Forces and Motion PDF

4 FORCES AND MOTION FORCES AND MOTION Speed is a term that is often used in everyday fife. Action films often feature...

4 FORCES AND MOTION FORCES AND MOTION Speed is a term that is often used in everyday fife. Action films often feature SPEED TRAP Suppose you want to find the speed of cars driving down your road. You may high-speed chases. Speed is a cause of fatal accidents on the road. Sprinters aim for greater speed in competition with other athletes. Rockets must reach have seen the police using a mobile speed camera to check that drivers are a high enough speed to put communications satellites in orbit around the keeping to the speed limit. Speed guns use microprocessors (computers on a Earth. This chapter will explain how speed is defined and measured and 'chip') to produce an instant reading of the speed of a moving vehicle, but you how distance-time graphs are used to show the movement of an object can conduct a very simple experiment to measure car speed. as time passes. We shall then look at changing speed - acceleration and Measure the distance between two points along a straight section of road with deceleration. We shall use velocity-time graphs to find the acceleration of an a tape measure or 'click' wheel. Use a stopwatch to measure the time taken object. We shall also find how far an object has travelled using its velocity-time for a car to travel the measured distance. Figure 1.4 shows you how to operate graph. You will fin d out about the difference between speed and velocity on your 'speed trap'. page 6. Measure 50 m from a start point along the side of the road. AVERAGE SPEED 2 Start a stopwatch when your partner signals that the car is passing the start point. A car travels 1 00 kilometres in 2 hours so the average speed of the car is.6. Figure 1.3 Astopwatch will measure the time KEY POINT taken for the vehicle to travel the distance. 3 Stop the stopwatch when the car passes you at the finish point. 50 km/h. You can work this out by doing a simple calculation using the Sometimes you may see ' d' used following definition of speed: as the symbol for distance travelled, but in this book 's' will be used to be average speed, v = distance moved, s consistent with the symbol used in time taken, t A level maths and physics. V=~ t The average speed of the car d uring the journey is the total distance travelled, d ivided by the time taken for the journey. ff you look at the speedometer in a car you will see that the speed of the car changes from instant to instant as the accelerator or brake is used. The speedometer therefore shows the instantaneous speed of the car. UNITS OF SPEED Typically the distance moved is measured in metres and time taken in seconds, so the speed is in metres per second (m/s). Other units can be used for speed, such as kilometres per hour (km/h), or centimetres per second.6. Figure 1.4 How to measure the speed of cars driving on the road (cm/s). In physics the units we use are metric, but you can measure speed in miles per hour (mph). Many cars show speed in both mph and kilometres per Using the measurements made with your speed trap, you can work out the hour (kph or km/h). Exam questions should be in metric units, so remember No measurements should be taken speed of the car. Use the equation: that m is the abbreviation for metres (and not miles). on the public road or pavement but it average speed, v = distance moved, s is possible to do so within the school time taken, t boundary within sight of the road. REARRANGING THE SPEED So if the time measured is 3.9 s, the speed of the car in this experiment is: ff you are g iven information about speed and time taken, you will be expected EQUATION to rearrange the speed equation to make the distance moved the subject: KfY POINT 50 m average speed, v = You can convert a speed in mis into a 3.9s distance moved, s = average speed, v x time, t speed in km/h. = 12.8 m/s and to make the t ime taken the subject if you are given the distance moved If the car travels 12.8 metres in one and speed: second it will travel DISTANCE-TIME GRAPHS time taken, t = distance moved, s 12.8 x 60 metres in 60 seconds (that is, average speed, v one minute) and 12.8 x 60 x 60 metres in 60 minutes REMINDER (that is, 1 hour). which is 46 080 metres in an hour or 46.1 km/h To use the triangle method to rearrange an equation, cover up the part of the triangle (to one decimal place). that you want to find. For example, in Figure 1.2, if you want to work