2 - Motion in One Dimension

# You will gain a great deal of experience in the use

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You will gain a great deal of experience in the use of these equations by solving a number of exercises and problems. Many times you will discover that more than one method can be used to obtain a solution. Remember that these equations of kinemat- ics cannot be used in a situation in which the acceleration varies with time. They can be used only when the acceleration is constant. x f x i v xi t 1 2 a x t 2 38 CHAPTER 2 Motion in One Dimension

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SECTION 2.5 One-Dimensional Motion with Constant Acceleration 39 Example 2.7 Carrier Landing If the plane travels much farther than this, it might fall into the ocean. The idea of using arresting cables to slow down landing aircraft and enable them to land safely on ships originated at about the time of the fi rst World War. The ca- bles are still a vital part of the operation of modern aircraft carriers. What If? Suppose the plane lands on the deck of the air- craft carrier with a speed higher than 63 m/s but with the same acceleration as that calculated in part (A). How will that change the answer to part (B)? Answer If the plane is traveling faster at the beginning, it will stop farther away from its starting point, so the answer to part (B) should be larger. Mathematically, we see in Equa- tion 2.11 that if v xi is larger, then x f will be larger. If the landing deck has a length of 75 m, we can fi nd the maximum initial speed with which the plane can land and still come to rest on the deck at the given acceleration from Equation 2.13: 68 m/s 0 2( 31 m/s 2 )(75 m 0) : v xi v xf 2 2 a x ( x f x i ) v 2 xf v 2 xi 2 a x ( x f x i ) A jet lands on an aircraft carrier at 140 mi/h ( 63 m/s). (A) What is its acceleration (assumed constant) if it stops in 2.0 s due to an arresting cable that snags the airplane and brings it to a stop? Solution We de fi ne our x axis as the direction of motion of the jet. A careful reading of the problem reveals that in ad- dition to being given the initial speed of 63 m/s, we also know that the fi nal speed is zero. We also note that we have no information about the change in position of the jet while it is slowing down. Equation 2.9 is the only equation in Table 2.2 that does not involve position, and so we use it to fi nd the acceleration of the jet, modeled as a particle: (B) If the plane touches down at position x i 0, what is the fi nal position of the plane? Solution We can now use any of the other three equations in Table 2.2 to solve for the fi nal position. Let us choose Equation 2.11: 63 m x f x i 1 2 ( v xi v xf ) t 0 1 2 (63 m/s 0)(2.0 s) 31 m/s 2 a x v xf v xi t 0 63 m/s 2.0 s Example 2.8 Watch Out for the Speed Limit! A car traveling at a constant speed of 45.0 m/s passes a trooper hidden behind a billboard. One second after the speeding car passes the billboard, the trooper sets out from the billboard to catch it, accelerating at a constant rate of 3.00 m/s 2 . How long does it take her to overtake the car?

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