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Applications in the real-world

What we have learnt in this chapter can be directly applied to road safety. We can analyse the relationship between speed and stopping distance. The following worked example illustrates this application.

A truck is travelling at a constant velocity of 10 m · s - 1 when the driver sees a child 50 m in front of him in the road. He hits the brakes to stop the truck. The truck accelerates at a rate of -1.25 m · s - 2 . His reaction time to hit the brakes is 0,5 seconds. Will the truck hit the child?

  1. It is useful to draw a timeline like this one:

    We need to know the following:

    • What distance the driver covers before hitting the brakes.
    • How long it takes the truck to stop after hitting the brakes.
    • What total distance the truck covers to stop.
  2. Before the driver hits the brakes, the truck is travelling at constant velocity. There is no acceleration and therefore the equations of motion are not used. To find the distance traveled, we use:

    v = D t 10 = d 0 , 5 d = 5 m

    The truck covers 5 m before the driver hits the brakes.

  3. We have the following for the motion between B and C:

    v i = 10 m · s - 1 v f = 0 m · s - 1 a = - 1 , 25 m · s - 2 t = ?

    We can use [link]

    v f = v i + a t 0 = 10 + ( - 1 , 25 ) t - 10 = - 1 , 25 t t = 8 s
  4. For the distance we can use [link] or [link] . We will use [link] :

    Δ x = ( v i + v f ) 2 t Δ x = 10 + 0 s ( 8 ) Δ x = 40 m
  5. The total distance that the truck covers is D A B + D B C = 5 + 40 = 45 meters . The child is 50 meters ahead. The truck will not hit the child.


  • A reference point is a point from where you take your measurements.
  • A frame of reference is a reference point with a set of directions.
  • Your position is where you are located with respect to your reference point.
  • The displacement of an object is how far it is from the reference point. It is the shortest distance between the object and the reference point. It has magnitude and direction because it is a vector.
  • The distance of an object is the length of the path travelled from the starting point to the end point. It has magnitude only because it is a scalar.
  • A vector is a physical quantity with magnitude and direction.
  • A scalar is a physical quantity with magnitude only.
  • Speed ( s ) is the distance covered ( D ) divided by the time taken ( Δ t ):
    s = D Δ t
  • Average velocity ( v ) is the displacement ( Δ x ) divided by the time taken ( Δ t ):
    v = Δ x Δ t
  • Instantaneous speed is the speed at a specific instant in time.
  • Instantaneous velocity is the velocity at a specific instant in time.
  • Acceleration ( a ) is the change in velocity ( Δ x ) over a time interval ( Δ t ):
    a = Δ v Δ t
  • The gradient of a position - time graph ( x vs. t ) give the velocity.
  • The gradient of a velocity - time graph ( v vs. t ) give the acceleration.
  • The area under a velocity - time graph ( v vs. t ) give the displacement.
  • The area under an acceleration - time graph ( a vs. t ) gives the velocity.
  • The graphs of motion are summarised in [link] .
  • The equations of motion are used where constant acceleration takes place:
    v f = v i + a t Δ x = ( v i + v f ) 2 t Δ x = v i t + 1 2 a t 2 v f 2 = v i 2 + 2 a Δ x