out how long (I) it takes to move a distance (s) at a given speed (v), covering I in Figure 1.2 leaves We have multiplied by 3600 (60 x 60) to convert from mis to m/h, then divided.6. Figure 1.2 You can use the triangle method f. or distance divided by speed. If an examination question asks you to write out by 1000 to convert from mlh to km/h for rearranging equations likes = v x t. the equation for calculating speed, distance or time, always give the actual equation (as there are 1000 m in 1 km). (such ass = v x t). You may not get the mark if you just draw the triangle. Rule: to convert mis to km/h simply multiply by 3.6..6. Figure 1.5 A car travelling at constant speed 6 FORCES AND MOTION FORCES AND MOTION Figure 1.5 shows a car travelling along a road. It shows the car at 0.5 second KEY POINT Displacement is an example of a vector. Vector quantities have magnitude intervals. The distances that the car has travelled from the start position after (size) and a specific direction. A vector is a quantity that has both size each 0.5 s time interval are marked on the picture. The picture provides a and direction. Displacement is distance Velocity is also a vector. Velocity is speed in a particular d irection. If a car record of how far the car has travelled as time has passed. The table below travelled in a particular direction. travels at 50 km/h around a bend, its speed is constant but its velocity will be shows the data for this car. You will be expected to plot a graph of the distance Force is another example of a vector changing for as long as the direction that the car is travelling in is changing. travelled (vertical axis) against time (horizontal axis) as shown in Figure 1.6. that you will meet in Chapter 2. The increase in displacement size of a force and the direction in average velocity = time taken Time from start/s 0.0 0.5 1.0 1.5 2.0 2.5 which it acts are both important. Distance travelled from start/m 0.0 6.0 12.0 18.0 24.0 30.0 Note that this graph slopes down Hil:iiii+ to the right. We call this a The global positioning system (GPS) in Figure 1.9 shows two points on a The distance-time graph tells us about how the car is travelling in a much NEGATIVE SLOPE or negative gradient. journey. The second point is 3 km north-west of the first. 30 more convenient fonm than the series of drawings in Figure 1.5. We can see a A walker takes 45 minutes to travel from the first point to the second. 24 that the car is travelling equal distances in equal time intervals - it is moving Calculate the average velocity of the walker. E at a steady or constant speed. This fact is shown immediately by the fact that b Explain why the average speed of the walker must be greater than this. QJ 18 the graph is a straight line. The slope or gradient of the line tells us the speed ~ of the car - the steeper the line the greater the speed of the car. So in this a Write down what you know: ·" 12 example: c " ~ increase in displacement is 3 km north-west 6 speed = gradient = distance = 30 m = 12 mis time taken is 45 min (45 min = 0.75 h) time 2.5s 1J ~ increase in displacement 0+-- ~ - ~ - ~ - - ~ a b C 'o ~ - - - - - - - - - -.......... average velocity = time taken 1b 0.0 0.5 1.0 1.5 2.0 2.5 time/s 1U_ time/s object A.6. Figure 1.8 In this graph displacement is = 4 km/h north-west.6. Figure 1.6 Distance-time graph for the { decreasing with time. travelling car in Figure 1.5 g b The walker has to follow the roads, so the distance walked is greater ~ KEY POINl than the straight-line distance between A and B (the displacement). The '6 Always show your working when walker's average speed (calculated using distance) must be greater than answering questions. You should show 1r·.. ,... _.. "'... -.'"~,... time/s time/s time/s his average velocity (calculated using displacement). your working by putting the values ~ KEY POINT given in the question into the equation. ~ A curved line on distance-time graphs means that the speed or velocity of the object is changing. To find the speed at a particular instant of time we would draw a tangent to the curve at that an aid to navigation that uses orbiting satellites to locate its position instant and find the gradient of the tangent. on the Earth's surface. time/s time/s.6. Figure 1.7 Examples of distance-time graphs In Figure 1.7a the distance is not changing with time - the line is horizontal. ACTIVITY 1 This means that the speed is zero. In Figure 1.7b the graph shows how two objects are moving. The red line is steeper