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Mathematics notes, chapter 9

Speed and Acceleration

Textbook Mathematics Form 2 (DLP), pages 168–187

What this chapter covers

  • Explain the meaning of speed as a rate involving distance and time (9.1.1)
  • Describe the differences between uniform and non-uniform speed (9.1.2)
  • Calculate speed and average speed, including unit conversion (9.1.3)
  • Solve problems involving speed (9.1.4)
  • Explain the meaning of acceleration and deceleration as a rate involving speed and time (9.2.1)
  • Perform calculations involving acceleration, including unit conversion (9.2.2)
  • Solve problems involving acceleration (9.2.3)

How fast is a car, a runner or a MotoGP rider? Speed tells you how much distance something covers in a fixed time, and acceleration tells you how quickly that speed is changing. In this chapter you will calculate speed and average speed, convert between units, read distance-time and speed-time graphs, and solve problems about acceleration and deceleration.

Galileo Galilei was the first scientist to measure speed as distance per unit of time. The same idea is used today in sports timing, road safety and automotive engineering.
EnglishBahasa Melayu
speed, distance, timelaju, jarak, masa
unitunit
average speedlaju purata
uniform speed, non-uniform speedlaju seragam, laju tak seragam
stationarypegun
acceleration, decelerationpecutan, nyahpecutan

9.1Speed

Speed as a rate

Speed is the rate at which distance is covered: distancetime. If you walk 10 km in 1 hour your speed is 10 km/h; if a particle moves 1 m in 1 s its speed is 1 m/s.
Triangle split into three parts: D (distance) on top, S (speed) bottom left and T (time) bottom right
Cover the quantity you want. D on top means distance = S × T. S beside T means speed = D ÷ T. T beside S means time = D ÷ S.
To findFormulaUnit
speedspeed = distance ÷ timekm/h, m/s, m/min
distancedistance = speed × timekm, m
timetime = distance ÷ speedh, min, s

Example 1 (like textbook p. 171)

Aida walks 100 m to a shop in 5 minutes. Find her speed.

  1. Speed = 100 m5 min = 20 m/min

Aida walks 20 metres every minute. The unit m/min is read "metre per minute".

Example 2 (like textbook p. 171)

In a 100 m race, Khairul finished in 10.18 s and Badrul in 10.25 s. Find the speed of each runner, correct to 2 decimal places.

  1. Khairul: 100 m10.18 s = 9.82 m/s
  2. Badrul: 100 m10.25 s = 9.76 m/s

The runner who takes the shorter time over the same distance has the greater speed.

Example 3: finding distance and time

A bus travels from Tanjung Malim to Muar, a distance of 272 km, at an average speed of 80 km/h. It leaves at 0830. At what time does it arrive? (24-hour system)

  1. Time = 272 km80 km/h = 3.4 h
  2. 0.4 h = 0.4 × 60 = 24 min, so 3.4 h = 3 h 24 min
  3. Arrival time = 0830 + 3 h 24 min = 1154
A decimal part of an hour is not minutes. 3.4 h is 3 h 24 min, not 3 h 40 min. Multiply the decimal by 60.
A cyclist rides 36 km in 2.4 hours. a) Find her speed. b) How far will she ride in 5 hours at the same speed?
Show answerHide answer
a) 36 ÷ 2.4 = 15 km/h
b) 15 × 5 = 75 km

Uniform and non-uniform speed

Marble A and marble B are rolled along a table. A photograph is taken every second and the distance from the start is read off a ruler.

Two rows of marbles at one-second intervals. Marble A is at 0, 2, 4, 6 and 8 cm. Marble B is at 0, 2, 3, 5 and 8 cm.
Marble A covers 2 cm every second. Marble B covers 2, 1, 2 and 3 cm in successive seconds.

Textbook figure, Mathematics Form 2 (DLP), p. 172.

Time (s)01234
Marble A: distance (cm)02468
Marble B: distance (cm)02358
Distance-time graph for marble A: a straight line through (0, 0) and (4, 8)
Marble A: a straight line.
Distance-time graph for marble B: points at (0, 0), (1, 2), (2, 3), (3, 5), (4, 8) joined by line segments of different steepness
Marble B: the line bends.
Uniform speed: equal distances are covered in equal intervals of time. Its distance-time graph is a straight line.
Non-uniform speed: unequal distances are covered in equal intervals of time. Its distance-time graph is not a straight line.
On a distance-time graph the gradient is the speed: change in distancechange in time. A steeper line means a greater speed and a horizontal line means the object is stationary (speed 0).

