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Chapter 22 Antihypertensive Drugs Hypertension Defined (JNC-8) Pharmacology Overview 7 main categories of drugs to treat HTN Adrenergic drugs (old friend) Angiotensin-converting enzyme (ACE) inhibitors Angiotensin II receptor blockers (ARBs) Calcium channel blockers (CCBs) Diuretics Vasodilators Direct renin inhibitors A. Adrenergic Drugs: 5 Subcategories and where they act A1. Adrenergic neuron blockers (central and peripheral)- we won’t talk about this A2. Alpha1 receptor blockers (peripheral) A3. Alpha2 receptor agonists (central) A4. Beta receptor blockers (peripheral) A5. Combined α and β receptor blockers (peripheral) A2. Peripherally Acting Adrenergic Drug Alpha1 Blockers (we’ve met these) Doxazosin, prazosin, alfuzosin Block alpha1-receptors which causes BP to decrease Reduces peripheral vascular resistance and BP by dilating both arterial and venous blood vessels Main Use: benign prostatic hyperplasia (BPH) Alpha1 Blockers REMEMBER Tamsulosin (Flomax)* is an α1 blocker BUT *Tamsulosin is not used to control BP, just for BPH. A3. Centrally Acting Adrenergic Drugs Alpha 2 agonist Clonidine and methyldopa 1- Stimulate alpha2-adrenergic receptors. in the brain Decreases sympathetic outflow from the CNS which decreases NE production 2. Stimulate alpha2-adrenergic receptors in kidneys remember alpha 2 opposes alpha 1 Dilates peripheral blood vessels → lowers peripheral resistance → Results in decreased BP So ….Clonidine (Catapres) Used primarily for its ability to decrease blood pressure in an urgent setting Also use in opioid withdrawal as previously discussed Oral (multiple times a day), and topical patch formulations Do not stop abruptly as it may lead to rebound hypertension In reality, Clonidine and methyldopa Not prescribed as first-line home antiHTN drugs High incidence of unwanted adverse effects: orthostatic hypotension, fatigue, and dizziness MIGHT be uses as adjunct drugs after other drugs have failed, in conjunction with other antiHTN such as diuretics A4. Adrenergic Drugs Selective Beta 1 Blockers Metoprolol, Atenolol Reduction of HR through β1 receptor blockade (remember adrenergic blocking of this receptor???) HR results in BP Cause reduced secretion of renin = BP A4. Adrenergic Drugs Selective Beta1 Blockers Nebivolol (Bystolic) Uses: hypertension and HF Action: blocks β1 receptors and produces vasodilatation, which results in a decrease in SVR High doses loses selectivity and blocks both β1 and β2 Less sexual dysfunction All BB- Do not stop abruptly; must be tapered over 1 to 2 weeks A4. Adrenergic Drugs NONSelective Beta Blockers Propranolol Acts equally on β1 and β2 Other uses include situational anxiety associated with public speaking, test taking As mentioned on previous slide, nebivolol at high doses becomes beta nonselective A5. Dual-Action Adrenergic Drugs α1 and β Receptor Blockers Dual antihypertensive effects of reduction in heart rate (beta1 receptor blockade) and vasodilation (alpha1 receptor blockade) Examples are carvedilol (common) and labetalol (not as common) A5. Dual-Action Adrenergic Drugs α1 and β Receptor Blockers Carvedilol (Coreg) Widely used drug that is well tolerated Uses: HTN, mild to moderate HF in conjunction with digoxin, diuretics, and ACE inhibitors Contraindications: severe bradycardia or unstable HF, bronchospastic conditions such as asthma, and various cardiac conduction problems Adrenergic Drugs Indications - HTN But also for Glaucoma (topical) BPH: doxazosin, prazosin, and terazosin (2 for 1) Management of severe HF when used with cardiac glycosides and diuretics Contraindications Acute HF- have to stabilize first MOAIs- yeah doesn’t everything interact with MAOIs? Peptic ulcers Severe liver/kidney disease Asthma (with beta blockers) Adrenergic Drugs: Adverse Effects Orthostatic hypotension 1st-dose syncope Rebound hypertension with abrupt discontinuation Most common: Dry mouth, drowsiness, constipation, sedation Interactions- always check for specific drug interactions