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2 digit times 1 digit multiplication word problems
Quiz by Cassie Blackwell
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The expression 2 + 4 1 + 2 is equal to (A) 0 (B) 1 (C) 2 (D) 4 (E) 5 2. The ones (units) digit of 542 is 2. When 542 is multiplied by 3, the ones (units) digit of the result is (A) 9 (B) 3 (C) 5 (D) 4 (E) 6 3. Some of the 1 × 1 squares in a 3 × 3 grid are shaded, as shown. What is the perimeter of the shaded region? (A) 10 (B) 14 (C) 8 (D) 18 (E) 20 4. If 3x + 4 = x + 2, the value of x is (A) 0 (B) −4 (C) −3 (D) −1 (E) −2 5. Which of the following is equal to 110% of 500? (A) 610 (B) 510 (C) 650 (D) 505 (E) 550 6. Eugene swam on Sunday, Monday and Tuesday. On Monday, he swam for 30 minutes. On Tuesday, he swam for 45 minutes. His average swim time over the three days was 34 minutes. For how many minutes did he swim on Sunday? (A) 20 (B) 25 (C) 27 (D) 32 (E) 37.5 7. For which of the following values of x is x 3 < x2 ? (A) x = 5 3 (B) x = 3 4 (C) x = 1 (D) x = 3 2 (E) x = 2112 years, Janice will be 8 times as old as she was 2 years ago. How old is Janice now? (A) 4 (B) 8 (C) 10 (D) 2 (E) 6 10. In the diagram, pentagon T P SRQ is constructed from equilateral 4 P T Q and square P QRS. The measure of ∠ST R is equal to (A) 10◦ (B) 15◦ (C) 20◦ (D) 30◦ (E) 45◦ Q P R S T Part B: Each correct answer is worth 6. 11. In the diagram, which of the following points is at a different distance from P than the rest of the points? (A) A (B) B (C) C (D) D (E) E y A x 2 2 4 4 6 8 6 8 B C D E P 12. If x = 2 and y = x 2 − 5 and z = y 2 − 5, then z equals (A) −6 (B) −8 (C) 4 (D) 76 (E) −4 13. In the diagram, P QR is a straight line segment. If x + y = 76, what is the value of x? (A) 28 (B) 30 (C) 35 (D) 36 (E) 38 x° x° x° y° y° P Q R 14. The line with equation y = 2x − 6 is reflected in the y-axis. What is the x-intercept of the resulting line? (A) −12 (B) 6 (C) −6 (D) −3 (E) 0 15. Amy bought and then sold 15n avocados, for some positive integer n. She made a profit of $100. (Her profit is the difference between the total amount that she earned by selling the avocados and the total amount that she spent in buying the avocados.) She paid $2 for every 3 avocados. She sold every 5 avocados for $4. What is the value of n? (A) 100 (B) 20 (C) 50 (D) 30 (E) 8 16. If 3x = 5, the value of 3x+2 is (A) 10 (B) 25 (C) 2187 (D) 14 (E) 45How is personal data collected? There are several ways that an unauthorised person can try and collect your data. These include: •phishing •smishing •vishing •pharming. Phishing Phishing is when a person sends a legitimate looking email to a user. The email contains a link to a website that also looks legitimate. The user is encouraged to click the link and to input personal data into a form on the website. The email could also simply ask the user to reply to the email with their personal data. The user is tricked into giving their personal data to a source that they believe is legitimate. However, both the email and the linked website are from a fake unauthorised source. The personal data that is input is then collected by an unauthorised person. This person can then use this data for criminal acts, for example, to commit fraud or steal the person's identity. Intimidation has become a common feature of phishing emails, threatening the user that they must click the link and rectify a situation immediately, or there will be a further issue. The aim of a phishing attack is to steal the user's personal data. Figure 5.1: Phishing. A real-life example of phishing PayPal have been the subject of several different phishing emails. Users receive an email that looks as though it has been sent from PayPal, as it has the PayPal branding. The email normally warns of an issue such as unexpected activity on their account, or that some kind of verification of their account is required. The user is then asked to click a link to log into their account and resolve the issue. The link takes them to a webpage that looks like the PayPal login page. If the user inputs their login details into this page, they will not be taken to their account. It is often at this stage that the user may realise that the email and webpage are fake. However, they have already given the unauthorised person their PayPal login details. Figure 5.2: An example of a phishing email claiming to be from PayPal. How to recognise phishing There are several guidelines to be aware of regarding emails to avoid being subjected to phishing. These include: •Don't even open an email that is not from a sender that you recognise or a trusted source. •Legitimate companies will never ask you for your personal data using email. Be immediately suspicious of any email that requests your personal data. •Legitimate companies will normally address you by your name. Be suspicious of any email that addresses you as ‘Dear Member' or ‘Dear Customer'. •Legitimate companies will send an email that uses their domain name. If you hover your mouse