Class 9 › Science › Chapter 4

Describing Motion Around Us

Position, distance and displacement, speed, velocity and acceleration, position-time and velocity-time graphs, kinematic equations, and uniform circular motion.

1. Motion and Describing Position

Scientists study complex motion first in idealised, simplified forms: linear (straight line), circular and oscillatory. This chapter covers linear motion and uniform circular motion.
Reference point: a fixed point from which position is described. Position = distance and direction from the reference point at an instant.
In motion: position with respect to the reference point changes with time. At rest: position does not change.
Origin (O): the reference point marked on a straight line with distances. Positions on one side (generally right) are positive, on the other side (left) negative; the sign gives the direction.
Instant vs interval: an instant is a single clock reading; a time interval is the duration between two instants (two clock readings).

2. Distance Travelled and Displacement

Distance travelled

Total path length

Has magnitude only (no direction). Never decreases while moving. SI unit: metre (m).

Displacement

Net change in position

Between two instants: final position − initial position. Needs magnitude and direction (from first position towards second). SI unit: metre (m).

Athlete example (Fig. 4.4)

O (0 m, $t=0$ s) → B (40 m, 4 s) → A (100 m, 10 s) → back to B (40 m, 16 s).

Distance travelled $= OA + AB = 100 + 60 = 160$ m.

Displacement $= 40$ m in the positive direction (net change from O to B).

Magnitude of displacement = straight-line distance between the two positions. It is less than or equal to the distance travelled, never greater.
Equal only if the object moves in a straight line without turning back (one direction).
Displacement can be zero while distance is not: a ball thrown up and caught back at the start; an athlete returning to the start. A ball rolling down a slope is not along a horizontal line, but its path can be straight (an inclined track).
Scalars have only a numerical value (distance, speed); vectors need magnitude and direction (displacement, velocity, acceleration). Vectors are studied in higher grades.
Fuel used depends on total distance travelled, not on displacement.

3. Average Speed and Average Velocity

Average speed (Eq. 4.1) $$\text{average speed} = \dfrac{\text{total distance travelled}}{\text{time interval}}$$ Average velocity (Eq. 4.2) $$v_{av} = \dfrac{\text{displacement}}{\text{time interval}} = \dfrac{s}{t}$$

	Average speed	Average velocity
Based on	Distance travelled	Displacement (change in position)
Type	Magnitude only	Magnitude + direction (same as displacement; sign in 1-D)
Can be zero?	Only if the object does not move	Yes, if displacement is zero
SI unit	metre per second (m s⁻¹ or m/s); also km h⁻¹

Unit conversion $$1\ \text{km h}^{-1} = \dfrac{1000\ \text{m}}{3600\ \text{s}} = \dfrac{5}{18}\ \text{m s}^{-1}$$ Eg: $36 \to 10$, $54 \to 15$, $72 \to 20$, $108 \to 30$ m s⁻¹.

Uniform motion in a straight line: equal distances in equal intervals of time (for all intervals); constant speed.
Non-uniform motion: unequal distances in equal intervals; speed increasing, decreasing or both.
Rate of change: ratio of change in a quantity to the corresponding change in time. Average velocity is the average rate of change of position with time.
For straight-line motion in one direction, average speed = magnitude of average velocity.
Instantaneous velocity: velocity at a particular instant (limit as the time interval becomes tiny). In this chapter 'velocity' means velocity at an instant. A speedometer reads (nearly) its magnitude.

Example 4.2: Sarang swims 25 m pool, to the other end and back in 50 s

Distance $= 50$ m, displacement $= 0$.

Average speed $= 50/50 = 1$ m s⁻¹; average velocity $= 0/50 = 0$ m s⁻¹.

Example 4.1 (Ganitakaumudi, 14th century CE): two postmen

Start 210 yojanas apart, walking towards each other at 9 and 5 yojanas/day. Combined rate $= 14$ yojanas/day.

Time to meet $= 210/14 = 15$ days (they cover 135 and 75 yojanas).

Speed as distance/time appears in Aryabhatiya (5th century CE).

Trip check: 200 km north in 3 h then 200 km south in 2 h: total distance 400 km in 5 h gives speed 80 km h⁻¹, displacement 0 gives velocity 0.

4. Average Acceleration

Average acceleration (Eq. 4.3) $$a = \dfrac{\text{change in velocity}}{\text{time interval}} = \dfrac{v - u}{t_2 - t_1}$$ $u$ = initial velocity (at $t_1$), $v$ = final velocity (at $t_2$). SI unit: m s⁻².