than the blue line because object ~ PRACTICAl: INVESTIGATE iTHE MOTION OF. EVERYDAY OBJECTS SUCH AS TOY CARS OR A is moving at a higher speed than object B. In Figure 1.7c the object is TENNIS BALLS speeding up (accelerating) shown by the graph line getting steeper (gradient getting bigger). In Figure 1.7d the object is slowing down (decelerating). You can use the following simple apparatus to investigate the motion of a toy car. You could use this to measure the average speed, v of the car for different values of h. Heavy wooden runways need THE DIFFERENCE BETWEEN SPEED AND VELOCITY to be stacked and moved carefully. They are best used Some displacement-time graphs look like the one shown in Figure 1. 7 e. It is at low level rather than being a straight line, showing that the object is moving with constant speed, but placed on benches or tables the line is sloping down to the right rather than up to the right. The gradient of where they may fall off. If such a line is negative because the distance that the object is from the starting heavy trolleys are used as point is now decreasing - the object is going back on its path towards the 'vehicles', a 'catch box' filled with bubble wrap or similar start. Displacement means 'distance travelled in a particular direction' from a material should be placed at specified point. So if the object w as originally travelling in a northerly direction, the end of the runway. the negative gradient of the graph means that it is now travelling south..6. Figure 1.t OInvestigating how a toy car rolls down a slope 8 FORCES AND MOTION FORCES AND MOTION You need to measure the height, h , of the raised end of the wooden track. The track must be securely clamped at ACCELERATION the height under test and h should be measured with a metre rule making sure that the rule is perpendicular to the bench surface. Make sure that you always measure to the same point or mark on the raised end of the track Figure 1.12 shows some objects whose speed is changing. The plane must (a fiducial mark). accelerate to reach take-off speed. In ice hockey, the puck (small disc that the player hits} decelerates only very slowly when it slides across the ice. When To find the average speed you will use the equation: the egg hits the ground it is forced to decelerate (decrease its speed} very d distance moved, s rapidly. Rapid deceleration can have destructive results. average spee ' v = lime taken, t so you will need to measure the distance AB with a metre rule and measure the time it takes for the car to travel this distance with a stop c lock. When timing with a stop clock, human reaction time will introduce measurement errors. To make these smaller the time to travel distance AB should, for a given value of h, be measured at least I three times and an average value found. Always start the car from the same point, A. If one value is quite different from the others it should be treated as anomalous (the result is not accurate) and ignored or repeated. l The results should be presented in a table like the one b e l o ~ You do not need to include these equations in your table headings but you may be asked to IDistance/m AB: I show how you did the calculations..& Figure 1.12 Acceleration...... constant speed...... and deceleration Height, h/m Time, t/s Average time, t/s - Average speed, v/m/s Acceleration is the rate at which objects change their velocity. It is defined as t = (t1 + t2 + t3)+ 3 V =AB + t t1 I t2 I t3 follows: I I acceleration, a= change in velocity or final velocity, v - initial velocity, u time taken, t time taken, t In a question you may be given a complete set of results or you may be required to complete the table by (v-u) doing the necessary calculations. You may be asked to plot a graph (see general notes above) and then draw a a= - t- conclusion. The conclusion you draw m ust be explained with reference to the graph, for example, if the best fit line through the plotted points is a straight line and it passes through the origin (the 0, 0 point) you can conclude that Why u? Simply because it comes before v! there is a proportional relationship between the quantities you have plotted on the graph. Acceleration, like velocity, is a vector because the direction in which the Some alternative methods acceleration occurs is important as well as the size of the acceleration. You could investigate the motion of moving objects using photographic methods either by: carrying out the experiment in a darkened room using a