# © Samir N. Shoukry, 2004, Dynamics MAE 242. Quiz 2 (5 minutes) A car accelerates according to the relation a=0.02s m/s 2. Determine its velocity when.

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© Samir N. Shoukry, 2004, Dynamics MAE 242

Quiz 2 (5 minutes) A car accelerates according to the relation a=0.02s m/s 2. Determine its velocity when s=100 m if s=v=0 when t=0.

© Samir N. Shoukry, 2004, Dynamics MAE 242 How Position May Be Described? New York Morgantown Position is defined relative to a point on which we set a coordinate system

© Samir N. Shoukry, 2004, Dynamics MAE 242 OR Position is defined by a vector “r” that originates from a reference point Remember: A vector has a length and direction New York Path N E W S r M/N Position Vector Position Vector New York Path Position Vector New York Path Position Vector r N/M N E W S r M/W r N/W

© Samir N. Shoukry, 2004, Dynamics MAE 242 r = x i + y j r = x i + y j + z k Position is defined by the distance from three perpendicular lines from an origin O. Rectangular Coordinates

© Samir N. Shoukry, 2004, Dynamics MAE 242 Rectangular Coordinates: Velocity & Acceleration r = x i + y j + z k v = v x i + v y j + v z k a = a x i + a y j + a z k d dt v= drdr dt = (xi)(xi) (yj)(yj) (zk)(zk) + d d + dx dt i + didi x d (xi)(xi)=== dx dt i vxivxi

© Samir N. Shoukry, 2004, Dynamics MAE 242 Example: Given, Calculate the magnitude and direction of the velocity and acceleration vectors. Calculate the magnitude of the position vector. y 2

© Samir N. Shoukry, 2004, Dynamics MAE 242 Motion of A Projectile

© Samir N. Shoukry, 2004, Dynamics MAE 242

Quiz 3 (5 minutes) A particle moves according to the relation r=5t 3 i +2t 2 j Calculate the magnitude and direction of its acceleration when t =0.5 second.

© Samir N. Shoukry, 2004, Dynamics MAE 242 Example: A particle accelerates according to the relation: Given the initial conditions at t=0: x, y, z =0 and v x =v y =1 & v z =0, calculate the particle velocity & position.

© Samir N. Shoukry, 2004, Dynamics MAE 242 Motion of A Projectile

© Samir N. Shoukry, 2004, Dynamics MAE 242 A skier leaves the ramp A at angle with the horizontal. He strikes the ground at point B, Determine his initial speed v A and the time of flight from A to B. -64

© Samir N. Shoukry, 2004, Dynamics MAE 242 x y

y 5 m

© Samir N. Shoukry, 2004, Dynamics MAE 242 The boy at A attempts to throw a ball over the roof of a barn such that it is launched at an angle of 40 degrees. Determine the speed v A at which he must throw the ball so that it reaches its maximum height at C. Also, find the distance d where the boy must stand so that he can make the throw. Given: Y 0 = 1m, At max. height, v Y = 0, Y = 8m and v 0x = v 0 cos40 0 ; v 0y = v 0 sin40 0 (1) X = v 0x t; (2) v x = v 0x (constant velocity in x-direction); (3) Y = Y 0 ­1/2gt 2 + v 0Y t(4) v Y = v 0Y ­ gt (constant acceleration -g in y-direction) At the top of the trajectory v y = 0 and from (4) we get: v 0y = gt Substituting the above into (3) we get 7 = ½ g t 2 and solving the above to get the time to reach maximum elevation. t = 1.195 seconds Substituting for v 0Y at the top of the trajectory, we get v 0 sin40 o = 9.81(1.195) which is solved for the initial velocity. v 0 = 18.24 m/sec Substituting into (1) we get the total distance X. X = 18.24(cos40)(1.195) = 16.7 m Which gives d = X ­ 4 = 12.70m

© Samir N. Shoukry, 2004, Dynamics MAE 242 The ball is thrown from the tower with a velocity of 20 ft/s as shown. Determine the x and y coordinates to where the ball strikes the slope. Also, determine the speed at which the ball hits the ground. Initial Conditions: v 0 = 20 ft/s v 0x = 20(3/5) = 12 ft/s v 0y = 20(4/5) = 16 ft/s theta = 26.6 0, X 0 = 0 Y 0 = 80 ft Motion in the x-direction at a constant velocity. (1) v x = v 0x = 12 ft/s (2) X = v x t = 12t Motion in the y-direction at a constant acceleration a = -g. Integrating for velocity (3) v y = v 0y - gt = 16 - 32.2(t) and the y-displacement is obtained by integrating the latter, which gives Y = -1/2gt 2 + v 0y t + Y 0 or (4) Y = -16.1t 2 +16t + 80 The x and y coordinates of the ball when it hits the ground can be written as. (5) x = 20 + Rcos26.6 o ; (6) y = Rsin26.6 o Equations (2), (4) (5) and (6) are four equations in four unknowns. They can be solved for t - the time when the ball hits the ground. Thus t = 2.7 sec. Substituting into (2) and (4) provides the x and y coordinates of the ball when it hits the ground. x = 32.4 ft; y = 6.56 ft To solve for the speed, we substitute t = 2.7s into equation (3) which gives. v x = 12 ft/sand v y = 70.9 ft/s The total speed is found from v = (v x 2 + v y 2 ) 1/2 ; v = 71.9ft/s

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