Speeding up (straight line): acceleration is in the direction of velocity (positive in the usual convention).
Slowing down: acceleration is opposite to velocity (negative sign), also called retardation.
Acceleration can come from a change in the magnitude of velocity, its direction, or both.
Acceleration depends on how quickly velocity changes, not on how fast the object moves. A bus moving at constant high velocity has zero acceleration.
Constant acceleration: velocity changes by equal amounts in equal intervals. Only constant acceleration is studied here.

Example 4.3: bus on a highway

(i) Accelerating: $u = 36$ km h⁻¹ $= 10$ m s⁻¹, $v = 54$ km h⁻¹ $= 15$ m s⁻¹, $t = 10$ s → $a = (15-10)/10 = 0.5$ m s⁻² (along velocity).

(ii) Braking: $u = 15$ m s⁻¹, $v = 0$, $t = 5$ s → $a = (0-15)/5 = -3$ m s⁻² (opposite to velocity).

Example 4.4 and acceleration due to gravity

A dropped object has $v = 0, 9.8, 19.6, 29.4, 39.2$ m s⁻¹ at $t = 0,1,2,3,4$ s. Acceleration in every 1 s interval $= 9.8$ m s⁻² (constant, in the direction of motion).

This is the acceleration due to Earth's gravity, $g = 9.8$ m s⁻².

Choice of origin and positive direction is yours (ground in Fig. 4.5, drop point and downward positive in Fig. 4.10), but do not change it while solving a problem.
Car specs: acceleration is often given as the time from 0 to 100 km h⁻¹; convert to m s⁻¹ and divide by that time.

5. Plotting and Position-Time Graphs

Plotting a graph (Activity 4.3)

1. Draw perpendicular axes meeting at the origin O: horizontal X-axis (time), vertical Y-axis (position).

2. Choose a suitable scale for each axis (eg: 5 divisions = 1 s; 5 divisions = 20 m).

3. Plot each (time, position) pair as a point; join them (straight line or smooth curve).

Table 4.3 (positions $0, 20, 40, \ldots, 120$ m at $0, 1, \ldots, 6$ s) gives a straight line. Table 4.4 (positions $0, 1, 4, 9, 16, 25, 36$ m at $0, 2, \ldots, 12$ s) gives a curve.
A graph is not a route map: it shows how position changes with time relative to the origin.
For motion in one direction with position zero at $t=0$, position-time = distance-time graph, and velocity-time = speed-time graph.

Shape of the position-time graph tells the nature of motion.

Position-time graph	Meaning
Straight line (sloping)	Equal displacement in equal time: constant velocity
Curve	Velocity not constant: accelerated motion
Straight line parallel to time axis	Position not changing: object at rest (eg: at 40 m from origin)
Steeper line	Greater velocity (compare two objects A and B)

Velocity from a position-time graph = slope $$v = \dfrac{s_2 - s_1}{t_2 - t_1} = \dfrac{BC}{CA}$$ Eg: points (2 s, 40 m) and (4 s, 80 m): $v = (80-40)/(4-2) = 20$ m s⁻¹.

Slope is the steepness of a line; it gives the rate of change of the Y quantity with respect to the X quantity.
Velocity at an instant on a curve can also be found geometrically (higher grades).

6. Velocity-Time Graphs

Velocity-time graph	Velocity	Acceleration
Straight line parallel to time axis	Constant	Zero (slope 0)
Straight line rising	Increasing by equal amounts	Constant, along velocity (positive slope)
Straight line falling	Decreasing by equal amounts	Constant, opposite to velocity (negative slope)

Two things a velocity-time graph gives $$a = \text{slope} = \dfrac{v-u}{t_2-t_1}$$ $$s = \text{area between the graph line and the time axis}$$

Worked values from the graphs

Slope: from 10 s to 20 s, velocity goes from 5 to 10 m s⁻¹: $a = 5/10 = 0.5$ m s⁻². For the falling graph the result is $-0.5$ m s⁻².

Area (constant velocity): rectangle: $20 \times 6 = 120$ m in 6 s.

Area (changing velocity), 10 s to 20 s: rectangle $5\times10 = 50$ m plus triangle $\tfrac12\times10\times5 = 25$ m, total $75$ m.

Tables 4.5 and 4.6: velocity rising from 0 to 15 m s⁻¹ in 30 s (a $= 0.5$) and falling from 15 to 0 in 30 s (a $= -0.5$).
For graphs of motion in one direction, area gives both distance and displacement.

7. Kinematic Equations (Constant Acceleration, Straight Line)

The three kinematic equations $$v = u + at \qquad (4.4a)$$ $$s = ut + \tfrac{1}{2}at^2 \qquad (4.4b)$$ $$v^2 = u^2 + 2as \qquad (4.4c)$$ $u$ initial velocity, $v$ final velocity, $a$ acceleration (constant), $t$ time interval, $s$ displacement.

