General Physics 1 Past Paper PDF

Module 1: Semester 1 — General Physics 1 Mr. Romo Sanvictores | S12-12 The metric system, on the other hand, uses a Introduction to Physics few core units and adds prefixes to denote bigger Physics is the study of t...

Module 1: Semester 1 — General Physics 1 Mr. Romo Sanvictores | S12-12 The metric system, on the other hand, uses a Introduction to Physics few core units and adds prefixes to denote bigger Physics is the study of the basic principles of the or smaller measurements. universe, such as motion, forces, energy, and ➔ Core Units – meter (m), gram (g), second more. It’s the most basic of the natural sciences. (s), liter (L) Some common prefixes (together with their The word “physics” comes from the Greek word numerical meanings) are: 12 physikos (φυσικός). ➔ tera- (T) – 10 9 ➔ giga- (G) – 10 Physics can generally be divided into classical 6 ➔ mega- (M) – 10 physics and modern physics. This course will 3 focus on classical physics. ➔ kilo- (k) – 10 2 ➔ hecto- (h) – 10 1 Measurement ➔ deka- (da) – 10 −1 ➔ deci- (d) – 10 Measurement is the process of comparing an −2 unknown quantity with a standard. ➔ centi- (c) – 10 −3 - It involves assigning a quantitative value ➔ milli- (m) – 10 to some data or phenomenon. −6 ➔ micro- (µ) – 10 Measurement includes magnitude, units, −9 and uncertainty using standard ➔ nano- (n) – 10 −12 instruments. ➔ pico- (p) – 10 There are two systems of measurement. ➔ The English units, also known as the Conversion imperial units, are historical units that Conversion involves dimensional analysis. were used. They are generally in use in the USA. To convert between units, we use conversion ➔ The SI units, also known as the Systéme factors. In general, the following rule is used. International, are the modern system of 𝑢𝑛𝑖𝑡𝑠 𝑦𝑜𝑢 𝑤𝑎𝑛𝑡 𝑢𝑛𝑖𝑡𝑠 𝑦𝑜𝑢 ℎ𝑎𝑣𝑒 * = 𝑢𝑛𝑖𝑡𝑠 𝑦𝑜𝑢 𝑤𝑎𝑛𝑡 units used around the world. It’s based on 𝑢𝑛𝑖𝑡𝑠 𝑦𝑜𝑢 ℎ𝑎𝑣𝑒 increments of 10. The fraction is what’s known as a conversion The English system uses the following units. factor, which will depend on the units used. ➔ Length – inch (in), foot (ft), yard (yd), mile Below are the conversion factors for English units (mi) and SI units. ➔ Mass – ounce (oz), pound (lb), ton (ton) ➔ English ➔ Capacity – teaspoon (tsp), cup (c), pint 12 in = 1 ft (pt), quart (qt), gallon (gal) 3 ft = 1 yd 5280 ft = 1 mi 16 oz = 1 lb 2000 lbs = 1 ton 48 tsp = 1 c Module 1: Semester 1 — General Physics 1 Mr. Romo Sanvictores | S12-12 2 c = 1 pt Rule 1: Numbers obtained through counting, 2 pt = 1 qt constants, and conversion factors have unlimited 4 qt = 1 gal significant figures. ➔ SI 12 𝑏𝑜𝑦𝑠 – unlimited SF Multiply the unit with the relevant 2 𝑔 = 9. 8 𝑚/𝑠 – unlimited SF numerical meaning. For example, 1000 1 𝑓𝑡 = 12 𝑖𝑛 – unlimited SF mg = 1 g because milli- tells us the −3 number is 10. Rule 2: All nonzero digits are significant. 22 𝑚 – 2 SF Let’s go through an example. 5423 – 4 SF 13 𝑖𝑛 – 2 SF 3 The density of the femur bone is 1. 