Example 4 (like textbook p. 172)

Farid drives a trailer 170 km in the first 2 hours and 190 km in the next 4 hours. Is the speed uniform?

  1. First 2 hours: 170 km2 h = 85 km/h
  2. Next 4 hours: 190 km4 h = 47.5 km/h
  3. The two speeds are different, so the speed is non-uniform.

Example 5 (like textbook p. 173)

A golf ball is hit from P through Q and R to S. P to Q is 40 m in 2 s, Q to R is 70 m in 3.5 s and R to S is 60 m in 3 s. Is its speed uniform?

  1. P to Q: 40 ÷ 2 = 20 m/s
  2. Q to R: 70 ÷ 3.5 = 20 m/s
  3. R to S: 60 ÷ 3 = 20 m/s
  4. All three speeds are equal, so the speed is uniform.
Compare equal intervals of time or compare the speed of each stage. A bigger distance does not mean a bigger speed if it also took longer.

Average speed

Average speed = total distancetotal time. It is used when the speed changes during a journey, such as a train or a bus. Time spent resting is part of the total time.

Example 6 (like textbook p. 174)

Amir cycles from his house to town A, a distance of 35 km. He leaves at 9:00 a.m., rests at a pizzeria on the way and arrives at 12:15 p.m. Find his average speed.

  1. Total time = 12:15 − 9:00 = 3 h 15 min = 3.25 h
  2. Average speed = 35 km3.25 h = 10.77 km/h

Example 7 (like textbook p. 174)

A bus leaves Puchong at 0825 and reaches Perai at 1345. The distance is 354 km. Find the average speed.

  1. Total time = 1345 − 0825 = 5 h 20 min = 5 13 h
  2. Average speed = 354 ÷ 5 13 = 66.38 km/h (2 decimal places)

If you round 5 h 20 min to 5.33 h first, you get 66.42 km/h. Keep the exact fraction (or more decimals) in your calculator and round only at the end.

Example 8: different units in one journey (like textbook p. 175)

Karmila won a triathlon: a 10 km run, a 1 500 m swim and a 40 km cycle, in 1 hour 56 minutes. Find the average speed in km/h.

  1. Total distance = 10 km + 1.5 km + 40 km = 51.5 km
  2. Total time = 1 h 56 min = 1 5660 h = 1.933 h
  3. Average speed = 51.5 km1.933 h = 26.64 km/h

Example 9: reading a distance-time graph (like textbook p. 176)

Distance-time graph of a school bus: a straight line from (0, 0) to B (2, 88), a horizontal rest from 2 h to 3 h, then a straight line to C (5, 203)

A school bus travels from school (O) to Kuala Lumpur (C) and stops for a rest from B. Find (a) the speed in the first 2 hours, (b) the average speed for the whole trip.

  1. (a) Speed = gradient = 88 km2 h = 44 km/h
  2. (b) Total distance = 203 km, total time = 5 h (the rest is included)
  3. Average speed = 203 km5 h = 40.6 km/h
Average speed is not the mean of the speeds. Go 120 km at 60 km/h and come back at 40 km/h: the times are 2 h and 3 h, so the average speed is 240 ÷ 5 = 48 km/h, not (60 + 40) ÷ 2 = 50 km/h.

Changing units of speed

1 km = 1 000 m1 m = 100 cm1 h = 60 min1 min = 60 s1 h = 3 600 s

To convert a speed, convert the distance unit and the time unit separately.

Example 10 (like textbook p. 175)

The speed limit on a road is 90 km/h. Convert it to (a) m/s, (b) km/min.

  1. (a) 90 km/h = 90 × 1 000 m3 600 s = 25 m/s
  2. (b) 90 km/h = 90 km60 min = 1.5 km/min
Shortcut: m/s × 3.6 = km/h, and km/h ÷ 3.6 = m/s. Check: 25 × 3.6 = 90.

Example 11 (like textbook p. 175)

Convert 120 m/s to km/min.