stroboscope to light up the object at regular known intervals (found from the frequency setting on the stroboscope} with the camera adjusted so that the shutter is UNITS OF ACCELERATION Velocity is measured in mis, so increase in velocity is also measured in mis. open for the duration of the movement, or Acceleration, the rate of increase in velocity with time, is therefore measured using a video camera and noting how far the object has travelled between each frame - the frame rate will allow in mls/s (read as ' metres per second per second'}. We normally write this as you to calculate the time between each image. mls2 (read as 'metres per second squared'}. Other units may be used - for example, cmls2. In either case a clearly marked measuring scale should be visible. Or you could use an electronically operated stop clock and electronic t iming gates. This will let you measure the time that it takes for the moving object to travel over a measured distance. This has the advantage of removing ·Fil:iiii- A car is travelling at 20 mis. It accelerates steadily for 5 s, after which time timing errors produced by human reaction time. it is travelling at 30 mis. Calculate its acceleration. You can also use timing gates to measure how the speed of the object changes as it moves. Write down what you know: ~ - timing gate initial or starting velocity, u = 20 mis card strip final velocity, v = 30 mis time taken, t = 5 s a= (v-u) t 30 mis - 20 mis.& Figure 1.11 Using a timing gate is a more accurate method for measuring time taken to travel a distance. 5s In this arrangement the stop clock will time while the card strip attached to the moving car passes through the 10mls timing gate. Measuring the length of the card strip and the time it takes for the card strip to pass through the timing It is good practice to include units in 5s gate allows you to calculate the average speed of the car as it passes through the timing gate. equations - this will help you to supply the answer with the correct unit. The car is accelerating at 2 mls2. 10 FORCES AND MOTION FORCES AND MOTION Galileo also noticed that the distance travelled by the ball increased in a DECELERATION Deceleration means slowing down. This means that a decelerating object will predictable way. He showed that the rate of increase of speed was steady or have a smaller final velocity than its starting velocity. If you use the equation uniform. We call this uniform acceleration. Most acceleration is non-uniform - for finding the acceleration of an object that is slowing down, the answer will that is, it changes from instant to instant - but we shall only deal with uniformly have a negative sign. A negative acceleration simply means deceleration. accelerated objects in this chapter. Ph/Hit VELOCITY-TIME GRAPHS An object hits the ground travelling at 40 mis. It is brought to rest in 0.02 s. What is its acceleration? The table below shows the distances between the bells in an experiment such Write down what you know: as Galileo's. initial velocity, u = 40 mis Bell 2 3 4 5 final velocity, v = 0 mis Time/s 0.5 1.0 1.5 2.0 2.5 time taken, t = 0.02 s o,stance of bell from start/cm 12 27 48 75 a = (v-u) t We can calculate the average speed of the ball between each bell by working Omls - 40m/s out the distance travelled between each bell, and the time it took to travel this 0.02s distance. For the first bell: -40mls distance moved, s 0.02s ve1ocity, v = time taken, t 2 =-2000 mls = 3 cm =6cmls 0.5s In Example 3, we would say that the object is decelerating at 2000 mls2. This is a very large deceleration. Later, in Chapter 3, we shall discuss the This is the average velocity over the 0.5 second time interval, so if we plot it on consequences of such a rapid deceleration! a graph we should plot it in the middle of the interval, at 0.25 seconds. Repeating the above calculation for all the results gives us the following table of results. We can use these results to draw a graph showing how the velocity MEASURING ACCELERATION When a ball is rolled down a slope it is clear that its speed increases as it rolls - of the ball is changing with time. The graph, shown in Figure 1.14, is called a that is, it accelerates. Galileo was interested in how and why objects, like the ball velocity-time graph. EXTENSION WORK rolling down a slope, speed up, and he created an interesting experiment to learn more about acceleration. A version of his experiment is shown in Figure 1.13. Time/s 0.25 0.75 1.25 1.75 2.25 Galileo was