Chapter 5 — ଗତି

A body is said to be in motion when its position changes with time, and at rest when its position does not change with time. But position must always be measured from something. Sit in a moving bus: the passenger next to you is at rest with respect to you, since the distance between you does not change, yet both of you are in motion with respect to a tree on the roadside. The tree, in turn, is at rest with respect to the road but is moving at about 30 km per second with respect to the Sun, along with the whole Earth. Rest and motion are therefore relative: the same body can be at rest for one observer and in motion for another. To describe motion we must first choose a reference point or origin — a fixed point from which positions are measured — and a set of axes fixed to it, called a frame of reference. In this chapter the reference point is usually the ground, a milestone or the starting point of the journey.

Motion may be perceived directly, as when we see a car move, or indirectly, as when we infer the motion of air from the swaying of leaves and the motion of the Earth from the rising and setting of the Sun. Some motions are so slow, like the growth of a plant, or so fast, like a bullet, that we cannot see them, but the definition holds: position changes with time.

The types of motion are classified by the path. Motion along a straight line — a train on a straight track, a stone dropped from a roof, an athlete on a 100-metre track — is rectilinear motion, and this chapter deals mainly with it. Motion along a circle — the tip of a clock hand, a stone whirled on a string, a satellite — is circular motion, treated at the end of the chapter. Motion about a fixed axis, like a spinning top or a fan blade, is rotational; to-and-fro motion about a mean position, like a swing or a pendulum, is oscillatory or periodic; a bouncing ball or a bird's wing shows a combination. Motion may also be classified by how the speed changes: uniform when equal distances are covered in equal intervals of time, non-uniform otherwise.

The physical quantities used to describe motion are of two kinds. A scalar quantity has magnitude only — distance, speed, time, mass. A vector quantity has both magnitude and direction — displacement, velocity, acceleration, force. The distinction runs through the whole chapter: 5 km is a distance, but 5 km due east is a displacement. To specify the motion of an object completely we need to say where it is (position), how far it has gone (distance or displacement), how fast (speed or velocity), and whether it is speeding up or slowing down (acceleration) — and each of these is defined in the topics that follow.

Two different quantities describe how far a body has moved, and confusing them is the commonest error in the chapter.

Distance is the total length of the path actually travelled by a body, irrespective of direction. It is a scalar quantity, always positive, and it can never decrease as the body moves. Its SI unit is the metre (m); kilometres and centimetres are also used.

Displacement is the shortest distance between the initial and final positions of the body, measured in a straight line and in a definite direction — from the starting point to the end point. It is a vector quantity; it may be positive, negative or zero; and its magnitude can never be greater than the distance travelled. Its unit is also the metre.

Consider a body that moves from O to A, 60 km due east, and then back from A to B, 35 km due west. The distance travelled is 60 + 35 = 95 km. The displacement is the straight line from O to B, which is 60 − 35 = 25 km east. If the body returns all the way to O, the distance is 120 km but the displacement is zero, because the initial and final positions coincide. So distance and displacement are equal in magnitude only when the body moves along a straight line without changing direction; in every other case the displacement is smaller.

Worked example 1. An athlete runs once round a circular track of radius 7 m (circumference 44 m). Distance = 44 m; displacement = 0, as she ends where she began. If she runs half the track, the distance is 22 m and the displacement is the diameter, 14 m.

Worked example 2. A farmer walks along the boundary of a square field of side 10 m in 40 s. What is the displacement after 2 minutes 20 seconds (140 s)? One round of 40 m takes 40 s, so in 140 s he completes 3.5 rounds. After 3 full rounds he is back at the start; the remaining half round takes him to the diagonally opposite corner. Displacement = diagonal = √(10² + 10²) = 10√2 = 14.1 m. Distance = 3.5 × 40 = 140 m.

Worked example 3. A man walks 3 km east and then 4 km north. Distance = 7 km. Displacement = √(3² + 4²) = 5 km, in the north-east direction.

Which of the following is true for displacement? It cannot be zero — false, it is zero for a round trip. Its magnitude is greater than the distance — false, it is always less than or equal to the distance. The correct statements are: displacement can be zero, negative or positive; its magnitude is at most the distance; and it depends only on the end points, not on the path. The odometer of a car measures distance; a straight-line map measurement from home to school gives the magnitude of displacement.

