, but the direction of motion changes continuously. Since velocity depends on direction, the velocity keeps changing even though speed is constant.