an Italian scientist who was born in 1564. He developed a telescope, which he used to study the M lffliffd & 6 18 30 42 54 movement of the planets and stars. He KEY POINT The graph in Figure 1.14 is a straight line. This tells us that the velocity of the also carried out many experiments on motion (movement). rolling ball is increasing by equal amounts in equal time periods. We say that The equations of motion we have learned work for uniform or constant the acceleration is uniform in this case. EXTENSION WORK acceleration only - therefore for objects 60 with velocity-time graphs that are Though Galileo did not have a clock or straight lines. watch (let alone an electronic timer), he 50 used his pulse (the sound of his heart) and a type of water clock to achieve j 40 timings that were accurate enough for.t. Figure 1.13 Galileo's experiment. A ball rolling down a slope, hitting small bells as it rolls } 30 his experiments. ~ ~ 20 Galileo wanted to discover how the distance travelled by a ball depends on the time it has been rolling. In this version of the experiment, a ball rolling down 10 a slope strikes a series of small bells as it rolls. By adjusting the positions of 0 ----,----.--,--,----, the bells carefully it is possible to make the bells ring at equal intervals of time 0.0 0.5 1.5 2.5 as the ball passes. Galileo noticed that the distances travelled in equal time time/s intervals increased, showing that the ball was travelling faster as t ime passed. Galileo did not have an accurate way of measuring time (there were no digital.t. Figure 1.14 Velocity-time graph for an experiment in which a ball is rolled down a slope. (Note stopwatches in seventeenth-century Italy!) but it was possible to judge equal that as we are plotting average velocity, the points are plotted in the middle of each successive time intervals accurately simply by listening. 0.5 s time interval.) 12 FORCES AND MOTION FORCES AND MOTION A MODERN VERSION OF GALILEO'S EXPERIMENT A cylinder vacuum cleaner (or similar) - ' , GRADIENT The results of the air-track experiments in Figure 1.1 6 show that the slope of the velocity-time g raph depends on the acceleration of the glider. The slope or gradient of a velocity-time graph is found by dividing the increase in the velocity by the time taken for the increase, as shown in Figure 1.17. In this example an object is travelling at u mis at the beginning and accelerates uniformly (at a constant rate) for ts. Its final velocity is v mis. Inc rease in velocity divided by time is, you will recall, the definition of acceleration (see used with the air-track should be placed on the floor as it may fall off a page 9), so we can measure the acceleration of an object by finding the slope bench or stool. Also, beware of any of its velocity-time graph. The meaning of the slope or gradient of a velocity- trailing leads. time graph is summarised in Figure 1.1 7. sloping air-track HINT When finding the gradient of a graph, draw a big triangle. 2 Choose a convenient number of units for the length of the base of the triangle to make the division easier. & Figure 1.15 Measuring acceleration time Today we can use data loggers to make accurate direct measurements that are & Figure 1.17 Finding the gradient of a velocity-time graph __ 't= ·~ collected and analysed by a computer. A spreadsheet program can be used to l~~ ll produce a velocity-time graph. Figure 1.15 shows a glider on a slightly sloping air-track. The air-track reduces friction because the glider rides on a cushion of air that is pushed continuously through holes along the air-track. As the glider accelerates down the sloping track the card stuck on it breaks a fight beam, and the time that the glider takes to pass is measured electronically. ff the length of the card is measured, and this is entered into the spreadsheet, the velocity of the I I I I glider can be calculated by the spreadsheet program using v = f· a shallow gradient - low acceleration b steep gradient - high acceleration c horizontal (zero gradient) - d negative gradient - negative acceleration no acceleration (deceleration) Figure 1.16 shows velocity-time graphs for two experiments done using the air-track apparatus. In each experiment the track was given a d ifferent slope. & Figure 1.18 The gradient of a velocity-time graph gives you information about the motion of an The steeper the slope of the air-track the greater the glider's acceleration. This object at a glance. is clear from the graphs: the greater