MATH 2203 Quiz 2 Solutions: Velocity, Speed, Acceleration, and Motion Equations - Prof. Se, Quizzes of Advanced Calculus

Download MATH 2203 Quiz 2 Solutions: Velocity, Speed, Acceleration, and Motion Equations - Prof. Se and more Quizzes Advanced Calculus in PDF only on Docsity! MATH 2203 Quiz 2 Solutions February 14, 2011 NAME_______________________________ 1) The position of a particle moving in the xy plane at time t is given by r (t) = cos (2t) i+ 3 sin (2t) j. Compute the velocity vector (v (t)), the speed (v (t)), the acceleration vector (a (t)). Also …nd an equation in x and y whose graph is the path of the particle. Answers: v (t) = v (t) = a (t) = Equation of path of motion: Show your work here: Solution: The velocity is v (t) = 2 sin (2t) i+ 6 cos (2t) j and the acceleration is a (t) = 4 cos (2t) i 12 sin (2t) j. The speed is v (t) = jv (t)j = q 4 sin2 (2t) + 36 cos2 (2t). To …nd an equation (in x and y) for the path of motion, note that parametric equations for the path of motion are x = cos (2t) y = 3 sin (2t) . 1 We can write these as x = cos (2t) y 3 = sin (2t) and observe that x2 + y 3 2 = cos2 (2t) + sin2 (2t) = 1. Thus the path along which the motion takes place is x2 + y2 9 = 1. (This is an ellipse.) 2) A projectile is …red with an initial speed of 500 meters per second at an angle of elevation of 45. When and how far away will the projectile strike? In answering this question, you must show all of your work beginning with the assumption that the acceleration of the projectile is given at all times by a (t) = gj where g = 9:8m= s2. Solution: I will skip showing the process of how to derive the equations of motion since we did this a few times in class. (Consult your class notes or ask me about this if you are unsure.) We end up obtaining the parametric equations x = v0 cos ( ) t y = v0 sin ( ) t 1 2 gt2 which (with the given information) becomes x = 500 cos (45) t = 250 p 2t y = 500 sin (45) t 1 2 gt2 = 250 p 2t 1 2 gt2. To …nd the time at which the projectile strikes the ground, we set y = 0: Using the factor that t 6= 0, we obtain 250 p 2 1 2 gt = 0 2