85 𝑔/𝑐𝑚. Given this, what is the density of the femur in Rule 3: Zeros at the end of a number are not 3 significant. 𝑘𝑔/𝑚 ? 100 – 1 SF 𝑔 1 𝑘𝑔 100 𝑐𝑚 100 𝑐𝑚 100 𝑐𝑚 1000 𝑔 – 1 SF 1. 85 * * * * 𝑐𝑚 3 1000 𝑔 1𝑚 1𝑚 1𝑚 2400 𝑚 – 2 SF 6 3 𝑘𝑔 10 3 = 1. 85 * 10 3 – Note: 3 = 10 𝑚 10 Rule 4: Zeros between nonzero numbers are 3 = 1850 𝑘𝑔/𝑚 significant. 101 – 3 SF Note that there are three 𝑐𝑚/𝑚 units to cancel 101201 𝐿 – 6 SF 3 the 𝑐𝑚 (𝑐𝑚 * 𝑐𝑚 * 𝑐𝑚). 403 𝑚 – 3 SF Given the same situation, what is the density of 3 Rule 5: Zeros to the right of a decimal point are the femur in 𝑚𝑔/𝑚𝑚 ? significant. 10. 10 – 4 SF 𝑔 1 𝑚𝑔 1 𝑐𝑚 3 1. 85 3 * 0.001 𝑔 * ( 10 𝑚𝑚 ) 105. 40 𝑔 – 5 SF 𝑐𝑚 1 𝑚𝑔 1 101. 240 𝑚 – 6 SF = 1. 85 * 0.001 * 3 1000 𝑚𝑚 = 1. 85 𝑚𝑔/𝑚𝑚 3 Rule 6: Zeros at the end of a number and to the right of the decimal point are significant. 100. 00 – 5 SF Significant Figures 1. 0 𝑔 – 2 SF Significant figures (SF) are digits of 500. 00 𝐿 – 5 SF measurement that represent meaningful measurements of data. Rule 7: Leading zeros are not significant. There are a few rules we should follow when 00023 – 2 SF considering significant figures. 000000001 𝑚𝐿 – 1 SF 0. 00000000000023 𝑚𝑔 – 2 SF Module 1: Semester 1 — General Physics 1 Mr. Romo Sanvictores | S12-12 Rule 8: Sums and differences are expressed with the least number of decimal places in the original Now, if there are errors in accuracy or precision, numbers. we call that an error. There are two types of 23. 24 − 20. 1 = 3. 1 (NOT 3. 14) errors. 23. 245 + 20. 20 = 43. 45 (NOT 43. 445) ➔ A random error causes each 23. 24 + 20. 1 = 43. 3 (NOT 43. 34) measurement to differ from other measurements by a random amount. It’s Rule 9: Products and quotients are expressed usually due to instrumental limitations, with the least number of SFs in the original environmental factors, or variations in numbers. procedure. Lacks precision, but clusters 20. 2 * 3 = 60 (NOT 60. 6) around the accurate value. 20. 42 * 3. 23 = 66. 0 (NOT 66. 9566) ➔ A systematic error causes each measurement to vary by a predictable Extension amount. It’s usually due to observational errors, calibration errors, or environmental When multiplying with conversion factors, interference. May be precise, but not express the answer with the least number of accurate. SFs. 12 𝑖𝑛 5 𝑓𝑡 * 1 𝑓𝑡 = 60 𝑖𝑛 Percentage Error Formula 12 𝑖𝑛 4 𝑓𝑡 * 1 𝑓𝑡 = 50 𝑖𝑛 (NOT 48 𝑖𝑛) 𝑒𝑥𝑝𝑒𝑟𝑖𝑚𝑒𝑛𝑡𝑎𝑙 𝑣𝑎𝑙𝑢𝑒 − 𝑡𝑟𝑢𝑒 𝑣𝑎𝑙𝑢𝑒 % 𝑒𝑟𝑟𝑜𝑟 = 𝑡𝑟𝑢𝑒 𝑣𝑎𝑙𝑢𝑒 × 100 Why do significant figures matter? They help determine whether a measurement is accurate or Example: A resistor’s color code tells you that it precise. should have a resistance of 240 Ohms. However, when you measure it, the meter reads 245 Ohms. Accuracy & Precision What is the % error of this measurement? Accuracy refers to how close the measurement experimental value: 245 Ohms is to the real value. accepted value: 240 Ohms Precision refers to how consistent the measurements are. 245−240 % 𝑒𝑟𝑟𝑜𝑟 = 240 × 100 In the test below, two students were asked to % 𝑒𝑟𝑟𝑜𝑟 = 2. 