2.1 2. Learning Standard 2.1.1 Describe the type of linear motion of an object in the following states (i) Stationary (ii) uniform velocity (iii) non-uniform velocity 2.1.2 Determine: (i) distance and displacement (ii) speed and velocity (iii) acceleration / deceleration 2.1.3 Solve problems involving linear motion using the following equations: (i) a = u + at (ii) s = ½ (u + v)t (iii) s = ut + ½ at2 (iv) v2 = u2 + 2as ************************************************************* Introduction Linear motion is a study of moving object in a straight line. How would you describe the motion of the runner in words? We describe linear motion in terms of distance, displacement, speed, velocity, acceleration and deceleration. Activity 1: Aim: To study distance and displacement 1. Your task is to put a pencil at point C. Diagram below shows the difference between distance and displacement. Distance = Length of the road Displacement = Length of the line AB 3. Every day Rahim walks from his house to the junction which is 1.5 km from his house. Then he turns back and stops at warung Pak Din which is 0.5 km from his house. (a) What is Rahim’s displacement from his house (i) when he reaches the junction? 1.5 km to the right / to East (ii) (b) 2. With a pencil take a ride from position A to B and to the final destination, C. Your total path of length = 1 + 2 = 3 m this value is known as distance 3. Repeat step 2 but now from A direct to the final destination of C. This is your shortest path to the final destination, C. Value = 1.73. This value is known displacement State the definition of distance and displacement Distance: total path length travel by an object. Displacement : distance in a specified direction. When he is at warung Pak Din? 0.5 km to the left / to West After breakfast, Rahim walks back to his house. When he reaches home, (i) what is the total distance traveled by Rahim? 1.5 + 1.5 + 0.5 + 0.5 = 4.0 km (ii) what was Rahim’s total displacement from his house? 1.5 + (-1.5) + (-0.5) + 0.5 = 0 km 4. In a jungle tracking activity, a scout is given a compass and a map. He starts his journey from station A and is required to walk to station B which is located 400 m to the east of station A. When he reached station B, he is ordered to go to station C which is 400 m north from station B. 3. Give definition of acceleration and its formula. Acceleration: Rate of change of velocity a = change in velocity = final velocity – initial velocity time time a=v–u t 4. Observe the motion of the car in Diagram (a), Diagram (b) and Diagram (c). (a) When the scout is at station C, (i) what is the distance he traveled from station A? 400 + 400 = 800 Rajah (a) Rajah (b) (ii) What is the scout’s displacement from station A? State the magnitude and direction of the scout from station A. 565.69 to the north east (b) Why does the answer in (i) and (ii) different? Displacement has magnitude and direction. Aktivity 2: Aim: speed, velocity, uniform velocity, non uniform velocity, acceleration, deceleration and zero acceleration 1. Difference between speed and velocity Speed: Velocity: Rate of change of Rate of change of distance displacement SI unit: ms-1 SI unit: ms-1 Derived quantity and scalar quantity Derived quantity and vector quantity Formula v = distance = s time t Formula v = displacement = s time t Rajah (c) Describe the changes in displacement and velocity of a car. Is the car accelerate or decelerate? Diagram Diagram Diagram (a) (b) (c) displacement Equal Increases decreases Velocity Equal Increases decreases Acceleration Zero Accelerate decelerate 5. A bicyclist starts from rest and increases his velocity at a constant rate until