A body is in uniform motion when it covers equal distances in equal intervals of time, however small the intervals. A car travelling on a straight highway at a steady 60 km/h covers 1 km every minute; a clock's second hand tip is in uniform (circular) motion. In non-uniform motion the body covers unequal distances in equal intervals of time — a bus in city traffic, a stone falling from a height, a cyclist starting from rest. Most real motion is non-uniform, and physics handles it with the ideas of average and instantaneous values.

Speed is the distance travelled per unit time: speed = distance ÷ time. It is a scalar quantity with the SI unit metre per second (m/s or m s⁻¹); km/h and cm/s are also used. The conversion between the two common units must be at the tips of the fingers: 1 km/h = 1000 m ÷ 3600 s = 5/18 m/s, so to convert km/h to m/s multiply by 5/18, and to convert m/s to km/h multiply by 18/5. Thus 36 km/h = 10 m/s, 54 km/h = 15 m/s, 72 km/h = 20 m/s, 90 km/h = 25 m/s, 108 km/h = 30 m/s.

For non-uniform motion the speed changes from moment to moment, so we use two ideas. The average speed over a journey is the total distance travelled divided by the total time taken, whatever happened in between. The instantaneous speed is the speed at a particular instant — what the speedometer of a vehicle shows. For uniform motion the two are equal.

Worked example 1. An object travels 16 m in 4 s and then 16 m in 2 s. Average speed = total distance ÷ total time = 32 ÷ 6 = 5.33 m/s.

Worked example 2. A car travels 100 km at 50 km/h and the next 100 km at 100 km/h. Time for the first part = 100 ÷ 50 = 2 h; for the second = 100 ÷ 100 = 1 h. Average speed = 200 ÷ 3 = 66.7 km/h — not the arithmetic mean 75 km/h, because more time is spent at the lower speed. This trap is a favourite of examiners.

Worked example 3. A bus starting from rest moves with uniform acceleration and reaches 72 km/h in 40 s; what is its speed in m/s? 72 × 5/18 = 20 m/s. A cheetah runs 100 m in 4 s: speed 25 m/s = 90 km/h.

A body's speed cannot be negative; it is the magnitude of how fast it is moving. The direction is added in the next topic, where speed becomes velocity. The examination expects a clear statement of the definition, the SI unit, and the km/h to m/s conversion, and a numerical on average speed for a journey in parts.

Speed tells how fast; it does not tell which way. To describe the motion of a body fully we must state both, and the quantity that does so is velocity: the displacement per unit time, or speed in a specified direction. Velocity = displacement ÷ time. It is a vector quantity, with the same SI unit as speed, m/s, and it may be positive or negative according to the direction chosen as positive. Two cars moving at 40 km/h, one north and one south, have the same speed but different velocities.

Velocity can change in three ways: when the speed changes (a car accelerating on a straight road), when the direction changes (a car turning a corner at constant speed), or when both change. This is why a body moving in a circle at constant speed has a changing velocity — an idea developed in the final topic.

Uniform velocity means equal displacements in equal intervals of time in the same direction — motion in a straight line at constant speed. Variable velocity means the speed or the direction or both change.

Average velocity = total displacement ÷ total time. When the velocity changes at a uniform rate — that is, for uniform acceleration — the average velocity is simply the arithmetic mean of the initial velocity u and the final velocity v: average velocity = (u + v) / 2. This formula is valid only for uniformly accelerated motion in a straight line, and it is the key to the second equation of motion later.

The difference between average speed and average velocity is best seen in a round trip. Usha swims in a 90 m long pool. She covers 180 m in one minute, swimming from one end to the other and back along the same straight path. Her average speed = 180 m ÷ 60 s = 3 m/s. Her average velocity = displacement ÷ time = 0 ÷ 60 = 0 m/s, because she finishes where she started. The magnitude of the average velocity is equal to the average speed only when the body moves in a straight line without reversing.

Worked example. A car moves 20 m east in 5 s, then 15 m west in 5 s. Average speed = 35 ÷ 10 = 3.5 m/s. Average velocity = (20 − 15) ÷ 10 = 0.5 m/s east.