08% measure the mass of a 50.0 g sample. Scientific Notation There are many cases where measurements are We can say that student A’s scale is precise, but either too big or too small to write completely not accurate. The values are close together, but without tiring out your hand or wasting space. they’re far off from the original measured mass. This is where scientific notation comes in. Student B’s scale, meanwhile, is accurate, but not precise. The values are closer to the original mass, but they’re fairly spread out. Module 1: Semester 1 — General Physics 1 Mr. Romo Sanvictores | S12-12 Scientific notation is a way of writing numbers Highlight to reveal them. in a way that only indicates significant figures. It’s a shorthand of sorts I. Convert the following units. Scientific notation takes the form of: 1. 48 𝑖𝑛 → 𝑓𝑡 | Answer: 4 ft 𝑛 2. 1 𝑘𝑔 → 𝑔 | Answer: 1000 g 𝑐 × 10 3. 9 𝑘𝑚/ℎ → 𝑚/𝑠 | Answer: 2.5 m/s 𝑐 – any number from 1 to 10, not including 10 II. Convert the following to scientific notation. 𝑛 – some integer −5 Let’s see two examples to understand how 1. 0.000023 | Answer: 2. 3 * 10 6 scientific notation works. 2. 5601000 | Answer: 5. 601 * 10 Convert 156000 to scientific notation. III. How many significant figures are in the numbers below? First, imagine a decimal point at the end. 156000. 1. 25000 | Answer: 2 2. 502.60 | Answer: 5 Now, move that decimal point until there is only 1 3. 0.000341 | Answer: 3 digit before the decimal point. 1. 56000 – The point moved 5 places to the left. IV. Express the following numbers in scientific notation with 3 significant figures. That decimal is now 𝑐. As for 𝑛, count how many −1 places the decimal point moved. Since the 1. 0.56815 | Answer: 5. 68 * 10 decimal point moved left, 𝑛 is positive. 3 5 2. 1079.97 | Answer: 1. 08 * 10 156000 = 1. 56 * 10 Convert 0. 000053 to scientific notation. Vectors Physical Quantities can be classified as either Here, we just go straight to moving the decimal vector or scalar point. 0000005. 3 – The decimal point moved 6 places Vector quantity is a quantity that is fully to the right. expressed by both magnitude and direction. The decimal is 𝑐. As for 𝑛, it’s still the number of places moved, however it must be negative. Scalar quantity is a quantity that is Since the decimal point moved right, 𝑛 is expressed by using magnitude alone. negative. −6 0. 000053 = 5. 3 * 10 magnitude refers to a number plus the unit of measurement. (e.g. 5.5m, 20kg) PRACTICE EXERCISE: The answers are in white text. Module 1: Semester 1 — General Physics 1 Mr. Romo Sanvictores | S12-12 Scalars Vectors distance displacement From here, we can see that a 20 mass force m long arrow can speed velocity be represented by a 5 cm long time acceleration arrow. work torque density momentum The magnitude Vector Representation and direction of – vector quantities are often represented by the vector must scaled vector diagrams. be clearly labeled. In this case, the – vector diagrams depict a vector by use of an diagram shows arrow drawn to scale in a