over the sender's name, it will show the email address that the email is sent from. If this does not look legitimate, for example, does not contain the correct domain name, then it is probably fake. For example, if the sender's email is user@paypal1.com rather than user@paypal.com, this is from an incorrect domain name. •Legitimate companies are protective of their professional reputation and thoroughly check any communications. They will make sure that all information given is grammatically and correctly spelt. Be suspicious of any email that contains bad grammar or spelling mistakes. •A link in an email from a legitimate company will also normally contain the domain name of the company. You can sometimes hover over the link, or right click and inspect the link, to see the address of the URL that is attached. If the URL does not contain the domain name, or also contains typical errors such as spelling mistakes, then be suspicious of this. PRACTICAL ACTIVITY 5.02 Ask a friend or a member of your family if they have ever received an email that they believed was a phishing email. Ask them how they identified it was phishing. Ask them if they know all of the given guidelines for identifying phishing emails. Smishing Smishing (or SMS phishing) is a variant of phishing that uses SMS text messages to lure the user into providing their personal details. The user is sent an SMS text message that either contains a link to a website, in the same way that phishing does, or it will ask the user to call a telephone number to resolve an urgent issue. The same advice can be followed for smishing as given for phishing. The user must question at all times any links that are sent from an unknown or suspicious user. It is advisable that if a user believes the message may be legitimate, to type in the domain name for the legitimate company website into their web browser, rather than following the link in the message. Users should block any numbers that they believe are suspicious to prevent any further risk of smishing from that number. Figure 5.3: Smishing. Vishing Vishing (or voice phishing) has the same aim as phishing, to obtain a user's personal details. The user receives a telephone call that could either be an automated system or could be a real person. An automated voice could speak to the user and advise them that an issue has occurred, such as there has been suspicious activity regarding their bank account. The user may then be asked to call another number, or just to simply press a digit and be directed to another automated system. This system will ask them to provide their bank account details to resolve the issue. The bank account details have then been obtained by the unauthorised user and can be used to commit a crime against the user. The automated system could be replaced by a real person who will try to do the same thing. They will try to convince the user that there has been an issue with an account they have and to provide the log-in details or PIN for the account to verify who they are so the issue can be resolved. The precaution to take for vishing is that no company will ever call you and ask you to provide any log-in details or PIN details over the telephone. They may ask you to provide other personal information, and if you are in doubt that the person on the other end of the phone is legitimate, it is always advisable to put the phone down and call the company back on a legitimate number that you may already know or can obtain. Figure 5.4: Vishing. Pharming Pharming is when an unauthorised user installs malicious code on a person's hard drive or server. The malicious code is designed to redirect a user to a fake website when they type in the address of a legitimate one. The fake website is designed to look like the legitimate one, to trick the user and make sure they are not aware that their request has been redirected. The user will then enter their personal details into the fake website, believing it is the legitimate one, and the unauthorised person will now have their personal data. A common technique used in pharming is called domain name server (DNS) cache poisoning. This technique exploits vulnerabilities in the DNS and diverts the internet traffic intended for a legitimate server toward a fake one instead. The unauthorised user needs to find a way to install the malicious code on the computer. They often hide the malicious code in an email attachment or link. When the user opens the email attachment or clicks the link, the malicious code is downloaded also. Figure 5.5: Pharming. The aim of a pharming attack is also to steal a user's personal data. A real-life example of pharming In 2007 50 different companies all over the world were subject to a pharming attack, these included PayPal, eBay, Barclays bank and American Express. Over a three-day period, hackers managed to infect over 1000 PCs a day with a malicious pharming code. When users who had been infected visited the websites of the different companies, they were redirected