Displacement = area under the line = rectangle + triangle.

Derivations

(4.4a): $a = \dfrac{v-u}{t} \Rightarrow v = u + at$. Graph: straight line starting at $u$, slope $= a$.

(4.4b): $s = $ rectangle $+$ triangle $= u\,t + \tfrac12\,t\,(v-u)$. Put $v-u = at$: $s = ut + \tfrac12 at^2$.

(4.4c): from (4.4a) $t = \dfrac{v-u}{a}$; substitute in (4.4b) and simplify: $2as = v^2 - u^2$, so $v^2 = u^2 + 2as$.

Two more (exercise): $s = vt - \tfrac12 at^2$ and $s = \tfrac12 (u+v)\,t$ (area of a trapezium).

Valid only for constant acceleration.
Which to use: no $s$ needed → (4.4a); no $v$ → (4.4b); no $t$ → (4.4c).
Signs: in motion with both directions, the signs of $u, v, a, s$ show direction. Convert km h⁻¹ to m s⁻¹ first.
Starting from rest: $u = 0$. Coming to a stop: $v = 0$ and $a$ is negative. Free fall: $a = g = 9.8$ m s⁻².

8. Braking Distance and Safety

Example 4.8: car braking at $a = -4$ m s⁻², $v = 0$

$0 = u^2 + 2(-4)s \Rightarrow s = \dfrac{u^2}{8}$.

(i) $u = 54$ km h⁻¹ $= 15$ m s⁻¹ → $s = 225/8 \approx 28.1$ m.

(ii) $u = 108$ km h⁻¹ $= 30$ m s⁻¹ → $s = 900/8 = 112.5$ m.

Doubling the speed makes the braking distance four times ($s \propto u^2$). Hence the safe following distance must increase with speed.
Stopping distance depends on: velocity when brakes are applied, road surface (wet/dry), braking capacity (negative acceleration), and the driver's reaction time (distance covered at constant velocity before braking begins).
V2V (vehicle-to-vehicle) communication lets vehicles exchange signals and warn drivers of possible collisions.

9. Motion in a Plane and Uniform Circular Motion

Motion in a plane (two dimensions): a vehicle overtaking, a kicked ball's path, a satellite orbiting. Motion in space (three dimensions): bird flying, car climbing a mountain road.
Circular motion: motion along a circular path. For a child on a merry-go-round going from A to C: distance = curve ABC; displacement = straight line AC (not equal).

One revolution (radius $R$, time $T$) $$\text{distance} = 2\pi R,\quad \text{displacement} = 0$$ $$v_{av} \text{ (speed)} = \dfrac{2\pi R}{T},\quad \text{average velocity} = 0$$

Uniform circular motion: motion in a circular path with constant speed. Speed is the same at every point, but the direction of velocity changes continuously.

Polygon argument: an athlete on a square or hexagonal track changes direction at every corner; as the number of sides increases, the track approaches a circle and the direction changes continuously.
Velocity is along the tangent to the circle at that point, in the direction of motion (a tangent meets the circle at exactly one point).
Marble-in-ring activity: when the ring is lifted, the marble moves off in a straight line along the direction of motion at that instant.
Uniform circular motion is accelerated because the direction of velocity keeps changing, even though the speed is constant. (Everyday use of 'accelerating' often misses this.)
Idealised model: real motion rarely has constant speed and a perfect circle, but the model is the foundation for planets around the Sun and a vehicle taking a circular turn.

10. Quick Revision

Quantity	Formula	SI unit	Scalar / vector
Distance travelled	Total path length	m	Scalar
Displacement	Final − initial position	m	Vector
Average speed	Total distance / time	m s⁻¹	Scalar
Average velocity	Displacement / time	m s⁻¹	Vector
Average acceleration	$(v-u)/t$	m s⁻²	Vector

Graph	Slope gives	Area gives
Position-time	Velocity	Not used
Velocity-time	Acceleration	Displacement

Equations

Kinematic set

$v = u + at$
$s = ut + \tfrac12 at^2$
$v^2 = u^2 + 2as$

Convert

km/h to m/s

Multiply by $5/18$ (or divide by $3.6$).

Remember

Circle

One revolution: distance $2\pi R$, displacement $0$, average velocity $0$.

Common mistakes:
• Using distance in place of displacement (or speed instead of velocity), especially for round trips.
• Forgetting to convert km h⁻¹ to m s⁻¹.
• Using kinematic equations when acceleration is not constant.
• Dropping the negative sign of acceleration when braking.
• Thinking a flat position-time graph means constant motion (it means at rest); a flat velocity-time graph means zero acceleration.
• Saying uniform circular motion has no acceleration.
• Reading a graph as the path of the object.
• Changing origin or positive direction midway through a problem.