the acceleration the steeper the gradient of the graph. AREA UNDER A VELOCITY-TIME GRAPH GIVES DISTANCE TRAVELLED Figure 1.19a shows a velocity-time graph for an object that travels with a The gradient of a velocity-time graph gives the acceleration. constant velocity of 5 mis for 10 s. A simple calculation shows that in this time the object has travelled 50 m. This is equal to the shaded (coloured) area under Air-track at 1.5° Air-track at 3.0° the graph. Figure 1.19b show s a velocity-time graph for an object that has 100 Timels Av Vel. Time/s Av Vel. accelerated at a constant rate. Its average velocity during this time is given by: ~ /cm/s /cm/s E 80. initial velocity + final velocity u+v 0.00 0.0 0.00 0.0 = ! 60 0.45 11.1 0.32 15.9 average velocity 2 or - 2 - " Q 40 1.35 33.3 0.95 47.6 fn this example the average velocity is, therefore: !I 2.25 55.6 1.56 79.4 20 average velocity = Omis + 1 O mis 3.15 77.8 2.21 111.1 2 2 3 time/s which works out to be 5 mi s. ff the object travels, on average, 5 metres in each second it will have travelled 20 metres in 4 seconds. Notice that this, too, is & Figure 1.16 Results of two air-track experiments. (Note, once again, that because we are equal to the shaded area under the graph (given by the area equation for a plotting average velocity in the velocity-time graphs, the points are plotted in the middle of each i t riangle: area = base x height). successive time interval - see page 11) 14 FORCES AND MOTION FORCES AND MOTION The area under a velocit y-time graph is equal to the distance travelled by HINT (displacement of) the object in a particular time interval. MiHii- l Find the distance travelled for more complicated velocity-time graphs by a b area of a triangle = !base x height A cylinder containing a vaccine is dropped from a helicopter hovering at l"------10s - a height of 200 m above the ground. The acceleration due to gravity is dividing the area beneath the graph 10+····················· ····· ··········, line into rectangles and triangles. Take 5- r 1O m/s2. Calculate the speed at which the cylinder will hit the ground. care that units on the velocity and time i area =5m/sx 10s i You are given the acceleration, a= 10 mls2, and the distance, s = 200 m, axes use the same units for time, for Sm/s example, m/s ands, or km/h and h. ~ ·g J =50m = distance travelled i Im/s through which the cylinder moves. The initial velocity, u, is not stated, but you assume it is O mis as the helicopter is hovering (staying in one place in the air). Substitute these values in the given equation: o+-~-..--~~~+-'- o 2 4 6 8 10 1 3 4 v 2 = u 2 + 2as time/s time/s = O m/s2 + (2 x 1 O mls2 x 200 m) Figure 1.19 a An object travelling at constant velocity; b An object accelerating at a constant rate = 4000 m 2/s2 therefore v = J(4000 m 2/s2) EQUATIONS OF UNIFORMLY ACCELERATED MOTION = 63.25 mis You must remember the equation: v-u a= - 1- and be able to use it to calculate the acceleration of an object. LOOKING AHEAD You may need to rearrange the equation to make another term the subject. The equations you have seen in this chapter are called the equations PiN!it of uniformly accelerated motion. This means that they will give you correct answers when solving any problems that have objects moving A stone accelerates from rest uniformly at 10 m/s2 when it is dropped with constant acceleration. In your exam you will only see p roblems down a deep well. It hits the water at the bottom of the well after 5 s. ~ ~ where this is the case or very nearly so. Examples in which objects Calculate how fast it is travelling when it hits the water. accelerate or decelerate (slow down) at a constant rate often have a You will need to make v the subject of this equation: constant acceleration due to the Earth's gravity (which we take as about ~~ v-u a= - 1- 10 m/s2). In real life, problems may not be quite so simple! Objects only fall with Figure 1.20 Cover v- utofind v- u= ax t constant acceleration if we ignore air resistance and the distance that You can use the triangle method to show that v- u = a x t they fall is quite small. then add u t o both sides of the equation to give: V = U +at These equations of uniformly accelerated motion are often called the 'suvat' equations, because they show how the terms s (distance moved), (In words this tells you that the final velocity is the initial velocity plus the increase in velocity after accelerating for t seconds.) u (velocity at the start), v (velocity at the finish), a (acceleration) and