to a legitimate-looking version of the site that was designed to steal their personal data. The original email, containing the malicious code, was set up to look like a shocking news story. Users were encouraged to click a link in the email to find out more information. The code was downloaded when the user clicked the link. This was quite a sophisticated attack that required legitimate looking websites to be set up for a large number of companies. It is not known how much money the hackers were able to retrieve as a result. How to prevent pharming All of the guidelines to avoid being subjected to phishing are also relevant for recognising pharming. There are also several other precautions that can be taken to check for pharming attacks. These include: •Have a firewall installed and operational. A firewall monitors incoming and outgoing traffic from your computer. It checks this traffic against set criteria and will flag and stop any traffic that does not meet the criteria. A firewall could detect and block suspicious traffic, such as a malicious code trying to enter your system. •Have an anti-virus program installed that is designed to detect malicious pharming code. You need to scan your computer on a regular basis to check for any malicious code. It is advisable to set up an automatic scan on a daily basis at a time when your computer will normally be switched on. •Be aware when using public Wi-Fi connections. A hacker could look to directly access your computer and install the malicious code if you are connected to a public Wi-Fi connection. It is often advisable to use a VPN when using public Wi-Fi. This will help shield your internet activity and personal details from a hacker, making it more difficult for them to access your computer. Smishing can also be used as a form of pharming. A user is sent a link, that when they click is designed to download malware onto their mobile device. Therefore, it is advisable to have security software installed on your mobile and also scan it regularly to detect any presence of malware.Introduction to Free Fall A free-falling object is an object that is falling under the sole influence of gravity. Any object that is being acted upon only by the force of gravity is said to be in a state of free fall. There are two important motion characteristics that are true of free-falling objects: • Free-falling objects do not encounter air resistance. • All free-falling objects (on Earth) accelerate downwards at a rate of 9.8 m/s/s (often approximated as 10 m/s/s for back-of-the-envelope calculations) Because free-falling objects are accelerating downwards at a rate of 9.8 m/s/s, a ticker tape trace or dot diagram of its motion would depict an acceleration. The dot diagram at the right depicts the acceleration of a free-falling object. The position of the object at regular time intervals - say, every 0.1 second - is shown. The fact that the distance that the object travels every interval of time is increasing is a sure sign that the ball is speeding up as it falls downward. Recall from an earlier lesson, that if an object travels downward and speeds up, then its acceleration is downward. Free-fall acceleration is often witnessed in a physics classroom by means of an ever-popular strobe light demonstration. The room is darkened and a jug full of water is connected by a tube to a medicine dropper. The dropper drips water and the strobe illuminate the falling droplets at a regular rate - say once every 0.2 seconds. Instead of seeing a stream of water free-falling from the medicine dropper, several consecutive drops with increasing separation distance are seen. The pattern of drops resembles the dot diagram shown in the graphic at the right. The Acceleration of Gravity It was learned in the previous part of this lesson that a free-falling object is an object that is falling under the sole influence of gravity. A free-falling object has an acceleration of 9.8 m/s/s, downward (on Earth). This numerical value for the acceleration of a free-falling object is such an important value that it is given a special name. It is known as the acceleration of gravity - the acceleration for any object moving under the sole influence of gravity. A matter of fact, this quantity known as the acceleration of gravity is such an important quantity that physicists have a special symbol to denote it - the symbol g. The numerical value for the acceleration of gravity is most accurately known as 9.8 m/s2. There are slight variations in this numerical value (to the second decimal place) that are dependent primarily upon on altitude. We will occasionally use the approximated value of 10 m/s2 in order to reduce the complexity of the many mathematical tasks that we will perform with this number. By so doing, we will be able to better focus on the conceptual nature of physics without too much of a sacrifice in numerical accuracy. g = 9.8 m/s2, downward Look It Up! Even on the surface of the Earth, there are local variations in the value of the