he reaches a speed of 4.0 m/s in 5.0 s. What is his average acceleration? (a) state the initial velocity = 0 (b) state the final velocity = 4.0 (c) state the time taken = 5.0 s 2. The speedometer reading for a car traveling north shows 80 km/hr. Another car traveling at 80 km/hr towards south. Is the speed of both cars same? Is the velocity of both cars same? Speed equal. Velocity different because opposite direction. (d) calculate acceleration = 4.0 – 0 = 0.8 ms-2 5.0 Exercise 1: 1. You walk along a long straight school corridor for 55 m, then you turn around and walk 30 m in the opposite. Finally, you turn again and walk 39 m in the original direction and stop. What is your displacement from your starting point? 4. Figure 2.6 shows Radzi’s run from A to B and then back to C. The total time taken is 20 s. Determine the (a) distance Distance = AB + BC = 100 + 20 = 120 m 2. A boy walks finish the following path AB. (c) speed of Radzi’s motion. Speed = 120/20 = 6 m/s (b) displacement 100 + (-20) = 80 to right (d) velocity velocity = 80/20 = 4 m/s 5. Find: (a) total distance traveled 5 + 7 + 5 + 10 + 10 + 10 + 10 = 57 m (b) displacement 7 + 10 + 10 = 27 m 3. Fill in the blanks: (a) Constant speed 10 m/s: A constant speed of 10 m/s: A distance of 10 m is traveled every second (b) Constant velocity 10 m/s: A steady velocity of 10 m/s: A distance of 10 m is traveled every second to the right. (c) Constant velocity – 10 m/s: A constant velocity of - 10 m/s: A distance of 10 m is traveled every second to the left. (d) Constant acceleration 4 ms-2 : Speed increases by 4 m/s every second. (e) Constant acceleration 4 ms-2 : A steady deceleration of 4 ms-2 : Speed decreases by 4 m/s every second. Muthu moves from O to B along the route OAB as shown in Figure 2.7. The time taken is 15 s. Determine the (a) distance 5 + 12 = 17 m (b) displacement 13 m (c) speed of Muthu’s motion. v = 17/15 = 1.13 m/s 6. (d) velocity v = 13/15 = 0.87 m/s After landing on the runway, a plane slows down so that its velocity reduces from 75 m s–1 to 5 m s–1 in 20 s. What is the acceleration of the plane? a = 5 – 75 = - 3.5 ms-2 20 Relating Displacement, Velocity, Acceleration and Time Ticker timer: • It is connected to an alternating current power supply of 50 Hz. When it is turned on, the iron strip will vibrate 50 times per second. • The time taken to make 50 dots on the ticker tape is 1 second. Hence, the time interval between 2 consecutive dots is 1/50 = 0.02 s. • Give definition of one tick. Time interval between two dots. Activity 2: Aim: Method of calculationn To investigate motion in laboratory mean to determine distance / displacement, speed/ velocity, time and acceleration/ deceleration STEP 1: Determination of time taken for 1 tick. (a) Time taken for 50 ticks = 1.0 s (b) Time taken for 1 tick = 0.02 s (c) Time taken from A to B = 10 ticks = 0.2 s STEP 2 : Determination of displacement The displacement of the object is determined by measuring the length of the ticker tape that is pulled through the ticker time Displacement from A to B = 8.0 cm STEP 3 : Determination of velocity Velocity, v = displacement Time = 8.0 = 40.0 cm/s 0.2 STEP 4 : Determination of acceleration The first strip: Initial velocity, u at AB = 1.5 = 7.5 0.2 The last strip: Final velocity, v at DE = 7.5 = 37.5 0.2 The time interval for, t = (4 – 1) x 0.2 = 0.6 s the change in the velocity Acceleration, a = 37.5 – 7.5 = 50 cms-2 0.6 2. Calculate the acceleration . u = 0.2 = 10 cm/s v = 1.4 = 70 cm/s 0.02 0.02 t = (5 – 1) x 0.02 = 0.08 s a = 70 – 10 = 750 cms-2 = 7.5 ms-2 0.08 3. A ticker tape below contains 5 ticks for every interval AB,BC,CD and DE. Calculate the acceleration. u = 8.0 = 80 cm/s v = 2.0 = 20 cm/s 0.1 0.1 t = (4 – 1) x 0.1 = 0.3 s a = 20 – 80 = - 200 cms-2 = - 2.0 ms-2 0.3 Making a speed-time graph Ticker tape gain from the experiment: 3. The diagram above shows a ticker tape chart for a moving trolley. The frequency of the ticker-timer used is 50 Hz. Each section has 10 dots-spacing. 