Chapter 2 · ApplicationsMotion in real lifeExplore online

Applications & Explore Online ചലനസമവാക്യങ്ങൾ

Displacement, velocity, acceleration, graphs and the three equations of motion are not only for the exam. They decide how far a car needs to stop before a zebra crossing, how long a train takes to cross the Bharathapuzha bridge, how the Kochi Metro runs from one station to the next, how a sprinter wins a 100 m race, why a speedometer does not show velocity and how high a ball goes. Every lab below uses only the ideas of this chapter.

📖 The ideas behind every lab

Sign convention everywhere on this page: forward, east and up are positive.

1. Stopping distance – why speed kills

When a driver sees a child step onto a zebra crossing, the car does not stop at once. First the driver needs a moment to see the danger and move the foot to the brake – the reaction time. During that time the car keeps its full speed and covers the reaction distance = u × reaction time. Only then do the brakes give a retardation a, and the car covers the braking distance. From v2 = u2 + 2as with v = 0, the braking distance is s = u2/2a.

Stopping distance = reaction distance + braking distance. The reaction distance grows in step with the speed, but the braking distance grows with the square of the speed: double the speed and the car needs four times the braking distance.

🚗 Stopping-distance lab: u × tr + u²/2a

The road is drawn to scale (one mark = 10 m); the car is drawn bigger than its real size so you can see it. The bars below use your reaction time and retardation at six different speeds; the purple bar is your speed. Reaction times and retardations are typical rough values (about 4 m/s² on a wet road, about 7 m/s² on a dry road with good tyres), used only to compare.

⚠ Near schools, crossings and in the monsoon: slow down. A car at 30 km/h stops in about 15 m; at 60 km/h it needs about 44 m – and still hits a child 40 m ahead at about 24 km/h. Never use a phone while driving or crossing the road.

📝 In the exam this looks like: “A car is moving at 72 km/h. The driver’s reaction time is 1 s and the brakes produce a retardation of 5 m/s2. Find the stopping distance.” → u = 72 × 5/18 = 20 m/s; reaction distance = 20 × 1 = 20 m; braking distance: 0 = 202 − 2 × 5 × s ⇒ s = 400/10 = 40 m; stopping distance = 20 + 40 = 60 m. At 36 km/h the braking distance is only 10 m – a quarter.

2. A train crossing a bridge or a platform

The opening question of the chapter: how long does a 200 m train at 20 m/s take to cross an 800 m bridge? The train has crossed only when its last coach leaves the bridge. By then the engine has travelled the bridge’s length plus the train’s own length: s = 800 + 200 = 1000 m, so t = s/v = 1000/20 = 50 s. Picture a train crossing the long bridge over the Bharathapuzha – or sliding past a station platform.

🚆 Train calculator: t = (length of train + length of bridge) ÷ v

Watch the moment the engine reaches the far end: the clock shows only (length of bridge) ÷ v, and part of the train is still on the bridge. Lengths are to scale; the picture shrinks to fit long bridges.

📝 In the exam this looks like: “A train 150 m long moving at 54 km/h crosses a platform 300 m long. Find the time taken.” → v = 54 × 5/18 = 15 m/s; s = 300 + 150 = 450 m; t = 450/15 = 30 s. Turned around: “A 200 m train crosses a 400 m bridge in 30 s. Find its speed.” → v = 600/30 = 20 m/s = 20 × 18/5 = 72 km/h. Passing a signal post, only the train’s length counts: 200 m at 20 m/s takes 10 s.

3. Kochi Metro – from one station to the next

A metro train between two stations has three stages: it accelerates uniformly from rest, cruises at a uniform velocity and then retards uniformly to stop at the next platform. Its velocity–time graph is a trapezium, and the area under it is the distance between the stations. The numbers in this lab are example values to learn from, not official Kochi Metro data.

🚇 Metro lab: accelerate – cruise – brake, area = distance

If the stations are too close for the train to reach its top speed, the graph becomes a triangle: it starts braking as soon as it must. The three coloured areas always add up to the distance between the stations.

📝 In the exam this looks like: “A metro train starts from rest and accelerates uniformly to 20 m/s in 20 s, runs at this velocity for 40 s and then stops uniformly in 20 s. Draw the velocity–time graph and find the total distance.” → areas: ½ × 20 × 20 = 200 m; 20 × 40 = 800 m; ½ × 20 × 20 = 200 m; total = 1200 m. Acceleration = 20/20 = 1 m/s2; retardation = 1 m/s2; average speed = 1200/80 = 15 m/s.