Can cause additive CNS depression with alcohol, benzodiazepines, opioids Question #1 When administering an alpha-adrenergic drug for hypertension, it is most important for the nurse to assess the patient for the development of what response? Hypotension Hyperkalemia Oliguria Respiratory distress Answer A Hypotension This is a key point in patient education These drugs have strong vasodilating properties and may cause severe hypotension, especially at the beginning of therapy. B. Angiotensin-Converting Enzyme Inhibitors aka ACE Inhibitors or ACEi Large group of safe and effective drugs Currently are 10 ACEi Often used as first-line drugs for HF and hypertension May be combined with a thiazide diuretic, loop diuretic, or Calcium Channel Blocker (CCB) You need to understand the basics ACE Inhibitors: Review RAAS ACE converts angiotensin I, formed through the action of renin, to angiotensin II Angiotensin 2 is a potent vasoconstrictor and also induces aldosterone secretion by the adrenal glands Aldosterone stimulates sodium resorption (H20 follows Na Both act to raise BP which causes kidneys to reduce renin production ACEi= Great drug to treat HTN BUT contraindicated in pregnancy (2nd,3rd trimester due to fetal renal damage) and breastfeeding first few weeks after birth B. ACE Inhibitors - PRIL Lisinopril (Prinivil) super common, often the 1st drug Enalapril (Vasotec) also common Captopril (Capoten) great if liver disease present Benazepril (Lotensin) Fosinopril (Monopril) Perindopril (Aceon) Quinapril (Accupril) Ramipril (Altace) Trandolapril (Mavik) Primary Effects of the ACE Inhibitors Prevent Na (and H2O) resorption by inhibiting aldosterone secretion (volume reduction) (GO BACK TO RAAS DIAGRAM) blood volume decreases work of the heart preload, or the left ventricular end-diastolic volume which is important in HF ACE SUMMARY OF ACTIVITY 1) Prevent vasoconstriction caused by angiotensin 2 (2) Prevent aldosterone secretion less sodium and water resorption Cardioprotective Effects of ACEi They slow progression of left ventricular hypertrophy (ventricular remodeling) after MI so considered cardioprotective ACE inhibitors have been shown to decrease morbidity and mortality in patients with HF Renal Protective Effects of ACEi ACE inhibitors: reduce glomerular filtration pressure by volume reduction Cardiovascular drug of choice for patients with diabetes since it helps protect kidneys by reducing pressure. Sometimes used low dose for kidney protection with DM without HTN B. ACEi Enalapril (Vasotec) Only ACEi available in both oral and IV Enalapril IV does not require cardiac monitoring Oral enalapril: prodrug (metabolized in liver) Improves patient’s chances of survival after an MI Reduces the incidence of HF B. ACEi Captopril (Capoten) Uses: prevention of ventricular remodeling after MI; reduce the risk of HF after MI Shortest half-life Must be administered multiple times throughout the day so this limits its use Not a prodrug so good for patient with liver disease Question #2 A patient with diabetes has a new prescription for the ACE inhibitor lisinopril. She questions this order because her provider has never told her that she has hypertension. What is the best explanation for this order? The doctor knows best The patient is confused This medication has cardioprotective properties This medication has a protective effect on the kidneys for patients with diabetes Answer D ACE inhibitors have been shown to have a protective effect on the kidneys because they reduce glomerular filtration pressure. This property makes them the cardiovascular drug of choice for patients with diabetes. Question #3 A patient with a history of pancreatitis and cirrhosis is also being treated for hypertension. Which drug will most likely be ordered for this patient? Clonidine Prazosin Diltiazem Captopril Answer D Captopril Captopril is not a prodrug; therefore, it does not need to be metabolized by the liver to be effective. This is an advantage in patients with liver disease. ACE Inhibitors: Adverse