acceleration of gravity (g). These variations are due to latitude, altitude and the local geological structure of the region. Recall from an earlier lesson that acceleration is the rate at which an object changes its velocity. It is the ratio of velocity change to time between any two points in an object's path. To accelerate at 9.8 m/s2 means to change the velocity by 9.8 m/s each second. If the velocity and time for a free-falling object being dropped from a position of rest were tabulated, then one would note the following pattern. Time (s) Velocity (m/s) 0 0 1 - 9.8 2 - 19.6 3 - 29.4 4 - 39.2 5 - 49.0 . Observe that the velocity-time data above reveal that the object's velocity is changing by 9.8 m/s each consecutive second. That is, the free-falling object has an acceleration of approximately 9.8 m/s2. Another way to represent this acceleration of 9.8 m/s2 is to add numbers to our dot diagram that we saw earlier in this lesson. The velocity of the ball is seen to increase as depicted in the diagram at the right. (NOTE: The diagram is not drawn to scale - in two seconds, the object would drop considerably further than the distance from shoulder to toes.) Representing Free Fall by Graphs • Early in Lesson 1 it was mentioned that there are a variety of means of describing the motion of objects. One such means of describing the motion of objects is through the use of graphs - position versus time and velocity vs. time graphs. In this part of Lesson 5, the motion of a free-falling motion will be represented using these two basic types of graphs. Representing Free Fall by Position-Time Graphs A position versus time graph for a free-falling object is shown below. Observe that the line on the graph curves. As learned earlier, a curved line on a position versus time graph signifies an accelerated motion. Since a free-falling object is undergoing an acceleration (g = 9.8 m/s/s), it would be expected that its position-time graph would be curved. A further look at the position-time graph reveals that the object starts with a small velocity (slow) and finishes with a large velocity (fast). Since the slope of any position vs. time graph is the velocity of the object (as learned in Lesson 3), the small initial slope indicates a small initial velocity and the large final slope indicates a large final velocity. Finally, the negative slope of the line indicates a negative (i.e., downward) velocity. Representing Free Fall by Velocity-Time Graphs A velocity versus time graph for a free-falling object is shown below. Observe that the line on the graph is a straight, diagonal line. As learned earlier, a diagonal line on a velocity versus time graph signifies an accelerated motion. Since a free-falling object is undergoing an acceleration (g = 9,8 m/s/s, downward), it would be expected that its velocity-time graph would be diagonal. A further look at the velocity-time graph reveals that the object starts with a zero velocity (as read from the graph) and finishes with a large, negative velocity; that is, the object is moving in the negative direction and speeding up. An object that is moving in the negative direction and speeding up is said to have a negative acceleration (if necessary, review the vector nature of acceleration). Since the slope of any velocity versus time graph is the acceleration of the object (as learned in Lesson 4), the constant, negative slope indicates a constant, negative acceleration. This analysis of the slope on the graph is consistent with the motion of a free-falling object - an object moving with a constant acceleration of 9.8 m/s/s in the downward direction. The Kinematic Equations The goal of this first unit has been to investigate the variety of means by which the motion of objects can be described. The variety of representations that we have investigated includes verbal representations, pictorial representations, numerical representations, and graphical representations (position-time graphs and velocity-time graphs). In Lesson 6, we will investigate the use of equations to describe and represent the motion of objects. These equations are known as kinematic equations. There are a variety of quantities associated with the motion of objects - displacement (and distance), velocity (and speed), acceleration, and time. Knowledge of each of these quantities provides descriptive information about an object's motion. For example, if a car is known to move with a constant velocity of 22.0 m/s, North for 12.0 seconds for a northward displacement of 264 meters, then the motion of the car is fully described. And if a second car is known to accelerate from a rest position with an eastward acceleration of 3.0 m/s2 for a time of 8.0 seconds, providing a final velocity of 24 m/s, East and an eastward displacement of 96 meters, then the motion of this car is fully described. These two statements provide a complete description of the motion of an object. However, such completeness is not always known. It is often the case that only a few parameters of