Worked example. A body starts at 10 m/s and accelerates uniformly to 30 m/s. Its average velocity during this time is (10 + 30) / 2 = 20 m/s; if the acceleration lasted 5 s the displacement was 20 × 5 = 100 m.

A table for revision: speed — distance/time, scalar, always positive, m/s; velocity — displacement/time, vector, positive, negative or zero, m/s. The speedometer measures speed; a navigation system that reports heading and speed measures velocity.

When the velocity of a body changes — in magnitude, in direction, or in both — the body is said to be accelerating. Acceleration is the rate of change of velocity: the change in velocity divided by the time taken for the change. If a body's velocity changes from u (initial) to v (final) in time t, then a = (v − u) / t. Acceleration is a vector quantity; its direction is the direction of the change in velocity. Its SI unit is metre per second squared, m/s² (or m s⁻²), which means that the velocity changes by so many metres per second every second.

The sign of acceleration matters. If the velocity increases in the direction of motion, the acceleration is positive, in the direction of the velocity: a car pulling away from a signal. If the velocity decreases, the acceleration is negative — opposite to the direction of motion — and is called retardation or deceleration: a car braking, a ball thrown upward. If the velocity is constant in a straight line, the acceleration is zero. A body may also have acceleration without any change in speed, when only the direction changes, as in circular motion.

Acceleration is uniform when the velocity changes by equal amounts in equal intervals of time — a freely falling body gains 9.8 m/s of velocity in every second, so its acceleration, called the acceleration due to gravity g, is uniform at 9.8 m/s². Acceleration is non-uniform when the velocity changes by unequal amounts in equal intervals — a bus in traffic. This chapter's equations apply to uniform acceleration only.

Worked example 1. A bus starting from rest reaches a velocity of 6 m/s in 30 s, then applies brakes and stops in 5 s. Acceleration in the first phase: u = 0, v = 6, t = 30; a = (6 − 0) / 30 = 0.2 m/s². In the braking phase: u = 6, v = 0, t = 5; a = (0 − 6) / 5 = −1.2 m/s², a retardation of 1.2 m/s².

Worked example 2. A bicycle accelerates uniformly from 18 km/h to 36 km/h in 10 s. Convert: 18 km/h = 5 m/s, 36 km/h = 10 m/s. a = (10 − 5) / 10 = 0.5 m/s². Always convert km/h to m/s before substituting; mixing units is the classic error.

Worked example 3. A car moving at 20 m/s comes to rest in 4 s under brakes. a = (0 − 20) / 4 = −5 m/s². The negative sign shows the acceleration opposes the motion.

Notice the pattern: the same formula gives both speeding up and slowing down, the sign taking care of the direction. And a body can have zero velocity at an instant yet non-zero acceleration — a ball at the top of its flight is momentarily at rest but is accelerating downward at 9.8 m/s²; it immediately starts falling. Velocity says how the position is changing; acceleration says how the velocity is changing; they are different quantities and one may be zero while the other is not.

A graph turns a table of numbers into a picture that shows the whole motion at a glance. In a distance-time graph time is plotted on the horizontal axis (x-axis) and distance on the vertical axis (y-axis), each with a suitable scale.

Uniform motion. If a body covers equal distances in equal times, the points lie on a straight line sloping upward through the origin. The steeper the line, the faster the body. The slope of the line gives the speed: take any two points A and B on the line, drop perpendiculars to the axes, and the speed = (s₂ − s₁) / (t₂ − t₁). For example, if the body is at 20 m at t = 2 s and at 60 m at t = 6 s, speed = (60 − 20) / (6 − 2) = 10 m/s. Two bodies moving uniformly are compared by their slopes: the line with the greater slope belongs to the faster body.

Body at rest. The distance does not change, so the graph is a horizontal straight line parallel to the time axis; the slope, and hence the speed, is zero.

Non-uniform motion. If the speed changes, the graph is a curve. For a body accelerating from rest — a train leaving a station — the distance covered in each second grows, and the curve bends upward, becoming steeper; the slope at any point (found by drawing a tangent there) gives the instantaneous speed at that time. For a body slowing down the curve flattens.

Reading a graph. A distance-time graph can tell the distance at any time, the time at which a given distance is reached, the speed in any interval, when the body stopped (horizontal portion), and when two bodies met (the point where their lines cross). A graph can be used to solve a problem without algebra: for two cars leaving different towns at different times, draw both lines on the same axes and read off where they intersect.