Effects *Dry, nonproductive cough, which reverses when therapy is stopped. This is a class effect Dizziness- Note: First-dose hypotensive effect may occur Headache & Fatigue Possible hyperkalemia ** Angioedema: rare but potentially fatal Not safe in pregnancy-are contraindicated during the second and third trimesters of pregnancy because of increased risk of fetal renal damage C. Angiotensin II Receptor Blockers (ARB) Considered an alternative to ACEi Less likely to cause a dry cough and hyper K+ that is common with ACE inhibitors Angiotensin II Receptor Blockers: Mechanism of Action Go back to RAAS diagram! ARBs affect primarily 2 places 1. Vascular smooth muscle - blocks vasoconstriction 2. Adrenal gland -Selectively blocks the binding of Ang 2 to certain Ang 2 receptors inhibiting secretion of aldosterone Lowers volume retention and BP Angiotensin II Receptor Blockers -ARTAN Losartan (Cozaar)- very common Eprosartan (Teveten) Valsartan (Diovan) Irbesartan (Avapro) Candesartan (Atacand) Olmesartan (Benicar) Telmisartan (Micardis) Azilsartan (Edarbi) C. ARB Losartan (Cozaar) Beneficial in patients with HTN and HF Used with caution in patients with kidney or liver dysfunction and in patients with renal artery stenosis ***Not safe for breastfeeding women and should not be used in pregnancy (Cat C 1st trimester, Cat D 2nd-3rd trimester), potential fetal toxicity Appear to be equally effective for the treatment of hypertension and well tolerated ARBs less likely to cause cough and hyperK+ but can still happen Evidence that ARBs are associated with lower mortality after MI than ACE inhibitors Never take ACEi and ARBs at the same time* 5. Calcium Channel Blockers (CCB) Primary use: HTN, angina, some dysrhythmias Cause smooth muscle relaxation by blocking the binding of calcium to its receptors, preventing muscle contraction Results in: Relaxed blood vessels to the heart Decreased peripheral smooth muscle tone Decreased SVResistance Decreased BP E. Diuretics First-line antiHTN in JNC 8 guidelines Decreases fluid volume The results from diuresis: preload, Peripheral resistance Overall effect Decreased workload of the heart and decreased BP Thiazide diuretics are the most commonly used diuretics for HTN Ie hydrochlorothiazide (HCTZ), chlorthalidone We will discuss diuretics further in the chapter on diuretics F. Vasodilators Directly relax arterial or venous smooth muscle (or both) Results in: Decreased SVR Decreased afterload Peripheral vasodilation Indicated for treatment of HTN May be used in combination with other drugs F. Vasodilators Hydralazine (Apresoline) Orally: routine cases of essential hypertension Injectable: hypertensive emergencies BiDil: specifically indicated as an adjunct for treatment of HF in African-American patients F. Vasodilators Sodium Nitroprusside (Nitropress) *Sodium nitroprusside and IV diazoxide are reserved for the management of hypertensive emergencies. Contraindications: severe HF, known inadequate cerebral perfusion (especially during neurosurgical procedures) F. Vasodilators Adverse Effects Hydralazine: dizziness, headache, tachycardia, edema, dyspnea, N/V/D, vitamin B6 deficiency, rash Sodium nitroprusside: hypotension, bradycardia, decreased platelet aggregation, rash G. Direct Renin Inhibitors Aliskirin (Tekturna) Blocks the RAS pathway at the point of activation. Inhibiting renin production prevents the downstream production of Ang II (potent vasoconstrictor) Adverse effects: N/V, severe hypotension, hyponatremia, hyperkalemia… Contraindicated in patients with DM taking ACEi or ARB Miscellaneous Antihypertensives Eplerenone (Inspra) Newer class of drugs called selective aldosterone blockers (remember RAAS?) Reduces BP by blocking the actions of aldosterone at its corresponding receptors in the kidney, heart, blood vessels, and brain Indications: routine treatment of hypertension and for post-MI HF Contraindicated if serum potassium levels are high (above 5.6 mEq/L) A Special Form of HTN Treatment of Pulmonary Hypertension Sildenafil and Tadalafil