an object's motion are known, while the rest are unknown. For example as you approach the stoplight, you might know that your car has a velocity of 22 m/s, East and is capable of a skidding acceleration of 8.0 m/s2, West. However you do not know the displacement that your car would experience if you were to slam on your brakes and skid to a stop; and you do not know the time required to skid to a stop. In such an instance as this, the unknown parameters can be determined using physics principles and mathematical equations (the kinematic equations). The BIG 4 The kinematic equations are a set of four equations that can be utilized to predict unknown information about an object's motion if other information is known. The equations can be utilized for any motion that can be described as being either a constant velocity motion (an acceleration of 0 m/s/s) or a constant acceleration motion. They can never be used over any time period during which the acceleration is changing. Each of the kinematic equations include four variables. If the values of three of the four variables are known, then the value of the fourth variable can be calculated. In this manner, the kinematic equations provide a useful means of predicting information about an object's motion if other information is known. For example, if the acceleration value and the initial and final velocity values of a skidding car is known, then the displacement of the car and the time can be predicted using the kinematic equations. Lesson 6 of this unit will focus upon the use of the kinematic equations to predict the numerical values of unknown quantities for an object's motion. The four kinematic equations that describe an object's motion are: There are a variety of symbols used in the above equations. Each symbol has its own specific meaning. The symbol d stands for the displacement of the object. The symbol t stands for the time for which the object moved. The symbol a stands for the acceleration of the object. And the symbol v stands for the velocity of the object; a subscript of i after the v (as in vi) indicates that the velocity value is the initial velocity value and a subscript of f (as in vf) indicates that the velocity value is the final velocity value. Each of these four equations appropriately describes the mathematical relationship between the parameters of an object's motion. As such, they can be used to predict unknown information about an object's motion if other information is known. In the next part of Lesson 6 we will investigate the process of doing this. Kinematic Equations and Problem-Solving The four kinematic equations that describe the mathematical relationship between the parameters that describe an object's motion were introduced in the previous part of Lesson 6. The four kinematic equations are: In the above equations, the symbol d stands for the displacement of the object. The symbol t stands for the time for which the object moved. The symbol a stand for the acceleration of the object. And the symbol v stands for the instantaneous velocity of the object; a subscript of i after the v (as in vi) indicates that the velocity value is the initial velocity value and a subscript of f (as in vf) indicates that the velocity value is the final velocity value. Problem-Solving Strategy In this part of Lesson 6 we will investigate the process of using the equations to determine unknown information about an object's motion. The process involves the use of a problem-solving strategy that will be used throughout the course. The strategy involves the following steps: 1. Construct an informative diagram of the physical situation. 2. Identify and list the given information in variable form. 3. Identify and list the unknown information in variable form. 4. Identify and list the equation that will be used to determine unknown information from known information. 5. Substitute known values into the equation and use appropriate algebraic steps to solve for the unknown information. 6. Check your answer to ensure that it is reasonable and mathematically correct. The use of this problem-solving strategy in the solution of the following problem is modeled in Examples A and B below. Example Problem A . Ima Hurryin is approaching a stoplight moving with a velocity of +30.0 m/s. The light turns yellow, and Ima applies the brakes and skids to a stop. If Ima's acceleration is -8.00 m/s2, then determine the displacement of the car during the skidding process. (Note that the direction of the velocity and the acceleration vectors are denoted by a + and a - sign.) The solution to this problem begins by the construction of an informative diagram of the physical situation. This is shown below. The second step involves the identification and listing of known information in variable form. Note that the vf value can be inferred to be 0 m/s since Ima's car comes to a stop. The initial velocity (vi) of the car is +30.0 m/s since this is the velocity at the beginning of the motion (the skidding motion). And