Worked example. A car's odometer readings are: 0 km at 8:00, 20 km at 8:30, 40 km at 9:00, 40 km at 9:30 (stopped), 70 km at 10:00. On the graph the line rises with slope 40 km/h from 8:00 to 9:00, is horizontal from 9:00 to 9:30, and rises more steeply with slope 60 km/h from 9:30 to 10:00. The average speed for the whole journey is 70 km ÷ 2 h = 35 km/h — the slope of the straight line from the first point to the last.

Two cautions. The distance-time graph of a real body can never slope downward, because distance cannot decrease; a graph that returns towards zero is a position-time or displacement-time graph, whose slope is velocity and can be negative. And the graph is never vertical, since that would mean covering a distance in zero time. The examination commonly presents a graph and asks for the speed in a marked interval, the interval during which the body was at rest, and which of two bodies is faster; the slope answers all three.

A velocity-time graph plots time on the x-axis and velocity on the y-axis. It carries more information than a distance-time graph: its slope gives the acceleration, and the area under it gives the displacement.

Uniform velocity. The velocity does not change, so the graph is a horizontal straight line parallel to the time axis at the height of the velocity. The slope is zero, so the acceleration is zero. The distance travelled in a time t is velocity × time, which is exactly the area of the rectangle under the line between t = 0 and t — height v, width t. This fact, that the area under a velocity-time graph equals the displacement, holds for every velocity-time graph, however shaped, and is the key to the equations of motion.

Uniform acceleration. If the velocity increases by equal amounts in equal times, the graph is a straight line sloping upward. The slope gives the acceleration: a = (v₂ − v₁) / (t₂ − t₁). A line starting at the origin means the body started from rest; a line starting at a height u means it started with velocity u. The displacement in a time t is the area under the line — a trapezium (or a triangle if u = 0) — which is the average velocity (u + v)/2 multiplied by t.

Uniform retardation. The velocity decreases uniformly, so the line slopes downward; its slope is negative, and where it meets the time axis the body has stopped.

Non-uniform acceleration. The graph is a curve; the acceleration at any instant is the slope of the tangent there, and the displacement is still the area under the curve, found by counting squares if necessary.

Worked example. A car starts from rest and reaches 20 m/s in 10 s with uniform acceleration, moves at 20 m/s for the next 20 s, then decelerates uniformly to rest in 10 s. The graph is a triangle rising from (0, 0) to (10, 20), a horizontal line from (10, 20) to (30, 20), and a triangle falling from (30, 20) to (40, 0). Acceleration in the first phase = 20/10 = 2 m/s²; in the third phase = −20/10 = −2 m/s². Total distance = area = ½ × 10 × 20 + 20 × 20 + ½ × 10 × 20 = 100 + 400 + 100 = 600 m. Average speed = 600/40 = 15 m/s.

Worked example. A body moves at 10 m/s for 5 s, then at 20 m/s for 5 s. Distance = area of two rectangles = 10 × 5 + 20 × 5 = 150 m; the vertical jump at t = 5 s is an idealisation — real velocity cannot change instantly.

Summary of what the two graphs give: on a distance-time graph the slope is speed; on a velocity-time graph the slope is acceleration and the area is displacement. The student should practise sketching the velocity-time graph for each of the standard motions — rest, uniform velocity, uniform acceleration from rest, uniform acceleration from an initial velocity, uniform retardation — since the examination asks both to draw them and to read numbers from them.

For a body moving in a straight line with uniform acceleration, three equations connect the five quantities: initial velocity u, final velocity v, acceleration a, time t and displacement s. Each equation omits one of the five, so that given any three the other two can be found. The equations are derived from the velocity-time graph, and the derivations are as examinable as the results.

Setting up the graph. Let a body start with velocity u at time 0 and, under uniform acceleration a, reach velocity v at time t. Its velocity-time graph is a straight line AB, starting at A at height u on the velocity axis and ending at B at height v above the point t on the time axis. Draw the perpendicular BC from B to the time axis (so BC = v and OC = t), and draw AD parallel to the time axis from A to meet BC at D (so OA = DC = u, and BD = v − u).