t (time) are related. State the things you have been told: initial velocity, u = o mis (It was stationary (standing still) at the start.) acceleration, a = 10 m/s2 time, t, of the acceleration = 5 s CHAPTER QUESTIONS More questions on speed and acceleration can be found at the end of Unit 1 Substitute these into the equation: v = 0 mis + (10 m/s2 x 5 s) on page 55. Then calculat e the result. The stone hit the water travelling at 50 m/s (downwards). mD PROBLEM SOLVING A sprinter runs 100 metres in 12.5 seconds. Calculate the speed in m /s. 2 A jet can travel at 350 m/ s. Calculate how far it will travel at this speed in: You will also be required to use the following equation of uniformly accelerated a 30 seconds motion: b 5 m inutes (final speed)2, v2 = (initial speed)2, u2 + (2 x acceleration, a x distance moved, s) c half an hour. v2 = u2 +2as 3 A snail crawls at a speed of 0.0 004 m/s. How long will it take to climb a garden stick 1.6 m high ? 16 FORCES AND MOTION FORCES AND MOTION..ANALYSIS 4 Look at the following distance-time graphs of moving objects. mD PROBLEM SOLVING 12 A plane starting from rest accelerates at 3 mls2 for 25 s. Calculate the 1L it:= 1l:1b increase in velocity after: a 1s b 5s C 25 S. time time time time A B C D.. ANALYSIS 13 Look at the following sketches of velocity-time graphs of moving objects. Identify in which graph the object is: a moving backwards b moving slowly c moving quickly d not moving at all. rL: iC t~ ,~ time A time B time C time D mD INTERPRETATION 5 Sketch a distance-time graph to show the motion of a person walking quickly, In which graph is the object: stopping for a moment, then continuing to walk slowly in the same direction. a not accelerating 6 Plot a distance-time graph using the data in the following table. Draw b accelerating from rest a line of best fit and use your graph to find the speed of the object c decelerating concerned. d accelerating at the greatest rate? l @Md o.oo 1.60 3.25 4.80 6.35 8.00 9.60 mD INTERPRETATION 14 Sketch a velocity-time graph to show how the velocity of a car travelling MtitM o.oo 0.05 0.10 0.15 0.20 0.25 0.30 along a straight road changes if it accelerates uniformly from rest for 5 s, mD PROBLEM SOLVING 7 The diagram below shows a trail of oil drips made by a car as it travels travels at a constant velocity for 1O s, then brakes hard to come to rest in 2 s. along a road. The oil is dripping from the car at a steady rate of one drip every 2.5 seconds. 15 a Plot a velocity-time graph using the data in the following table: oil drips on the road I\ MWIM o.o 2.5 5.o 7.5 10.0 10.0 10.0 10.0 10.0 10.0 I \ MiitM o.o 1.0 2.0 3.o 4.o 5.o 6.o 1.0 a.o 9.o Draw a line of best fit and use your graph to find: b the acceleration during the first 4 s a Describe the way the car is moving. c the distance travelled in: i the first 4 s of the motion shown b The distance between the first and the seventh drip is 135 metres. ii the last 5 s of the motion shown Determine the average speed of the car. d the average speed during the 9 seconds of motion shown. 8 A car is travelling at 20 mis. It accelerates uniformly at 3 m/s2 for 5 s. a Sketch a velocity-time graph for the car during the period that it is mD CRITICAL THINKING 16 The dripping car from Question 7 is still on the road! It is still dripping oil but now at a rate of one drop per second. The trail of drips is shown on the accelerating. Include numerical detail on the axes of your graph. diagram below as the car travels from left to right. b Calculate the distance the car travels while it is accelerating. mD INTERPRETATION 9 Explain the difference between the following terms: a average speed and instantaneous speed b speed and velocity. Describe the motion (the way the car is moving) using the information in mD PROBLEM SOLVING 10 A sports car accelerates uniformly from rest to 24 mis in 6 s. Calculate the this diagram. acceleration of the car. mD PROBLEM SOLVING 17 This question uses the equation v2 = u2 + 2as. mD..,.., i: 1 a Explain what each of the terms in this equation represents. ,.., INTERPRETATION 11 Sketch velocity-time graphs for an object:.'ti:.. a moving with a constant velocity of 6 mis b A ball is thrown vertically upwards at 25 mis. Gravity causes the ball to decelerate at 10 mls2. Calculate the maximum height the ball will reach. b accelerating uniformly from rest at 2 mls2 for 1O s c decelerating to rest at 4 mls2 for 5 s. Include numbers and units on the velocity and time axes in each case.