End of chapter exercises: motion in one dimension

  1. Give one word/term for the following descriptions.
    1. The shortest path from start to finish.
    2. A physical quantity with magnitude and direction.
    3. The quantity defined as a change in velocity over a time period.
    4. The point from where you take measurements.
    5. The distance covered in a time interval.
    6. The velocity at a specific instant in time.
  2. Choose an item from column B that match the description in column A. Write down only the letter next to the question number. You may use an item from column B more than once.
    Column A Column B
    a. The area under a velocity - time graph gradient
    b. The gradient of a velocity - time graph area
    c. The area under an acceleration - time graph velocity
    d. The gradient of a displacement - time graph displacement
  3. Indicate whether the following statements are TRUE or FALSE. Write only 'true' or 'false'. If the statement is false, write down the correct statement.
    1. A scalar is the displacement of an object over a time interval.
    2. The position of an object is where it is located.
    3. The sign of the velocity of an object tells us in which direction it is travelling.
    4. The acceleration of an object is the change of its displacement over a period in time.
  4. [SC 2003/11] A body accelerates uniformly from rest for t 0 seconds after which it continues with a constant velocity. Which graph is the correct representation of the body's motion?
    (a) (b) (c) (d)
  5. [SC 2003/11] The velocity-time graphs of two cars are represented by P and Q as shown
    The difference in the distance travelled by the two cars (in m) after 4 s is ...
    1. 12
    2. 6
    3. 2
    4. 0
  6. [IEB 2005/11 HG] The graph that follows shows how the speed of an athlete varies with time as he sprints for 100 m.
    Which of the following equations can be used to correctly determine the time t for which he accelerates?
    1. 100 = ( 10 ) ( 11 ) - 1 2 ( 10 ) t
    2. 100 = ( 10 ) ( 11 ) + 1 2 ( 10 ) t
    3. 100 = 10 t + 1 2 ( 10 ) t 2
    4. 100 = 1 2 ( 0 ) t + 1 2 ( 10 ) t 2
  7. [SC 2002/03 HG1] In which one of the following cases will the distance covered and the magnitude of the displacement be the same?
    1. A girl climbs a spiral staircase.
    2. An athlete completes one lap in a race.
    3. A raindrop falls in still air.
    4. A passenger in a train travels from Cape Town to Johannesburg.
  8. [SC 2003/11] A car, travelling at constant velocity, passes a stationary motor cycle at a traffic light. As the car overtakes the motorcycle, the motorcycle accelerates uniformly from rest for 10 s. The following displacement-time graph represents the motions of both vehicles from the traffic light onwards.
    1. Use the graph to find the magnitude of the constant velocity of the car.
    2. Use the information from the graph to show by means of calculation that the magnitude of the acceleration of the motorcycle, for the first 10 s of its motion is 7,5 m · s - 2 .
    3. Calculate how long (in seconds) it will take the motorcycle to catch up with the car (point X on the time axis).
    4. How far behind the motorcycle will the car be after 15 seconds?
  9. [IEB 2005/11 HG] Which of the following statements is true of a body that accelerates uniformly?
    1. Its rate of change of position with time remains constant.
    2. Its position changes by the same amount in equal time intervals.
    3. Its velocity increases by increasing amounts in equal time intervals.
    4. Its rate of change of velocity with time remains constant.
  10. [IEB 2003/11 HG1] The velocity-time graph for a car moving along a straight horizontal road is shown below.
    Which of the following expressions gives the magnitude of the average velocity of the car?
    1. Area A t
    2. Area A + Area B t
    3. Area B t
    4. Area A - Area B t
  11. [SC 2002/11 SG] A car is driven at 25 m · s - 1 in a municipal area. When the driver sees a traffic officer at a speed trap, he realises he is travelling too fast. He immediately applies the brakes of the car while still 100 m away from the speed trap.
    1. Calculate the magnitude of the minimum acceleration which the car must have to avoid exceeding the speed limit, if the municipal speed limit is 16.6 m · s - 1 .
    2. Calculate the time from the instant the driver applied the brakes until he reaches the speed trap. Assume that the car's velocity, when reaching the trap, is 16.6 m · s - 1 .
  12. A traffic officer is watching his speed trap equipment at the bottom of a valley. He can see cars as they enter the valley 1 km to his left until they leave the valley 1 km to his right. Nelson is recording the times of cars entering and leaving the valley for a school project. Nelson notices a white Toyota enter the valley at 11:01:30 and leave the valley at 11:02:42. Afterwards, Nelson hears that the traffic officer recorded the Toyota doing 140 km · hr - 1 .
    1. What was the time interval ( Δ t ) for the Toyota to travel through the valley?
    2. What was the average speed of the Toyota?
    3. Convert this speed to km · hr - 1 .
    4. Discuss whether the Toyota could have been travelling at 140km · hr - 1 at the bottom of the valley.
    5. Discuss the differences between the instantaneous speed (as measured by the speed trap) and average speed (as measured by Nelson).
  13. [IEB 2003/11HG] A velocity-time graph for a ball rolling along a track is shown below. The graph has been divided up into 3 sections, A, B and C for easy reference. (Disregard any effects of friction.)
    1. Use the graph to determine the following:
      1. the speed 5 s after the start
      2. the distance travelled in Section A
      3. the acceleration in Section C
    2. At time t 1 the velocity-time graph intersects the time axis. Use an appropriate equation of motion to calculate the value of time t 1 (in s).
    3. Sketch a displacement-time graph for the motion of the ball for these 12 s. (You do not need to calculate the actual values of the displacement for each time interval, but do pay attention to the general shape of this graph during each time interval.)
  14. In towns and cities, the speed limit is 60 km · hr - 1 . The length of the average car is 3.5 m, and the width of the average car is 2 m. In order to cross the road, you need to be able to walk further than the width of a car, before that car reaches you. To cross safely, you should be able to walk at least 2 m further than the width of the car (4 m in total), before the car reaches you.
    1. If your walking speed is 4 km · hr - 1 , what is your walking speed in m · s - 1 ?
    2. How long does it take you to walk a distance equal to the width of the average car?
    3. What is the speed in m · s - 1 of a car travelling at the speed limit in a town?
    4. How many metres does a car travelling at the speed limit travel, in the same time that it takes you to walk a distance equal to the width of car?
    5. Why is the answer to the previous question important?
    6. If you see a car driving toward you, and it is 28 m away (the same as the length of 8 cars), is it safe to walk across the road?
    7. How far away must a car be, before you think it might be safe to cross? How many car-lengths is this distance?
  15. A bus on a straight road starts from rest at a bus stop and accelerates at 2 m · s - 2 until it reaches a speed of 20 m · s - 1 . Then the bus travels for 20 s at a constant speed until the driver sees the next bus stop in the distance. The driver applies the brakes, stopping the bus in a uniform manner in 5 s.
    1. How long does the bus take to travel from the first bus stop to the second bus stop?
    2. What is the average velocity of the bus during the trip?