Derivation of v = u + at. The acceleration is the slope of the line AB. Slope = (change in velocity) / (time taken) = BD / AD = (v − u) / t. Since the slope is the acceleration a, we have a = (v − u) / t, which gives at = v − u, that is v = u + at. This is the velocity-time relation. It is also simply the definition of acceleration rearranged: the final velocity is the initial velocity plus the velocity gained (or lost) in time t. For a body starting from rest u = 0 and v = at; for a body coming to rest v = 0 and u = −at, the acceleration being negative.

Worked example 1. A bus starts from rest with a uniform acceleration of 0.1 m/s². What is its velocity after 2 minutes? u = 0, a = 0.1, t = 120 s. v = 0 + 0.1 × 120 = 12 m/s.

Worked example 2. A train moving at 90 km/h is brought to rest in 10 s by brakes. Find the retardation. u = 90 × 5/18 = 25 m/s, v = 0, t = 10. a = (0 − 25) / 10 = −2.5 m/s²; retardation 2.5 m/s².

Worked example 3. A car accelerates uniformly at 2 m/s² from 10 m/s. How long does it take to reach 30 m/s? t = (v − u) / a = (30 − 10) / 2 = 10 s.

Worked example 4. A stone is thrown vertically upward at 19.6 m/s. Taking g = 9.8 m/s² acting downward, how long does it take to reach the top? At the top v = 0; a = −9.8 (opposing the upward motion). 0 = 19.6 − 9.8 t, so t = 2 s.

Method for full marks: list the known quantities with units, convert km/h to m/s, choose the equation that contains the three known quantities and the one required, substitute and compute, and give the answer with its unit and, for a and v, its sign or direction. The first equation is used whenever s is neither given nor required.

The second equation gives the displacement in terms of the initial velocity, the acceleration and the time, and is called the position-time relation. It is derived from the same velocity-time graph by using the fact that the area under the graph equals the displacement.

Derivation. Refer to the graph of the previous topic: the line AB from (0, u) to (t, v), with BC perpendicular to the time axis and AD parallel to it, meeting BC at D. The displacement s in time t is the area under AB between 0 and t — the area of the figure OABC. This figure is a trapezium, which can be split into the rectangle OADC and the triangle ABD.

Area of rectangle OADC = OA × OC = u × t.

Area of triangle ABD = ½ × AD × BD = ½ × t × (v − u).

So s = ut + ½ t (v − u). From the first equation, v − u = at. Substituting: s = ut + ½ t × at, that is s = ut + ½ at².

The equation reads naturally: the displacement is the distance the body would have covered at its initial velocity alone (ut) plus the extra distance due to the acceleration (½ at²). For a body starting from rest it reduces to s = ½ at², and for a freely falling body dropped from rest to s = ½ gt² = 4.9 t² metres — 4.9 m in the first second, 19.6 m in two seconds, 44.1 m in three. For retardation a is negative.

Worked example 1. A train starts from rest and accelerates uniformly at 0.5 m/s². How far does it travel in 2 minutes? u = 0, a = 0.5, t = 120. s = 0 + ½ × 0.5 × 120² = 0.25 × 14400 = 3600 m = 3.6 km.

Worked example 2. A car moving at 20 m/s brakes with a retardation of 4 m/s² and stops. How far does it go before stopping? First find the time: 0 = 20 − 4t, t = 5 s. Then s = 20 × 5 + ½ × (−4) × 25 = 100 − 50 = 50 m. (The third equation gives this in one step.)

Worked example 3. A ball is dropped from a building 78.4 m high. How long does it take to reach the ground (g = 9.8 m/s²)? u = 0, s = 78.4, a = 9.8. 78.4 = ½ × 9.8 × t² = 4.9 t², so t² = 16 and t = 4 s.

Worked example 4. A body starts at 5 m/s with acceleration 2 m/s². Distance in the fifth second alone? Distance in 5 s = 5 × 5 + ½ × 2 × 25 = 50 m; distance in 4 s = 20 + 16 = 36 m; so the distance covered during the fifth second = 50 − 36 = 14 m.

The second equation is used whenever the final velocity v is neither given nor required. A common slip is to square only a and not t, or to forget the half; writing the formula in full before substituting prevents both.