the acceleration (a) of the car is given as - 8.00 m/s2. (Always pay careful attention to the + and - signs for the given quantities.) The next step of the strategy involves the listing of the unknown (or desired) information in variable form. In this case, the problem requests information about the displacement of the car. So d is the unknown quantity. The results of the first three steps are shown in the table below. Diagram: Given: Find: vi = +30.0 m/s vf = 0 m/s a = - 8.00 m/s2 d = ?? The next step of the strategy involves identifying a kinematic equation that would allow you to determine the unknown quantity. There are four kinematic equations to choose from. In general, you will always choose the equation that contains the three known and the one unknown variable. In this specific case, the three known variables and the one unknown variable are vf, vi, a, and d. Thus, you will look for an equation that has these four variables listed in it. An inspection of the four equations above reveals that the equation on the top right contains all four variables. vf2 = vi2 + 2 • a • d Once the equation is identified and written down, the next step of the strategy involves substituting known values into the equation and using proper algebraic steps to solve for the unknown information. This step is shown below. (0 m/s)2 = (30.0 m/s)2 + 2 • (-8.00 m/s2) • d 0 m2/s2 = 900 m2/s2 + (-16.0 m/s2) • d (16.0 m/s2) • d = 900 m2/s2 - 0 m2/s2 (16.0 m/s2)*d = 900 m2/s2 d = (900 m2/s2)/ (16.0 m/s2) d = (900 m2/s2)/ (16.0 m/s2) d = 56.3 m The solution above reveals that the car will skid a distance of 56.3 meters. (Note that this value is rounded to the third digit.) The last step of the problem-solving strategy involves checking the answer to assure that it is both reasonable and accurate. The value seems reasonable enough. It takes a car a considerable distance to skid from 30.0 m/s (approximately 65 mi/hr) to a stop. The calculated distance is approximately one-half a football field, making this a very reasonable skidding distance. Checking for accuracy involves substituting the calculated value back into the equation for displacement and insuring that the left side of the equation is equal to the right side of the equation. Indeed it is! Example Problem B Ben Rushin is waiting at a stoplight. When it finally turns green, Ben accelerated from rest at a rate of a 6.00 m/s2 for a time of 4.10 seconds. Determine the displacement of Ben's car during this time period. Once more, the solution to this problem begins by the construction of an informative diagram of the physical situation. This is shown below. The second step of the strategy involves the identification and listing of known information in variable form. Note that the vi value can be inferred to be 0 m/s since Ben's car is initially at rest. The acceleration (a) of the car is 6.00 m/s2. And the time (t) is given as 4.10 s. The next step of the strategy involves the listing of the unknown (or desired) information in variable form. In this case, the problem requests information about the displacement of the car. So d is the unknown information. The results of the first three steps are shown in the table below. Diagram: Given: Find: vi = 0 m/s t = 4.10 s a = 6.00 m/s2 d = ?? The next step of the strategy involves identifying a kinematic equation that would allow you to determine the unknown quantity. There are four kinematic equations to choose from. Again, you will always search for an equation that contains the three known variables and the one unknown variable. In this specific case, the three known variables and the one unknown variable are t, vi, a, and d. An inspection of the four equations above reveals that the equation on the top left contains all four variables. d = vi • t + ½ • a • t2 Once the equation is identified and written down, the next step of the strategy involves substituting known values into the equation and using proper algebraic steps to solve for the unknown information. This step is shown below. d = (0 m/s) • (4.1 s) + ½ • (6.00 m/s2) • (4.10 s)2 d = (0 m) + ½ • (6.00 m/s2) • (16.81 s2) d = 0 m + 50.43 m d = 50.4 m The solution above reveals that the car will travel a distance of 50.4 meters. (Note that this value is rounded to the third digit.) The last step of the problem-solving strategy involves checking the answer to assure that it is both reasonable and accurate. The value seems reasonable enough. A car with an acceleration of 6.00 m/s/s will reach a speed of approximately 24 m/s (approximately 50 mi/hr) in 4.10 s. The distance over which such a car would be displaced during this time period would be approximately one-half a football field, making this a very reasonable distance. Checking for accuracy involves substituting the calculated value back into the equation for displacement and insuring that the left side of the equation is equal to the right side of the equation. Indeed, it is! The two example problems above illustrate how the kinematic equations can be combined with