  1. 120 m in 1 s means 120 × 60 = 7 200 m in 1 min
  2. 7 200 m = 7 200 ÷ 1 000 = 7.2 km/min

The textbook shows 7.06 km/min because it rounds 1 ÷ 60 to 0.017 too early. The exact answer is 7.2 km/min.

a) Convert 50 km/h to m/min. b) Convert 0.8 m/s to km/h. c) A leopard sprints at 25.9 m/s. State this speed in km/h. (Textbook p. 179)
Show answerHide answer
a) 50 × 1 000 ÷ 60 = 833.33 m/min
b) 0.8 × 3.6 = 2.88 km/h
c) 25.9 × 3.6 = 93.24 km/h

Solving problems involving speed

Use four steps: understand the problem, plan the strategy, carry it out, and write a conclusion.

Example 12 (like textbook p. 176)

Khairul Idham Pawi won the Moto3 race at the German Grand Prix. The track is 40.38 km and he took 47 minutes 8 seconds. Find (a) his average speed in km/h; (b) the difference in time between him and the second place rider, whose speed was 0.85 km/min.

  1. (a) 47 min 8 s = 47 860 min = 2 8283 600 h = 0.7856 h
  2. Speed = 40.380.7856 = 51.40 km/h
  3. (b) Second place time = 40.38 km0.85 km/min = 47.51 min, which is about 47 min 30 s
  4. Difference = 47 min 30 s − 47 min 8 s = about 22 s

Example 13 (like textbook p. 177)

Mr Tan drove from Kuala Lumpur to Skudai in 3 hours 7 minutes at an average speed of 103 km/h. On the way back he took 34 of an hour longer. Find his average speed for the return journey.

  1. Time to Skudai = 3 h 7 min = 3 760 h = 3.1167 h
  2. Distance = 103 × 3.1167 = 321.02 km
  3. Return time = 3.1167 + 0.75 = 3.8667 h
  4. Return speed = 321.023.8667 = 83.02 km/h

Example 14: finding a missing stage

A taxi drives 60 km from Ipoh to Tapah at 100 km/h. The whole trip from Ipoh to Kuala Lumpur is 220 km at an average speed of 110 km/h. Find the average speed from Tapah to Kuala Lumpur.

  1. Total time = 220 ÷ 110 = 2 h. Time Ipoh to Tapah = 60 ÷ 100 = 0.6 h
  2. Tapah to Kuala Lumpur: distance = 220 − 60 = 160 km, time = 2 − 0.6 = 1.4 h
  3. Speed = 160 ÷ 1.4 = 114.29 km/h
Subtract distances and times of the stages you know. Do not subtract the speeds.

9.2Acceleration

Acceleration and deceleration

Acceleration is the rate of change of speed with time for an object moving in a straight line: acceleration = final speed − initial speedtime.
If speed increases there is acceleration. If speed decreases the acceleration is negative, which is called deceleration. An object with uniform speed has zero acceleration.
Speed unitTime unitAcceleration unit
m/ssecondm/s² (m/s per second)
km/hhourkm/h² (km/h per hour)
km/hsecondkm/h per second

Example 15 (like textbook p. 180)

A motorcycle starts from rest and accelerates uniformly to 20 m/s in 5 s. Find its acceleration.

  1. Acceleration = (20 − 0) m/s5 s = 4 m/s²

Every second the speed goes up by 4 m/s.

Example 16 (like textbook p. 181)

The rider brakes and the motorcycle slows down uniformly from 20 m/s to a stop in 4 s. Find its acceleration.

  1. Acceleration = (0 − 20) m/s4 s = −5 m/s²

The deceleration is 5 m/s². Write either "acceleration = −5 m/s²" or "deceleration = 5 m/s²", but not "deceleration = −5 m/s²".

Final speed minus initial speed, in that order. A negative answer means the object slows down.

Example 17 (like textbook p. 180)

A racing car accelerates from rest to 120 km/h in 6 s. Find the acceleration in km/h per second.

  1. Change of speed = 120 − 0 = 120 km/h
  2. Acceleration = 120 km/h6 s = 20 km/h per second

In m/s² this is 20 ÷ 3.6 = 5.56 m/s² (see Example 19 for the conversion).

Speed-time graph

Speed-time graph: speed rises from 0 to 12 m/s in 6 s, stays at 12 m/s until 16 s, then falls to 0 at 20 s. The area under the graph is shaded.
On a speed-time graph: the gradient is the acceleration and the area under the graph is the distance travelled.
StageTimeAccelerationDistance (area)
speeding up0 s to 6 s12 ÷ 6 = 2 m/s²½ × 6 × 12 = 36 m
constant speed6 s to 16 s010 × 12 = 120 m
slowing down16 s to 20 s(0 − 12) ÷ 4 = −3 m/s²½ × 4 × 12 = 24 m
whole journey20 s36 + 120 + 24 = 180 m
The whole shape is a trapezium: ½ × (parallel sides) × height = ½ × (10 + 20) × 12 = 180 m.
A horizontal line on a speed-time graph means constant speed. On a distance-time graph a horizontal line means the object is at rest. Check the axis label.

