# chap4 - CHAPTER 4 MOTION IN MORE THAN ONE DIMENSION Problem...

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CHAPTER 4 MOTION IN MORE THAN ONE DIMENSION ActivPhysics can help with these problems: All Activities in Section 3, Projectile Motion Section 4-1: Velocity and Acceleration Problem 1. A skater is gliding along the ice at 2.4 m/s, when she undergoes an acceleration of magnitude 1.1 m/s 2 for 3.0 s. At the end of that time she is moving at 5.7 m/s. What must be the angle between the acceleration vector and the initial velocity vector? Solution For constant acceleration, Equation 4-3 shows that the vectors v 0 , a Δ t and v form a triangle as shown. The law of cosines gives v 2 = v 2 0 + ( a Δ t ) 2 2 v 0 a Δ t × cos(180 θ 0 ). When the given magnitudes are substituted, one can solve for θ 0 : (5 . 7 m/s) 2 = (2 . 4 m/s) 2 + (1 . 1 m/s 2 ) 2 (3 . 0 s) 2 + 2(2 . 4 m/s) × (1 . 1 m/s 2 )(3 . 0 s) cos θ 0 , or cos θ 0 = 1 . 00 (exactly), and θ 0 = 0 . Since v 0 and a are colinear, the change in speed is maximal. Problem 1 Solution. Problem 2. In the preceding problem, what would have been the magnitude of the skater’s ±nal velocity if the acceleration had been perpendicular to her initial velocity? Solution When v 0 and a Δ t are perpendicular, Equation 4-3 and the Phythagorean theorem imply v = r v 2 0 + ( a Δ t ) 2 = R (2 . 4 m/s) 2 + (1 . 1 m/s 2 )(3 . 0 s) 2 = 4 . 08 m/s . Problem 3. An object is moving in the x direction at 1.3 m/s when it is subjected to an acceleration given by a= 0 . 52 ˆ m/s 2 . What is its velocity vector after 4.4 s of acceleration? Solution From Equation 4-3, v =v 0 +a Δ t = (1 . 30 m/s) ˆ ı + (0 . 52 ˆ m/s 2 )(4 . 4s) = (1 . 30 ˆ ı+ 2 . 29 ˆ ) m/s . Problem 4. An airliner is ²ying at a velocity of 260 ˆ ı m/s, when a wind gust gives it an acceleration of 0 . 38 ˆ ı+ 0 . 72 ˆ m/s 2 for a period of 24 s. (a) What is its velocity at the end of that time? (b) By what angle has it been de²ected from its original course? Solution (a) Equation 4-3 gives v = 260 ˆ ı m/s + (0 . 38 ˆ ı+ 0 . 72 ˆ )(m / s 2 )(24 s) = (269 ˆ ı+ 17 . 3 ˆ ) m/s . (b) Since v 0 is along the x -axis, the angular de²ection is just tan 1 ( v y /v x ) = tan 1 (17 . 3 / 269) = 3 . 67 . Section 4-2: Constant Acceleration Problem 5. The position of an object as a function of time is given by r = (3 . 2 t + 1 . 8 t 2 ) ˆ ı+ (1 . 7 t 2 . 4 t 2 ) ˆ m, where t is the time in seconds. What are the magnitude and direction of the acceleration? Solution One can always ±nd the acceleration by taking the second derivative of the position, a ( t ) = d 2 r ( t ) /dt 2 . However, collecting terms with the same power of t, one can write the position in meters as r ( t ) = (3 . 2 ˆ ı+ 1 . 7 ˆ ) t + (1 . 8 ˆ ı - 2 . 4 ˆ ) t 2 . Comparison with Equation 4-4 shows that this represents motion with constant acceleration equal to twice the coe³cient of the t 2 term, or a = (3 . 6 ˆ ı - 4 . 8 ˆ ) m/s 2 . The magnitude and direction of a are r (3 . 6) 2 + ( 4 . 8) 2 m / s 2 = 4 . 49 m/s 2

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## This note was uploaded on 10/31/2010 for the course PHYSICS 69054 taught by Professor Huang during the Fall '09 term at Berkeley.

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chap4 - CHAPTER 4 MOTION IN MORE THAN ONE DIMENSION Problem...

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