a simple problem-solving strategy to predict unknown motion parameters for a moving object. Provided that three motion parameters are known, any of the remaining values can be determined. In the next part of Lesson 6, we will see how this strategy can be applied to free fall situations. Or if interested, you can try some practice problems and check your answer against the given solutions. Kinematic Equations and Free Fall As mentioned in Lesson 5, a free-falling object is an object that is falling under the sole influence of gravity. That is to say that any object that is moving and being acted upon only be the force of gravity is said to be "in a state of free fall." Such an object will experience a downward acceleration of 9.8 m/s/s. Whether the object is falling downward or rising upward towards its peak, if it is under the sole influence of gravity, then its acceleration value is 9.8 m/s/s. Like any moving object, the motion of an object in free fall can be described by four kinematic equations. The kinematic equations that describe any object's motion are: The symbols in the above equation have a specific meaning: the symbol d stands for the displacement; the symbol t stands for the time; the symbol a stands for the acceleration of the object; the symbol vi stands for the initial velocity value; and the symbol vf stands for the final velocity. Applying Free Fall Concepts to Problem-Solving There are a few conceptual characteristics of free fall motion that will be of value when using the equations to analyze free fall motion. These concepts are described as follows: • An object in free fall experiences an acceleration of -9.8 m/s/s. (The - sign indicates a downward acceleration.) Whether explicitly stated or not, the value of the acceleration in the kinematic equations is -9.8 m/s/s for any freely falling object. • If an object is merely dropped (as opposed to being thrown) from an elevated height, then the initial velocity of the object is 0 m/s. • If an object is projected upwards in a perfectly vertical direction, then it will slow down as it rises upward. The instant at which it reaches the peak of its trajectory, its velocity is 0 m/s. This value can be used as one of the motion parameters in the kinematic equations; for example, the final velocity (vf) after traveling to the peak would be assigned a value of 0 m/s. • If an object is projected upwards in a perfectly vertical direction, then the velocity at which it is projected is equal in magnitude and opposite in sign to the velocity that it has when it returns to the same height. That is, a ball projected vertically with an upward velocity of +30 m/s will have a downward velocity of -30 m/s when it returns to the same height. These four principles and the four kinematic equations can be combined to solve problems involving the motion of free-falling objects. The two examples below illustrate application of free fall principles to kinematic problem-solving. In each example, the problem solving strategy that was introduced earlier in this lesson will be utilized. Example Problem A Luke Autbeloe drops a pile of roof shingles from the top of a roof located 8.52 meters above the ground. Determine the time required for the shingles to reach the ground. The solution to this problem begins by the construction of an informative diagram of the physical situation. This is shown below. The second step involves the identification and listing of known information in variable form. You might note that in the statement of the problem, there is only one piece of numerical information explicitly stated: 8.52 meters. The displacement (d) of the shingles is -8.52 m. (The - sign indicates that the displacement is downward). The remaining information must be extracted from the problem statement based upon your understanding of the above principles. For example, the vi value can be inferred to be 0 m/s since the shingles are dropped (released from rest; see note above). And the acceleration (a) of the shingles can be inferred to be -9.8 m/s2 since the shingles are free-falling (see note above). (Always pay careful attention to the + and - signs for the given quantities.) The next step of the solution involves the listing of the unknown (or desired) information in variable form. In this case, the problem requests information about the time of fall. So t is the unknown quantity. The results of the first three steps are shown in the table below. Diagram: Given: Find: vi = 0.0 m/s d = -8.52 m a = - 9.8 m/s2 t = ?? The next step involves identifying a kinematic equation that allows you to determine the unknown quantity. There are four kinematic equations to choose from. In general, you will always choose the equation that contains the three known and the one unknown variable. In this specific case, the three known variables and the one unknown variable are d, vi, a, and t. Thus, you will look for an equation that has these four variables listed in it. An inspection of the four equations above reveals that