4. The 100 m sprint – acceleration, then uniform velocity

A sprinter starts from rest in the blocks, accelerates for the first few seconds and then runs at a nearly uniform velocity to the finish. On a position–time graph the line first curves upwards (the slope – the velocity – is growing) and then becomes a sloping straight line. On a velocity–time graph it is a rising line followed by a horizontal line. The area under the v–t graph is exactly 100 m.

🏃 Sprint lab: position–time and velocity–time graphs

Simple model: uniform acceleration until the top speed, then uniform velocity. The purple dots mark the runner’s position every second – their gaps grow while accelerating and become equal at uniform velocity (compare Car A and Car B in Fig 2.5).

📝 In the exam this looks like: “A sprinter starts from rest, accelerates uniformly and reaches 10 m/s in 4 s, then runs with uniform velocity. Find (a) the acceleration (b) the distance covered in the first 4 s (c) the time taken for the 100 m race.” → a = 10/4 = 2.5 m/s2; s = ½ × 2.5 × 42 = 20 m (area of the triangle ½ × 4 × 10); the remaining 80 m at 10 m/s takes 8 s, so the race takes 4 + 8 = 12 s.

5. Speedometer vs velocity – a car on a ghat road

A car climbs a hill road with hairpin bends, like the Thamarassery churam. The driver keeps the speedometer steady at 36 km/h. Is the car’s velocity uniform? No! A speedometer shows only the speed – a scalar. The velocity also has a direction, and on every bend the direction of motion (ചലനദിശ) changes. A changing direction means a changing velocity – non-uniform velocity, and so there is an acceleration even though the speed is constant (just like Car C on the circular track). The odometer adds up the distance; the displacement is only the straight line from the start.

🛣️ Ghat-road lab: same speed, changing velocity

Map scale: the whole road is about 3 km long (north is up). The blue arrow is the velocity – always along the road, same length (same speed), changing direction. The red dashed arrow is the displacement from the start.

📝 In the exam this looks like: “A car on a ghat road shows a steady 40 km/h on its speedometer. Is its velocity uniform? Does it have an acceleration?” → No: its direction of motion changes at every bend, so its velocity changes (non-uniform velocity) and it has an acceleration, although its speed is constant. With numbers (Let us assess Q5): half a round of a 440 m track (radius 70 m) in 40 s – speed = 220/40 = 5.5 m/s, but velocity = 140/40 = 3.5 m/s along the diameter.

6. A ball thrown straight up – the v–t line through zero

Throw a ball straight up at u. Gravity gives it a uniform acceleration of about 10 m/s2 downwards all the time, so taking up as positive, a = −10 m/s2. Going up it slows down (retardation), stops for an instant at the top, and comes back down faster and faster. Its velocity v = u − 10t is one straight line on the v–t graph: from +u, through zero at the top, to −u when it is back in the hand. The area above the time axis is the height; the area below is the same size but negative, so the total displacement is zero.

⚾ Throw-up lab: v = u − 10t

Air resistance is ignored and g is taken as 10 m/s², as in the textbook. Once the ball is back in the hand (t = 2u/10) the clock stops.

📝 In the exam this looks like: extended activity 2 – “An object thrown up at 30 m/s returns to the same place. Draw its v–t graph.” → a straight line from (0, 30 m/s) through (3 s, 0) to (6 s, −30 m/s); maximum height = area of the first triangle = ½ × 3 × 30 = 45 m; total displacement = 0. Let us assess Q12: thrown up at 20 m/s – maximum height 20 m; after 3 s it is 15 m high, having travelled 25 m.

✅ Check yourself

🌐 Explore online – resources hub

Everything in this guide works offline. The links below are extras: simulations, short videos and articles from trusted sources (PhET and others), grouped by topic. They need internet and open in a new tab, so you will not lose your place here. Ask a parent or teacher before you go online. The PhET simulations run directly in your browser – nothing to install.

Distance, displacement and vectors

🌐 Distance, displacement and vectors needs internet · opens in a new tab

Speed and velocity

🌐 Speed and velocity needs internet · opens in a new tab

Acceleration and retardation

🌐 Acceleration and retardation needs internet · opens in a new tab

Position–time and velocity–time graphs

🌐 Position–time and velocity–time graphs needs internet · opens in a new tab

Equations of motion

🌐 Equations of motion needs internet · opens in a new tab

Throwing up and falling down

🌐 Throwing up and falling down needs internet · opens in a new tab

Road safety, signs and markings

🌐 Road safety, signs and markings needs internet · opens in a new tab

Malayalam explainers

🌐 Malayalam explainers needs internet · opens in a new tab

More to explore

🌐 More to explore needs internet · opens in a new tab
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Now pull the whole chapter together: Summary & Quick Quiz ▶