specific direction. the magnitude is 20 m and the direction is (30 Components of Vector Diagram degrees West of 1. A vector is represented by an arrow. The North) arrow has 3 important parts: arrowhead – indicated the direction of the vector. the length of the arrow – Vector Addition represents the magnitude of the vector. Resultant Vector (R) tail – represents the origin of the - the sum of two or more vectors that are vector. represented by a singel vector. 2. Devise a suitable scale. 1. Graphical Method PRACTICE EXERCISE: Head-to-Tail Method A student walks 20 m from his classroom to the Parallelogram Method next building in a direction of 30° West of North. How will you represent his displacement using a 2. Analytical/Mathematical Method vector diagram? Pythagorean Theorem Component Method Solution: 3. Experimental Method To represent 20 m long arrow in a paper, we can use a scale of 1 cm = 4 m. We can use Vector Addition - Graphical Method dimensional analysis, to arrive at the actual length. PRACTICE EXERCISE Module 1: Semester 1 — General Physics 1 Mr. Romo Sanvictores | S12-12 HEAD-TO-TAIL METHOD two vectors, so we have finished placing arrows tip to tail. A person walks 9 blocks east and 5 blocks north. Using the head-to-tail method justifies that the Step 4. Draw an arrow from the tail of the first resultant displacement/vector is 10.3 blocks at an vector to the head of the last vector. This is the angle of 29.10 north of east. resultant, or the sum, of the other vectors. Step 1. Draw an arrow to represent the first vector (9 blocks to the east) using a ruler and protractor. Step 5. To get the magnitude of the resultant, Step 2. Now draw an arrow to represent the measure its length with a ruler. (Note that in most second vector (5 blocks to the north). Place the calculations, we will use the Pythagorean tail of the second vector at the head of the first theorem to determine this length.) vector. Step 6. To get the direction of the resultant, measure the angle it makes with the reference frame using a protractor. (Note that in most calculations, we will use trigonometric relationships to determine this angle.) Remember that the reference direction is the last direction in the phrase *direction* of *direction*. (e.g. you started at the south and went to the west, therefore the direction is “west of south”) Step 3. If there are more than two vectors, ANSWER: Using the graphical method, the continue this process for each vector to be head-to-tail method, the resultant vector is 10.3 added. Note that in our example, we have only units north of east. Module 1: Semester 1 — General Physics 1 Mr. Romo Sanvictores | S12-12 Vector Addition - Analytical Method PRACTICE EXERCISE PYTHAGOREAN THEOREM USING PYTHAGOREAN THEOREM TO DETERMINE THE MAGNITUDE: Eric leaves the base camp and hikes 11 km, Using the sine function, the resultant north and then hikes 11 km east. Determine displacement is 15. 