a Use Eq 2 12a to determine the duration of that acceleration acc 2 1389m s 0m

A use eq 2 12a to determine the duration of that

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a Use Eq. 2-12a to determine the duration of that acceleration. 0 0 acc 2 13.89m s 0m s 6.94s 2.0m s v v v v at t a The distance traveled during that time is found from Eq. 2-12b. 2 2 2 1 1 0 0 acc acc 2 2 acc 0 2.0m s 6.94s 48.2m x x v t at Since 6.94 s have elapsed, there are 13 6.94 = 6.06 s remaining to clear the intersection. The car travels another 6.06 s at a speed of 13.89 m/s, covering a distance of constant speed avg x v t  13.89m s 6.06 s 84.2m. Thus the total distance is 48.2 m + 84.2 m = 132.4 m. No , the car cannot make it through all three lights without stopping. The car has to travel another 32.6 m to clear the third intersection, and is traveling at a speed of 13.89 m/s. Thus the care would enter the intersection a time 32.6m 2.3s 13.89m s x t v after the light turns red.
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79. In putting, the force with which a golfer strikes a ball is planned so that the ball will stop within some small distance of the cup, say 1.0 m long or short, in case the putt is missed. Accomplishing this from an uphill lie (that is, putting the ball downhill, see Fig. 2 48) is more difficult than from a downhill lie. To see why, assume that on a particular green the ball decelerates constantly at 2 1.8 m/s going downhill, and constantly at 2 2.8 m/s going uphill. Suppose we have an uphill lie 7.0 m from the cup. Calculate the allowable range of initial velocities we may impart to the ball so that it stops in the range 1.0 m short to 1.0 m long of the cup. Do the same for a downhill lie 7.0 m from the cup. What in your results suggests that the downhill putt is more difficult? First consider the “uphill lie,” in which the ball is being putted down the hill. Choose 0 0 x to be the ball’s original location, and the direction of the ball’s travel as the positive direction. The final velocity of the ball is 0m s , v the acceleration of the ball is 2 1.8m s , a   and the displacement of the ball will be 0 6.0m x x for the first case and 0 8.0m x x for the second case. Find the initial velocity of the ball from Eq. 2-12c. 2 2 2 2 0 0 0 0 2 0 2 1.8m s 6.0m 4.6m s 2 2 0 2 1.8m s 8.0m 5.4m s v v a x x v v a x x The range of acceptable velocities for the uphill lie is 4.6m s to 5.4m s , a spread of 0.8 m/s. Now consider the “downhill lie,” in which the ball is being putted up the hill. Use a very similar set -up for the problem, with the basic difference being that the acceleration of the ball is now 2 2.8m s . a   Find the initial velocity of the ball from Eq. 2-12c. 2 2 2 2 0 0 0 0 2 0 2 2.8m s 6.0m 5.8m s 2 2 0 2 2.8m s 8.0m 6.7m s v v a x x v v a x x The range of acceptable velocities for the downhill lie is 5.8m s to 6.7m s , a spread of 0.9 m/s.
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Because the range of acceptable velocities is smaller for putting down the hill, more control in putting is necessary, and so putting the ball downhill (the “uphill lie”) is more difficult.
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