The third equation connects the velocities with the displacement and the acceleration without involving time, and is called the position-velocity relation. It is derived from the graph by expressing the area of the trapezium in a different way, or by eliminating t from the first two equations.

Derivation from the graph. The displacement s is the area of the trapezium OABC. The area of a trapezium is half the sum of its parallel sides multiplied by the distance between them. The parallel sides are OA = u and BC = v, and the distance between them is OC = t. So s = ½ (u + v) × t. From the first equation, t = (v − u) / a. Substituting: s = ½ (u + v)(v − u) / a = (v² − u²) / 2a, since (u + v)(v − u) = v² − u². Hence 2as = v² − u², that is v² = u² + 2as.

The intermediate result s = ½ (u + v) t — displacement equals average velocity times time — is itself useful and is sometimes called the fourth equation.

Derivation by algebra. From v = u + at, t = (v − u)/a. Put this in s = ut + ½ at²: s = u(v − u)/a + ½ a (v − u)²/a² = [2u(v − u) + (v − u)²] / 2a = (v − u)(2u + v − u) / 2a = (v − u)(v + u) / 2a = (v² − u²) / 2a, the same result.

Worked example 1. A car accelerates uniformly from 18 km/h to 36 km/h in 5 s. Find the acceleration and the distance covered. u = 5 m/s, v = 10 m/s, t = 5. a = (10 − 5)/5 = 1 m/s². s = (v² − u²) / 2a = (100 − 25) / 2 = 37.5 m. Check with the second equation: s = 5 × 5 + ½ × 1 × 25 = 25 + 12.5 = 37.5 m.

Worked example 2. A stone is thrown vertically upward at 5 m/s. What is the maximum height, taking g = 10 m/s²? At the top v = 0; a = −10. 0 = 25 − 2 × 10 × s, so s = 25/20 = 1.25 m.

Worked example 3. A car moving at 20 m/s stops in 50 m under uniform braking. Find the retardation. 0 = 400 + 2a × 50, so a = −400/100 = −4 m/s².

Worked example 4. A trolley on an incline has acceleration 2 cm/s²; what is its velocity 3 s after the start from rest? Here the first equation is right: v = 0 + 2 × 3 = 6 cm/s. A racing car with uniform acceleration 4 m/s² covers what distance in 10 s from rest? s = ½ × 4 × 100 = 200 m. A stone thrown upward at 5 m/s: height 1.25 m, time to the top 0.5 s (from v = u + at with g = 10).

Choice of equation: v = u + at when s is absent; s = ut + ½at² when v is absent; v² = u² + 2as when t is absent. All three assume uniform acceleration in a straight line, and all three are consistent with each other, so any answer can be checked by a second route.

This topic gathers the standard problems of the examination and works each in full, since the marks are for the method as much as the answer. Use g = 9.8 m/s² unless the question says 10.

Problem 1. A bus starting from rest moves with uniform acceleration of 0.1 m/s² for 2 minutes. Find the speed acquired and the distance travelled. u = 0, a = 0.1, t = 120 s. v = u + at = 0 + 0.1 × 120 = 12 m/s. s = ut + ½at² = 0 + ½ × 0.1 × 14400 = 720 m.

Problem 2. A train is travelling at 90 km/h. Brakes produce a uniform acceleration of −0.5 m/s². How far does the train go before it stops? u = 25 m/s, v = 0, a = −0.5. v² = u² + 2as: 0 = 625 + 2(−0.5)s = 625 − s, so s = 625 m.

Problem 3. A trolley on an inclined plane has an acceleration of 2 cm/s². What is its velocity 3 s after the start? v = 0 + 2 × 3 = 6 cm/s.

Problem 4. A racing car has a uniform acceleration of 4 m/s². What distance will it cover in 10 s after the start? s = ½ × 4 × 100 = 200 m.

Problem 5. A stone is thrown vertically upward with a velocity of 5 m/s. If g = 10 m/s² downward, what is the height attained and how long does it take to reach there? Height: 0 = 25 − 20s, s = 1.25 m. Time: 0 = 5 − 10t, t = 0.5 s.

Problem 6. A ball is gently dropped from a height of 20 m. If its velocity increases uniformly at 10 m/s², with what velocity will it strike the ground, and after what time? v² = 0 + 2 × 10 × 20 = 400, v = 20 m/s. t = v/a = 20/10 = 2 s.