6 km,45 North of East ◦ Eric's resulting displacement. PRACTICE EXERCISE 2 Given: d1 = 11km, North d2 = 11km, East A motorboat leaves the shore at a velocity of 4m/s due east. The water current in the river moves due north at a velocity of 3m/s. What would be the resultant velocity of the boat? Solution: 2 2 2 𝑐 =𝑎 +𝑏 2 2 2 𝑐 = (11𝑘𝑚) + (11𝑘𝑚) 2 2 2 𝑐 = 121𝑘𝑚 + 121𝑘𝑚 2 2 𝑐 = 242 Given: 𝑉𝑏𝑜𝑎𝑡 = 4𝑚/𝑠, 𝐸𝑎𝑠𝑡 2 𝑐 = 242𝑘𝑚 2 𝑉𝑟𝑖𝑣𝑒𝑟 𝑐𝑢𝑟𝑟𝑒𝑛𝑡 = 3𝑚/𝑠, 𝑁𝑜𝑟𝑡ℎ 𝑐 = 15. 6 𝑅 = 15. 6𝑘𝑚 magnitude: Eric is 15.6 km away from his original position. USING TRIGONOMETRIC FUNCTIONS 2 2 2 TO DETERMINE THE DIRECTION 𝑐 =𝑎 +𝑏 2 2 2 𝑐 = (4. 0𝑚/𝑠) + (3. 0𝑚/𝑠) 2 2 2 2 2 𝑐 = 16𝑚 /𝑠 + 9𝑚 /𝑠 2 2 2 𝑐 = 25𝑚 /𝑠 2 2 2 𝑐 = 25𝑚 /𝑠 𝑐 = 5. 0 𝑅 = 5. 0𝑚/𝑠 Continuation of Eric’s direction: Module 1: Semester 1 — General Physics 1 Mr. Romo Sanvictores | S12-12 direction: 𝑡𝑎𝑛θ = (𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒/𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡) 𝑡𝑎𝑛θ = (3/4) −1 θ = 𝑡𝑎𝑛 (3/4) ◦ θ = 36. 9 Therefore the resultant velocity of the boat is x-components y-components 5.0m/s, 36.9° from the +x axis or North of East. This is the effect of the river current upon the water boat. Vector Addition - Analytical Method PRACTICE EXERCISE 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑐𝑜𝑠θ = 𝑠𝑖𝑛θ = 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 COMPONENT METHOD ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒 ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒 ◦ 𝑑1𝑥 used for adding two or more vectors 𝑐𝑜𝑠 34 = 36𝑚 ◦ 𝑑1𝑦 ◦ 𝑠𝑖𝑛 34 = 36𝑚 ◦ 𝑑1𝑥 = (3𝑔𝑚)(𝑐𝑜𝑠 34 ) ◦ Given: 𝑑1 = 36𝑚, 34 𝑁 𝑜𝑓 𝐸 𝑑1𝑦 = (36𝑚)(𝑠𝑖𝑛 34 ) 𝑑1𝑥 = 29. 8𝑚 ◦ 𝑑1𝑦 = 20. 1𝑚 𝑑2 = 23𝑚, 64 𝑊 𝑜𝑓 𝑁 this x-component is aimed at the right, so, the y-component goes Solution: it is positive. up so it’s positive. Step 1: Determine the size and direction of the Vector 𝑑2 vectors that you want to add. Step 2: Find the x- and y-components for the first x-components y-components and second vectors. That is, by using SOH-CAH-TOA trigonometric functions. Vector 𝑑1 Module 1: Semester 1 — General Physics 1 Mr. Romo Sanvictores | S12-12 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 |Σ𝑦| 𝑡𝑎𝑛θ = 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 or 𝑡𝑎𝑛θ = |Σ𝑥| 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 |30.1𝑚| 𝑠𝑖𝑛θ = ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒 𝑐𝑜𝑠θ = ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒 𝑡𝑎𝑛θ = |9.2𝑚| ◦ 𝑑2𝑥 ◦ 𝑑2𝑦 −1 |30.1| 𝑠𝑖𝑛 64 = 23𝑚 𝑐𝑜𝑠 64 = 23𝑚 θ = 𝑡𝑎𝑛 |9.2| ◦ ◦ 𝑑2𝑥 = (23𝑚)(𝑠𝑖𝑛 64 ) 𝑑2𝑦 = (23𝑚)(𝑐𝑜𝑠 64 ) ◦ θ = 73. 0 , 𝑄𝑢𝑎𝑑𝑟𝑎𝑛𝑡1 𝑑2𝑥 = 20. 6𝑚 𝑑2𝑦 = 10. 0𝑚 this x-component the y-component goes points to the left so it up, so it is positive. is negative. Step 3: Add all the x- components and add all the y-components of your two vectors, but do not add an x- component to the y-component. Step 4: Use the Pythagorean theorem to get the magnitude of the total 2D displacement. 2 2 𝑑𝑇 = (𝑑𝑇𝑥) + (𝑑𝑇𝑦) 2 2 𝑑𝑇 = (9. 2𝑚) + (30. 1𝑚) 2 2 𝑑𝑇 = 84. 6𝑚 + 906. 0𝑚 2 𝑑𝑇 = 990. 6𝑚 𝑑𝑇 = 31. 