5. To identify the types of motion Distance between the dots: equal Type of motion: Constant velocity Distance between the dot increases Uniformly The velocity is of the object is increasing uniformly (a) What is the time between two dots? 0.02 s (b) What is the time for one strips. 10 x 0.02 = 0.2 s (c) What is the initial velocity 2.0 = 10 cm/s 0.2 (d) What is the final velocity. 12.0 = 60 cm/s 0.2 The object is moving at a uniform acceleration Distance between the dots decreases uniformly The velocity of the object is decreasing uniformly The object is (e) What is the time interval to change from initial velocity to final velocity? t = (11 – 1) x 0.2 = 2.0 s (f) What is the acceleration of the object? a = 60 – 10 = 25 cms-2 2.0 4. The figure shows a tape chart. Calculate (a) the acceleration (b) the average velocity (a) u = 4/0.2 = 20 cm/s v = 24/0.2 = 120 cm/s t = (6 – 1) x 0.2 = 1.0 a = 120 – 20 = 100 cms-2 = 1 ms-2 1.0 (b) total distance = 4 + 8 + 12 + 18 + 20 + 24 = 84 Average speed = 84 = 70.0 cm/s 1.2 experiencing uniform decelaration. Activity 4: To determine displacement, average velocity and acceleration Apparatus: Ticker timer, trolley, 12 V power supply, runway, ticker tape, ruler Procedure: 1. Raise one end of runway to a reasonable height. 2. Pass the ticker tape through the ticker timer and attach it to a trolley at the top of the runway. 3. Switch on the ticker timer and release the trolley. 4. When the trolley comes to a stop, cut the tape. 5. 6. Mark and cut the tape into 10-tick strips from the start of the first clear dot Paste the 10-tick strips side-by-side on a paper to make a tape chart. Find 1st strip Last strip The displacement of the 10-tick 16.1 cm 39.3 cm The time covered for the 10-tick strip 0.2 s 0.2 s Average velocity over the 10-tick Strip 16.1 0.2 = 80.5 cm/s 39.3 0.2 = 196.5 cm/s Change in velocity between the two 10tick strip Time taken for the change in velocity 196.5 – 80.5 = 116.0 cm/s (5 – 1) x 0.2 = 0.8 s Acceleration 116.0 = 145.0 cms-2 0.8 Conclusion: The acceleration of the trolley is 145.0 cms-2 Activity 5: Aim: Derive linear motion equations and solve problems First linear motion equation a=v–u t v – u = at v = u + at (1) Second linear motion equation Displacement = average velocity x time 1. A car accelerates from rest to 25 m s-1 in 4 s. Find the acceleration of the car. u = 0, v = 25, t = 4, a = ? v = u + at a = (25 – 0) = 6.25 ms-2 4 2. When a racing car passes through on a straight track, its velocity is 40 ms-1. After 3 seconds, the racing car achieves the speed 50 ms-1. Calculate the displacement travelled. u = 40, v = 50, t = 3, s = ? s = ½ (u + v)t = ½ (40 + 50)3 = 135 m 3. The 100 m men world record holder, Usian Bolt starts his run from rest and achieved his maximum velocity after he accelerating uniformly for 9.58 s. Determine his acceleration. s = 100 m u = 0 t = 9.58 s a = ? s = ut + ½ at2 100 = ½ a(9.58)2 a = 2.18 ms-2 4. Maria rides a bicycle at a velocity of 8 m s–1. She brakes suddenly and stops after a distance of 2 m. What is the acceleration of Maria and her bicycle? u = 12 v = 0 s = 2 a = ? v2 = u2 + 2as 0 = 144 + 2(a)2 4a = - 144 a = -16 ms-2 5. A car moving along a straight road at a velocity of 30 m s–1 reduces its velocity at a constant rate until it stops after 5 s. What is the acceleration of the car? u=0 