Unit conversion for acceleration

Convert the distance unit and each time unit one at a time. An acceleration has the time unit twice.

Example 18 (like textbook p. 181)

Convert 420 m/min per minute to km/min².

  1. 420 m/min² = 420 ÷ 1 000 kmmin² = 0.42 km/min²

Example 19: m/s² and km/h per second

Convert 5 m/s² to km/h per second.

  1. Speed gain each second = 5 m/s = 5 × 3.6 = 18 km/h
  2. So 5 m/s² = 18 km/h per second

In the same way 1 m/s² = 3.6 km/h per second.

Example 20 (like textbook p. 182)

Samy increases his speed from 70 km/h to 100 km/h in 30 minutes. Find the acceleration in (a) km/h per hour, (b) km/h per second.

  1. Change of speed = 100 − 70 = 30 km/h
  2. (a) 30 min = 0.5 h, acceleration = 30 ÷ 0.5 = 60 km/h²
  3. (b) 30 min = 1 800 s, acceleration = 30 ÷ 1 800 = 0.0167 km/h per second
a) A boat slows from 70 km/h to 40 km/h in 5 minutes. Find its acceleration in km/h per hour. b) Vinod speeds up from 8 m/s to 10 m/s in 4 s. Find the acceleration in m/s². (Textbook pp. 184–185)
Show answerHide answer
a) Change = 40 − 70 = −30 km/h, time = 5 ÷ 60 = 112 h. Acceleration = −30 ÷ 112 = −360 km/h², a deceleration of 360 km/h².
b) (10 − 8) ÷ 4 = 0.5 m/s²

Solving problems involving acceleration

Example 21 (like textbook p. 183)

Lisnah drives at 100 km/h and then accelerates at 4 km/h per second while overtaking. Find her speed after 5 s.

  1. Increase in speed = 4 × 5 = 20 km/h
  2. Speed after 5 s = 100 + 20 = 120 km/h

Example 22 (like textbook p. 183)

A motorcycle at 40 km/h reduces its speed by 20% of its beginning speed within 40 s. Find the acceleration.

  1. End speed = 80% × 40 = 32 km/h
  2. Acceleration = (32 − 40) km/h40 s = −0.2 km/h per second (deceleration 0.2 km/h per second)

Example 23: two trips (like textbook p. 182)

Rani cycles for 34 of an hour with a speed change of 18 km/h. The return trip takes 40% less time at the same speed change. Find the difference in the two accelerations.

  1. Trip 1: 18 ÷ 0.75 = 24 km/h²
  2. Trip 2 time = 0.75 × 0.6 = 0.45 h, acceleration = 18 ÷ 0.45 = 40 km/h²
  3. Difference = 40 − 24 = 16 km/h²

Example 24: braking distance

A car travelling at 72 km/h brakes uniformly and stops in 8 s. Find (a) the deceleration in m/s², (b) the distance covered while braking.

  1. 72 km/h = 72 ÷ 3.6 = 20 m/s
  2. (a) Acceleration = 0 − 208 = −2.5 m/s², so the deceleration is 2.5 m/s²
  3. (b) The speed-time graph is a triangle: distance = ½ × 8 × 20 = 80 m

Summary

IdeaFormula or rule
Speeddistance ÷ time
Distance, timedistance = speed × time; time = distance ÷ speed
Average speedtotal distance ÷ total time (include rests)
Uniform speedequal distances in equal times; straight line on a distance-time graph
Acceleration(final speed − initial speed) ÷ time
Decelerationnegative acceleration; the speed decreases
Graphsdistance-time: gradient = speed. Speed-time: gradient = acceleration, area = distance
Unitsm/s × 3.6 = km/h; 1 m/s² = 3.6 km/h per second
  • Make the units agree before you calculate. Change minutes to hours by dividing by 60.
  • Average speed is total distance over total time, not the mean of the speeds.
  • Acceleration needs the final speed minus the initial speed, then divide by the time taken.