phyfinal2 - Linear motion topics 15 Linear motion -...

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Unformatted text preview: Linear motion topics 15 Linear motion - problems A student driving up Gaglardi way at 72 km/hr is late for class. When he is 100 m from the Curtis/Gagliardi stoplight, the light turns yellow. He hits the gas and accelerates at 3 m/s 2 . But after 2 seconds, he decides that he can't make the light in time and jams on the brakes. He decelerates uniformly, coming to a stop at the light. What is his deceleration? During accelerat ion v i = 72 km/hr = 72 x 10 3 / 3.6 x 10 3 = 20 m/s a = 3 m/s t = 2 s The distance he covers during acceleration is d = v i t + at 2 /2 = 20 2 + 0.5 3 2 2 = 46 m and his final velocity is v f = v i + at = 20 + 3 2 = 26 m/s. During deceleration the distance he covers is 100 - 46 = 54 m His deceleration is therefore a = ( v f 2- v i 2 )/2a = (0 - 26 2 ) / (2 54) = -6.3 m/s 2 . Linear motion topics 16 The geometry of the "shoot-the-can" demonstration done in class is approximately tin can blow tube 10 m line of sight 30 o The projectile is aimed at an angle of 30 o with respect to the horizontal, and the horizontal distance between the end of the blow tube and the can is 10 m. How far will the can fall before the projectile strikes it, assuming that the can starts to fall as soon as the projectile leaves the tube? The speed of the projectile when it leaves the tube is 24 m/s (neglect air resistance). The horizontal velocity, v x , of the projectile as it leaves the tube is v x = 24 cos 30 o = 24 0.866 = 20.8 m/s. Thus, the projectile covers the horizontal distance to the falling can in a time t , given by t = d / v x = 10 / 20.8 = 0.48 s. Since the can starts off with zero velocity, then during time t , the can has fallen a height given by h = - gt 2 / 2, where g is the acceleration due to gravity. Thus, h = - 9.8 (0.48) 2 / 2 = - 1.13 m. Linear motion topics 17 The space shuttle orbiter is moving in a circular orbit with a speed of 7 km/s and a period of 80 minutes. In order to return to Earth, the orbiter fires its engines opposite to its direction of motion. The engines provide a deceleration in this direction of 20 m/s 2 . What is the magnitude and direction of the total acceleration of the orbiter when the engines are first ignited? The centripetal acceleration is given by a c = 2 v / T = 2 7x10 3 / (80 60) = 9.16 m/s 2 Because a c is perpendicular to the direction of motion, the total acceleration can be found from pythagoras theorem to be a tot = ( a c 2 + a tan 2 ) 1/2 = (9.16 2 + 20 2 ) 1/2 = 22.0 m/s 2 . The angle of the acceleration vector is away from the direction of motion, in the general direction of the Earth, making an angle with respect to the vertical obtained from tan = 20.0 / 9.16 = 2.18 or = 65.4 o . Linear motion topics 18 An object moves horizontally with a position given as a function of time t by x ( t ) = at 3- bt , where a and b are positive constants....
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This note was uploaded on 06/14/2011 for the course PHYS 101 taught by Professor Vighen during the Fall '08 term at Simon Fraser.

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phyfinal2 - Linear motion topics 15 Linear motion -...

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