a=4 t=5 v=? v = u + at = 0 + (4)(5) = 20 cm/s 6. A boy is cycling down a hill. His initial velocity is 4 ms-1. After he moves 35 m, his velocity becomes 10 m s-1. (a) What is the time for the boy to travel at a distance 35 m? u = 4, v = 10, s = 35 t = ? s = (u + v) t 2 35 = (4 + 10) t 2 t=5s displacement = [initial velocity + final velocity] x time 2 s = ½ (u + v) t Third linear motion equation v = u + at (1) s = ½ [u + v]t (2) Substitute equation (1) into equation (2) s = ½ [u + u + at]t s = ½ [2u + at]t s = ½ [2ut + at2] s = ut + ½ ut2 (3) Fourth linear motion equation Square equation (1) v = u + at v2 = (u + at)2 v2 = u2 + 2uat + a2t2 v2 = u2 + 2a(ut + ½ at2) at2 daripada (3) s = ut + ½ v2 = u2 + 2as (4) v = u + at .(1) Where s :displacement s= u+v t 2 .(2) s = ut + ½ at2 .(3) u :initial velocity v :final velocity a :acceleration v2 = u2 + 2as .(4) t :time (b) What it is his acceleration? u = 4, v = 10, t = 5, a = ? v = u + at a = (10 – 4) 5 a = 1.2 ms-2 7. A car accelerates from rest at 3 ms-2 along a straight road. How far has the car traveled after 4 s? u = 0, a = 3, t = 4, s = ? s = ut + ½ at2 = 0 + ½ (3)(16) = 24 m 8. A car is traveling at 20 m/s along a straight road. The driver puts the brakes on for 5 s. It this causes a deceleration of 3 m s-2 , what is the car’s final velocity? u = 20, t = 5, a = -3, v = ? v = u + at = 20 + (-3)(5) = 5 m/s 9. A car moving with constant velocity of 40 ms-1. The driver saw and obstacle in front and he immediately stepped on the brake pedal and managed to stop the car in 8 s. The distance of the obstacle from the car when the driver spotted it was 180 m. How far is the obstacle from the car after it stops. u = 40, v = 0 , t = 8 , s = ? s = (40 + 0) 8 = 160 m 2 the obstacle from the car after it stops: 180 – 160 = 20 m 10. You can determine your reaction time by catching a ruler between your fingers which is released. Ready to catch the ruler by opening your fingers at the ‘zero’ mark of the ruler. Catch the ruler as soon as it is released. Mark the position where you catch the ruler. The ruler falls with acceleration 10 m s-2, calculate your reaction time. u = 0, a = 10, s = 20 cm = 0.2 m t = ? s = ut + ½ at2 0.2 = 0 + ½ (10)t2 t2 = 0.2/5 = 0.04 t = 0.2 s TUTORIAL 2.1 1. Which physical quantity is equal to displacement? Time A. Speed C. Distance B. Velocity D. Acceleration 2. The acceleration of a car moving with a constant velocity will be A. increased B. constant C. zero D. decreased 3. Muthu cycles from his house to the shop. While coming back, he stops at Ahmad’s house. What is his displacement from his house? A. 300 m B. 400 m C. 1 100 m 4. Which of the following is true regarding the motion of an object having zero acceleration? A. The object is not moving B. The object is moving with uniform velocity C. The object is at rest or moves with uniform velocity D. The object is moving with maximum velocity. 5. A tick from a ticker timer is A. time interval between two consecutive dots on a ticker tape. B. distance between two consecutive dots on a ticker tape. C. frequency of vibration from the ticker timer D. velocity of vibration from the ticker timer. 6. Which statement is true about the ticker tape shown below? A. B. C. D. Velocity between CD is low. Velocity between DE and velocity between AB are equal. Frequency between BC is higher than DE Time between AB is equal with the time between DE A. 20 cm s-2 B. 200 cm s-2 C. 240 cm s-2 D. 400 cm s-2 u = 4/0.1 = 40 v = 24/0.1 = 240 t = (6 – 1) x 0.1 = 0.5 s a = (240 – 40)/0.5 = 400 cms-2 7. The diagram shows a ticker tape which is pulled by a trolley through a ticker-timer of frequency 50 Hz. The average speed of the trolley’s motion is A. 0.2 ms-1 C. 0.3 ms-1 B. 0.4 ms-1 D. 0.5 ms-1 Average speed = 4/0.1 = 40 cms-1 8. Diagram below shows the path travelled by a car from P to S. Average velocity from P to Q in the ticker tape below is A. B. 9. 