the equation on the top left contains all four variables. d = vi • t + ½ • a • t2 Once the equation is identified and written down, the next step involves substituting known values into the equation and using proper algebraic steps to solve for the unknown information. This step is shown below. -8.52 m = (0 m/s) • (t) + ½ • (-9.8 m/s2) • (t)2 -8.52 m = (0 m) *(t) + (-4.9 m/s2) • (t)2 -8.52 m = (-4.9 m/s2) • (t)2 (-8.52 m)/(-4.9 m/s2) = t2 1.739 s2 = t2 t = 1.32 s The solution above reveals that the shingles will fall for a time of 1.32 seconds before hitting the ground. (Note that this value is rounded to the third digit.) The last step of the problem-solving strategy involves checking the answer to assure that it is both reasonable and accurate. The value seems reasonable enough. The shingles are falling a distance of approximately 10 yards (1 meter is pretty close to 1 yard); it seems that an answer between 1 and 2 seconds would be highly reasonable. The calculated time easily falls within this range of reasonability. Checking for accuracy involves substituting the calculated value back into the equation for time and insuring that the left side of the equation is equal to the right side of the equation. Indeed it is! Example Problem B Rex Things throws his mother's crystal vase vertically upwards with an initial velocity of 26.2 m/s. Determine the height to which the vase will rise above its initial height. Once more, the solution to this problem begins by the construction of an informative diagram of the physical situation. This is shown below. The second step involves the identification and listing of known information in variable form. You might note that in the statement of the problem, there is only one piece of numerical information explicitly stated: 26.2 m/s. The initial velocity (vi) of the vase is +26.2 m/s. (The + sign indicates that the initial velocity is an upwards velocity). The remaining information must be extracted from the problem statement based upon your understanding of the above principles. Note that the vf value can be inferred to be 0 m/s since the final state of the vase is the peak of its trajectory (see note above). The acceleration (a) of the vase is -9.8 m/s2 (see note above). The next step involves the listing of the unknown (or desired) information in variable form. In this case, the problem requests information about the displacement of the vase (the height to which it rises above its starting height). So d is the unknown information. The results of the first three steps are shown in the table below. Diagram: Given: Find: vi = 26.2 m/s vf = 0 m/s a = -9.8 m/s2 d = ?? The next step involves identifying a kinematic equation that would allow you to determine the unknown quantity. There are four kinematic equations to choose from. Again, you will always search for an equation that contains the three known variables and the one unknown variable. In this specific case, the three known variables and the one unknown variable are vi, vf, a, and d. An inspection of the four equations above reveals that the equation on the top right contains all four variables. vf2 = vi2 + 2 • a • d Once the equation is identified and written down, the next step involves substituting known values into the equation and using proper algebraic steps to solve for the unknown information. This step is shown below. (0 m/s)2 = (26.2 m/s)2 + 2 •(-9.8m/s2) •d 0 m2/s2 = 686.44 m2/s2 + (-19.6 m/s2) •d (-19.6 m/s2) • d = 0 m2/s2 -686.44 m2/s2 (-19.6 m/s2) • d = -686.44 m2/s2 d = (-686.44 m2/s2)/ (-19.6 m/s2) d = 35.0 m The solution above reveals that the vase will travel upwards for a displacement of 35.0 meters before reaching its peak. (Note that this value is rounded to the third digit.) The last step of the problem-solving strategy involves checking the answer to assure that it is both reasonable and accurate. The value seems reasonable enough. The vase is thrown with a speed of approximately 50 mi/hr (merely approximate 1 m/s to be equivalent to 2 mi/hr). Such a throw will never make it further than one football field in height (approximately 100 m), yet will surely make it past the 10-yard line (approximately 10 meters). The calculated answer certainly falls within this range of reasonability. Checking for accuracy involves substituting the calculated value back into the equation for displacement and insuring that the left side of the equation is equal to the right side of the equation. Indeed, it is! Kinematic equations provide a useful means of determining the value of an unknown motion parameter if three motion parameters are known. In the case of a free-fall motion, the acceleration is often known. And in many cases, another motion parameter can be inferred through a solid knowledge of some basic kinematic principles.Make 10 word problems that are multiplication with 2 digits times one digit2 digits number time to one digit number2 digit by 1 digit multiplication2-digit subtraction with and without regrouping2 digit addition with regrouping