5𝑚 Step 5: Use the tangent function to get the angle. Module 2: Semester 1 — General Physics 1 Mr. Romo Sanvictores | S12-12 Given: Motion in One Dimension 𝑑1 = + 45𝑚 assuming that the 5 Quantities to describe original direction is to motion quantitatively: the right or going East 1. Distance 𝑑2 = − 25𝑚 opposite direction 2. Displacement 3. Speed 𝑑3 = + 37𝑚 original direction 4. Velocity 5. Acceleration Solution: Distance & Displacement a. total displacement (consider the direction) Distance Total Displacement = + 45𝑚 + (− 25𝑚) + 37𝑚 a scalar quantity (doesn’t follow the Total Displacement = + 57𝑚 𝑜𝑟 57𝑚, 𝐸𝑎𝑠𝑡 operations of signed numbers) b. total distance (disregard the direction) refers to the total length of the path taken by an object from its origin or initial Total Distance = 45𝑚 + 25𝑚 + 37𝑚 position to the final position. Total Distance = 107𝑚 Displacement a vector quantity (follows the operations Speed & Velocity of signed numbers) Speed a scalar quantity refers to the straight-line distance between the initial and final position. a measure of how fast an object is moving, or the rate at which distance is PRACTICE EXERCISE traveled You walk along a straight sidewalk for 45m, then Velocity you turn around and walk 25m in the opposite a vector quantity direction. Finally, you turn again and walk 37m in the original direction and stop. the speed in a given direction or the rate at which displacement is traveled a. What is your displacement from your starting point? average speed = total distance/total time average veocity = total displacement/total time b. What is the total distance covered while you are walking? Module 2: Semester 1 — General Physics 1 Mr. Romo Sanvictores | S12-12 PRACTICE EXERCISE Motion Graphs Motion may be defined as continuous change of position with respect to a certain reference point. A reference frame is a physical entity such as the earth’s surface, the deck of a From the diagram, what will be the coach’s ship, or a moving vehicle, to which the average speed and average velocity as he position and motion of an object is moves from positon A to D? relative. Solution: TYPES OF MOTION 1. vibratory motion 𝐴𝑣𝑒𝑟𝑎𝑔𝑒 𝑆𝑝𝑒𝑒𝑑 = 𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑇𝑟𝑎𝑣𝑒𝑙𝑒𝑑 2. linear motion *our focus for now* (motion 𝑇𝑖𝑚𝑒 𝑜𝑓 𝑇𝑟𝑎𝑣𝑒𝑙 in a straight line) 𝑑1 (𝐴 − 𝐵) + 𝑑2 (𝐵 − 𝐶) + 𝑑3(𝐶 − 𝐷) 𝐴𝑣𝑒𝑟𝑎𝑔𝑒 𝑆𝑝𝑒𝑒𝑑 = 𝑇𝑜𝑡𝑎𝑙 𝑡𝑖𝑚𝑒 3. curvilinear motion 𝐴𝑣𝑒𝑟𝑎𝑔𝑒 𝑆𝑝𝑒𝑒𝑑 = 35𝑚 + 20𝑚 + 40𝑚 4. oscillatory motion 10 𝑚𝑖𝑛𝑠 5. circular motion 𝐴𝑣𝑒𝑟𝑎𝑔𝑒 𝑆𝑝𝑒𝑒𝑑 = 9. 5𝑚/𝑚𝑖𝑛 6. rotatory motion ∆𝑝𝑜𝑠𝑖𝑡𝑖𝑜𝑛 𝑑𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑚𝑒𝑛𝑡 𝐴𝑣𝑒𝑟𝑎𝑔𝑒 𝑉𝑒𝑙𝑜𝑐𝑖𝑡𝑦 = 𝑡𝑖𝑚𝑒 = 𝑡𝑖𝑚𝑒 𝑑1(−𝐴+𝐵) + 𝑑2(−𝐵+𝐶) + 𝑑3(−𝐶+𝐷) 𝐴𝑣𝑒𝑟𝑎𝑔𝑒 𝑉𝑒𝑙𝑜𝑐𝑖𝑡𝑦 = 𝑇𝑜𝑡𝑎𝑙 𝑡𝑖𝑚𝑒 (−35) + 20𝑚 + (−40𝑚) 𝐴𝑣𝑒𝑟𝑎𝑔𝑒 𝑉𝑒𝑙𝑜𝑐𝑖𝑡𝑦 = 10 𝑚𝑖𝑛𝑠 𝐴𝑣𝑒𝑟𝑎𝑔𝑒 𝑉𝑒𝑙𝑜𝑐𝑖𝑡𝑦 = − 5. 5𝑚/𝑚𝑖𝑛, 𝑊𝑒𝑠𝑡 Acceleration it is the change in velocity per unit time ∆𝑣 𝑉𝑓−𝑉𝑖 𝑎 = 𝑎𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛 = 𝑡 = 𝑡 There are three cases when an object experiences an acceleration: - change in speed - change in direction - change in both speed and direction