Commonly used for erectile dysfunction Used for pulmonary hypertension but with different trade names Sildenafil: Revatio* (Viagra for ED) Tadalafil: Adcirca* (Cialis for ED)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.1. What is the meaning of the word "Izhaar"? A) To hide the sound B) To make it clear C) To change the sound D) To merge two letters 2. Which part of the body is used to pronounce Izhaar Halqi letters? A) The lips B) The tongue only C) The throat D) The nose 3. How many letters are there for Izhaar Halqi? A) 4 letters B) 6 letters C) 15 letters D) 2 letters 4. When do we apply the rule of Izhaar Halqi? A) When any letter comes after Meem Sakinah B) When an Izhaar letter comes after Noon Sakinah or Tanween C) When we see a Shaddah D) Only at the end of a Surah 5. Which of the following is NOT an Izhaar Halqi letter? A) Hamzah (أ) B) Haa (هـ) C) Baa (ب) D) 'Ayn (ع) 6. Which pair of letters comes from the deepest part of the throat (closest to the chest)? A) ع and ح B) غ and خ C) ء and هـ D) ق and ك 7. When you do Izhaar, do you make a long Ghunnah (nasal sound)? A) Yes, a very long one B) No, we pronounce the Noon clearly without extra Ghunnah C) Only if we want to D) Yes, for 2 counts 8. Which letter comes from the top part of the throat (closest to the mouth)? A) Khaad (خ) B) Haa (ح) C) Hamzah (أ) D) Meem (م) 9. What are the middle throat letters? A) ء and هـ B) ع and ح C) غ and خ D) ت and د 10. In the phrase "مَنْ عَمِلَ" (Man 'Amila), which rule is applied? A) Idghaam B) Ikhfaa C) Izhaar Halqi D) Iqlaab 11. Why do we do Izhaar in "مَنْ عَمِلَ"? A) Because the letter 'Ayn (ع) comes after Noon Sakinah B) Because it is easy to say C) Because Meem has a Fathah D) Because the Noon has a Shaddah 12. What does "Noon Sakinah" mean? A) A Noon with a Fathah B) A Noon with a Kasrah C) A Noon with no vowel (has a Sukoon) D) A double Noon 13. What is Tanween? A) A double vowel sign (Fathatain, Kasratain, Dammatain) at the end of a word D) A small Meem on top of a letter C) A stretching sign D) A stop sign 14. Can Izhaar Halqi happen within a single word? A) No, never B) Yes, it can happen in one word or between two words C) Only in short words D) Only in Surah Al-Fatihah 15. Look at the word "وَانْحَرْ" (Wanhar). What is the Izhaar letter here? A) Waw (و) B) Noon (ن) C) Haa (ح) D) Raa (ر) 16. In the Quran, what sign is usually placed on the Noon Sakinah to show it is Izhaar? A) A small circle or head of a Khaa (Sukoon sign) B) A Shaddah C) Nothing at all D) A little Meem 17. What happens to the Tanween vowels when there is Izhaar? A) They are written far apart from each other B) They are aligned perfectly parallel above/below each other C) One vowel is deleted D) They change color 18. Which of the following words contains an Izhaar Halqi rule? A) أَنْعَمْتَ B) مَنْ يَقُولُ C) مِنْ بَعْدِ D) كُنْتُمْ 19. Choose the group that contains ONLY Izhaar Halqi letters: A) ي ، ر ، م ، ل B) ء ، هـ ، ع ، ح ، غ ، خ C) ك ، ق ، ج ، د D) ب ، ت ، ث 20. In the phrase "عَذَابٌ أَلِيمٌ" ( 'Adhaabun Aleem), why is there Izhaar? A) Because Tanween is followed by Hamzah (أ) B) Because it ends with Meem C) Because the word is long D) Because of the letter Laam 21. What is the correct way to read "مِنْ حَكِيمٍ"? A) Mi---hakeem (hide the Noon) B) Min Hakeem (read Noon clearly and quickly) C) Mih-hakeem (mix them together) D) Mim-hakeem (change Noon to Meem) 22. "Ghain" (غ) and "Khaa" (خ) come from which part of the throat? A) Deep throat B) Middle throat C) Top throat D) The lips 23. If a Noon Sakinah is followed by the letter "هـ" (Haa), how do we pronounce it? A) Clear Noon B) Hidden Noon C) Double Noon D) Silent Noon 24. Which of these is a middle throat letter? A) ء B) خ C) ح D) هـ 25. Complete the sentence: Izhaar Halqi means to pronounce the Noon Sakinah or Tanween cleanly from its articulation point without any ________. A) Breathing B) Vowel (Harakah) C) Extra Ghunnah (nasalization) D) Stoppingmĩ thuậtMĩ thuật ứng dụngMĩ thuật việt nam thời kì tiền sửMĩ thuật 6Mĩ thuật 6 bài 1