Problem 7. A car moving at 30 m/s is brought to rest in 6 s. Find the retardation and the distance travelled during braking. a = (0 − 30)/6 = −5 m/s². s = ½ (u + v) t = ½ × 30 × 6 = 90 m.

Problem 8. A motorcycle accelerates from 36 km/h to 72 km/h in 10 s. Find the acceleration and the distance covered. u = 10, v = 20, t = 10. a = 1 m/s². s = ½ (10 + 20) × 10 = 150 m.

Problem 9. A body covers 20 m in the third second and 28 m in the fifth second of uniformly accelerated motion. Find u and a. Distance in the nth second = u + ½ a (2n − 1). Third second: u + 2.5a = 20. Fifth second: u + 4.5a = 28. Subtracting: 2a = 8, a = 4 m/s²; u = 20 − 10 = 10 m/s.

Problem 10. An object thrown upward returns to the thrower after 4 s (g = 10). Find the velocity of projection and the maximum height. Time up = 2 s; at the top v = 0, so 0 = u − 10 × 2, u = 20 m/s. Height = ½ × 10 × 4 = 20 m (using the fall of 2 s from rest), or from v² = u² − 2gs: 0 = 400 − 20s, s = 20 m.

Points that cost marks when forgotten: convert km/h to m/s; take a negative for retardation and for upward throws against gravity; write the equation in symbols before substituting; state the unit. Where two equations could be used, use one to solve and the other to check.

So far the motion has been along a straight line. Now consider a body moving along a circular path at constant speed — the tip of a clock hand, a stone whirled on a string, the Moon around the Earth, a satellite around the Moon, an athlete on a circular track, a cyclist on a roundabout. This is uniform circular motion: motion in a circle with uniform speed.

The surprising fact is that uniform circular motion is accelerated motion, although the speed does not change. Velocity is a vector; it has direction as well as magnitude. On a circle the direction of motion is along the tangent at every point, and the tangent changes direction continuously as the body goes round. So the velocity is changing — in direction if not in magnitude — and a change of velocity means acceleration. The acceleration is always directed towards the centre of the circle, and the force that provides it (the tension in the string, gravity for the Moon, friction for a car turning) is studied in the next chapter.

A simple activity shows that the velocity is along the tangent. Tie a stone to a thread and whirl it in a horizontal circle; when the thread is released the stone flies off in a straight line along the tangent at the point of release, not outward along the radius and not along the circle. An athlete throwing a hammer or a discus spins in a circle and lets go at the moment when the tangent points along the field.

Consider how a body moving along a closed path changes direction. Along a rectangular track of length l and breadth b, the athlete changes direction four times per round, at the corners. Along a hexagonal track, six times; along an octagonal track, eight times. As the number of sides increases, the polygon approaches a circle, and the number of direction changes becomes infinite — the direction changes continuously at every point of the circle. This is the argument the textbook uses to introduce the idea.

Speed in uniform circular motion. If a body moves once round a circle of radius r in time t, the distance covered in one revolution is the circumference 2πr, so the speed is v = 2πr / t. The time for one revolution is the time period T, and the number of revolutions per second is the frequency. The direction of the velocity at any instant is along the tangent.

Worked example 1. An artificial satellite moves in a circular orbit of radius 42,250 km and completes one revolution in 24 hours. Its speed = 2 × 3.14 × 42250 / 24 = 265,330 / 24 ≈ 11,055 km/h ≈ 3.07 km/s.

Worked example 2. The tip of a second hand of length 7 cm: distance in one revolution = 2 × 22/7 × 7 = 44 cm in 60 s, so speed = 0.73 cm/s. Its average velocity over one full minute is zero, since the displacement is zero.

Worked example 3. A cyclist completes a circular track of radius 100 m in 2 minutes: speed = 2 × 3.14 × 100 / 120 = 5.23 m/s; after half a round, distance 314 m, displacement 200 m.

The examination asks for the definition, why it is accelerated motion, the direction of the velocity, the tangent activity, and a speed calculation; the essential sentence is that in uniform circular motion the speed is constant but the velocity changes continuously because the direction changes, so the body is accelerating.