Rubric3A

Rubric3A - Quiz 3A DL Sec Grading: Name First three letters...

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Unformatted text preview: Quiz 3A DL Sec Grading: Name First three letters of your family name Last 6 digits of student ID: 1. A car is traveling at a constant speed of 40mph on a curvy road shown below. (a) Draw position vectors and velocity vectors with respect to the origin shown, when the car reaches each of the two points : point A and point B. Label your vectors clearly. (b) In the space given on the right, draw and label the change in the velocity vector between the point A and the point B, i.e., ∆vAB. Additionally, draw vector subtraction that gives ∆vAB as well. A vA FINISH Show ∆vAB and vector subtraction here : ∆vAB = vB – vA –vA CAR RA RB ∆RAB START B vB ∆vAB vB Three letters are assigned for this problem. One for position vectors, one for velocity vectors and one for the graphical calculation of ∆vAB. 3 A correct 2 B minor mistake 1 C major mistakes or several minor * Plus additional 1 point for attendance Position: Two vectors, from origin to points A, B, labeled appropriately, with arrow heads on the vectors. Velocity: Two vectors from (more or less) points A,B, directed tangentially along the path, labeled appropriately, with arrow heads on the vectors. ∆vAB: Redraw both velocity vectors on the right in configuration for subtraction - either head to tail method or parallelogram method. Find difference vector graphically and label all. Exponential change model dy/dt = given by y(t) = yoe ±kt ±ky, i.e. k is the constant of proportionality. A solution to this equation is 2. In a simple model, the quantity of a drug in a patient’s body decreases at a rate proportional to the quantity remaining in the body. In this problem, 20 milligram of a drug was administered to a patient at t = 0 hours. The table below summarizes the amount of the drug remaining in the body at later time. Time (hours) t = 0 hours t = 2.25 hours Amount of the drug 20 milligram 1/e ( = 0.368) of the administered amount remaining in the body (a) Draw a curve that describes this decay on the graph on the right. Extend the curve accurately to t = 10 hours. Label the half-life and the 1/e life (i.e., time constant) clearly. Show your work below. The decay of the amount of the drug remaining in the body is described by the exponential change model with the time constant of 2.25 hours. At t = 2.25 hours, y(t) = (1/e)*20mg = 7.36mg At t = 4.5 hours, y(t) = (1/e)*7.36mg = 2.7mg At t = 6.75 hours, y(t) = (1/e)*2.7mg = 0.99mg Draw the curve using above data points, and then estimate the half-life (≈ 1.6hours). OR alternatively, you can find the function that describes the quantity of the drug in the body and calculate above data points. After 2.25 hours, the amount of drug decreases to 1/e of the original amount, y(t = 2.25hours) = 20mg x e = 20 mg x (1/e) Therefore, k * (2.25 hours) = 1, ∴ k = 0.44 – 0.44 t ∴ y(t) = 20mg x e Now find Half-life(t1/2), – 0.44 t1/2 y(t1/2) =10mg = 20mg x e ∴e – 0.44 t1/2 0.44 t1/2 – k(2.25hours) Half-life =1.57 hours 1/e life (Time constant) = = 2 hours 15 min. = 1/2 e =2 (Take natural log of both sides) 0.44 * t1/2 = ln 2, t1/2 = 1.57 hours At t= 10 hours, y(t = 10 hours) = 20mg x e use above data points to draw the curve. – 4.4 = 0.25mg (b) Write down the function that describes the amount of the drug remaining in the body at a time t. From (a), y(t) = 20mg x e – 0.44 t Exponential change model dy/dt = given by y(t) = yoe ±kt ±ky, i.e. k is the constant of proportionality. A solution to this equation is First letter is a categorization by graph. 5 A 4 B 3.5 D 3 C 2 E 2 F basically correct go roughly through point (2.25 hours, 12.64mg) go roughly through point (2.25 hours, 10mg) go roughly through point (2.25 hours, 0.368mg) other not a graph or more than one Second letter is for *labeling* 1/e life and 1/2 life. Half-life could be calculated or estimated from the graph. Just dots are not ok. Since correctness here overlaps strongly with the first letter, we look for consistency. 2 A both correct/consistent 1.5 B one correct/consistent 0.5 C neither correct/consistent Third letter is for the formula. Look for correct yo, k and sign of exponent. Values must be inserted into formula or appear defined under part b. 3 A 2.6 B 2.0 C 1 D 0 E all three correct two correct one correct zero correct blank Exponential change model dy/dt = given by y(t) = yoe ±kt ±ky, i.e. k is the constant of proportionality. A solution to this equation is ...
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This note was uploaded on 10/26/2008 for the course PHY 7b taught by Professor Taylor during the Spring '08 term at UC Davis.

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