85 cms-1 C. 170 cms-1 -1 200 cms D. 240 cms-1 Total distance = 2 + 3 + 4 + 2 + 6 = 17 cm Total time taken = 5 x 0.02 = 0.1 s v = 17/0.1 = 170 cms-1 Calculate the acceleration. A. B. 10. 11. - 50 ms-2 C. - 500 ms-2 -2 50 ms D. 500 ms-2 u = 2.5/0.02 = 125 v = 0.5/0.02 = 25 t = (11 – 1) x 0.02 = 2.0 s a = (25 – 125)/2.0 = - 50 cms-2 = - 0.5 ms-2 The diagram below shows a tape chart which is produced by a moving trolley. The frequency of the ticker-tape timer is 50 Hz and each stripe of ticker tape contains 5 ticks. What is the trolley’s acceleration? (1994) What is the displacement of the car? (2007) A. 5.0 km B. 6.8 km C. 8.2 km D. 9.0 km 12. Diagram 2 shows Ali stands at O. He walks towards A, then moves towards B and stops at B. What is the displacement of Ali? [2012] A. 2 m towards west B. 5 m towards east C. 7 m towards east PAPER 2, SECTION A 1. Question 1: Melaka Mid 08 Ali walks to the north for a distance of 300 m. He then turns west and walks for another distance of 400 m to reach Pak Kassim’s stall. Diagram 1 shows the path taken by Ali. (a) Distance is a ……… Tick (√) the correct answer in the box below √ 3. SPM 2016 Diagram 1 shows a ticker tape with 5 ticks obtained from an experiment. Scalar quantity Vector quuantity (b) (c) (d) On Diagram 1, mark an arrow to show the displacement made by Ali [1 mark] What is the total distance travelled by Ali? 300 + 400 = 700 [1 mark] Ali then walks home for another distance of 500 m. What is the total displacement made by Ali. [1 mark] 0 2. SPM 2011 Question 2 Diagram 1.1 shows a trolley moving down an inclined plane. The ticker timer vibrates at frequency 50 Hz. Diagram 1.2 shows the ticker tape produced by the motion of the trolley. (a) One tick is the time taken from P to Q [1 mark] (b) Complete the following sentence by ticking (√) the correct answer in the box provided. The ticker tape can be analysed to determine ……√…. Distance and speed of an object ………. Force and momentum of an object [1 mark] (c) (i) What is the type of motion shown by the ticker tape in Diagram 1? Constant velocity [1 mark] (ii) Give one reason for your answer in 1(c)(i). The distance between the ticks are equal. [1 mark] 4. SPM 2019 Diagram 5.1 and Diagram 5.2 show two ticker tapes with different lengths produced by the motion of two trolleys. Diagram 1.1 Diagram 5.1 Diagram 5.2 Diagram 1.2 (a) Underline the correct answer in the bracket to complete the sentence below. The type of current used in the ticker timer is (direct current, alternating current). [1 mark] (b) (i) Based on Diagram 1.2, one tick is represented by the time taken from point P to point Q [ 1 mark] (ii) Compare the distance between PQ and ST. ST > PQ 1 mark] (iii) State the type of motion of the trolley. Accelerate [1 mark] (a) What is the meaning of length? Distance between two points. [1 mark] (b) Based on Diagram 5.1 and Diagram 5.2, compare (i) the number of tickes Equal (ii) the length of the ticker tapes 5.2 > 5.1 (iii) average speed 5.2 > 5.1 [3 marks] (c) Using your answer in 5(b), state the relationship between the length of a ticker tape and average speed. Directly proportional [1 mark] (d) Diagram 5.3 shows a ticker tape chart produced from the motion of a trolley. Diagram 5.3 Based on Diagram 5.3, explain the type of motion from W to Y. • • • • Velocity increase constant // uniform acceleration, a Increasing in length uniformly zero acceleration // constant velocity uniform length