Questions & Answers

how do you translate this in Algebraic Expressions
linda Reply
Need to simplify the expresin. 3/7 (x+y)-1/7 (x-1)=
Crystal Reply
. After 3 months on a diet, Lisa had lost 12% of her original weight. She lost 21 pounds. What was Lisa's original weight?
Chris Reply
what's the easiest and fastest way to the synthesize AgNP?
Damian Reply
types of nano material
abeetha Reply
I start with an easy one. carbon nanotubes woven into a long filament like a string
many many of nanotubes
what is the k.e before it land
what is the function of carbon nanotubes?
I'm interested in nanotube
what is nanomaterials​ and their applications of sensors.
Ramkumar Reply
what is nano technology
Sravani Reply
what is system testing?
preparation of nanomaterial
Victor Reply
Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
Himanshu Reply
good afternoon madam
what is system testing
what is the application of nanotechnology?
In this morden time nanotechnology used in many field . 1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc 2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc 3- Atomobile -MEMS, Coating on car etc. and may other field for details you can check at Google
anybody can imagine what will be happen after 100 years from now in nano tech world
after 100 year this will be not nanotechnology maybe this technology name will be change . maybe aftet 100 year . we work on electron lable practically about its properties and behaviour by the different instruments
name doesn't matter , whatever it will be change... I'm taking about effect on circumstances of the microscopic world
how hard could it be to apply nanotechnology against viral infections such HIV or Ebola?
silver nanoparticles could handle the job?
not now but maybe in future only AgNP maybe any other nanomaterials
I'm interested in Nanotube
this technology will not going on for the long time , so I'm thinking about femtotechnology 10^-15
can nanotechnology change the direction of the face of the world
Prasenjit Reply
At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
Ali Reply
the Beer law works very well for dilute solutions but fails for very high concentrations. why?
bamidele Reply
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Smarajit Reply
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The fundamental frequency of a sonometer wire streached by a load of relative density 's'are n¹ and n² when the load is in air and completly immersed in water respectively then the lation n²/na is
Mukesh Reply
Properties of longitudinal waves
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Source:  OpenStax, Siyavula textbooks: grade 10 physical science [caps]. OpenStax CNX. Sep 30, 2011 Download for free at http://cnx.org/content/col11305/1.7
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