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Example 2: Two football players separated by 42 m run directly toward each other. Football player Allen starts from rest and accelerates at 2.4 m/s ! [right], and football player Bob moves uniformly at 5.4 m/s [left]. (a) How long does it take for the players to collide? (b) How far does each player move? (c) How fast is football player Allen moving when they collide? 42 m G: !! = 0.0 m/s, ! = 2.4 m/s ! [right], ! = 5.4 m/s [left], ! = 0.0 m/s ! Δ ! = Δ ! = Δ Δ ! = 42 m R: (a) Δ (b) Δ ! and Δ ! (c) !! using time from part (a) A: (a) and (b) Δ ! = !! Δ + ! ! ! Δ ! Δ ! = !! Δ + ! ! ! Δ ! = ! ! ! Δ ! = !! Δ Δ ! = Δ ! + Δ ! Δ ! = ! ! ! Δ ! + !! Δ (c) !! = ! Δ + !! = ! Δ S: (a) 42 m = ! ! (2.4 m/s ! )Δ ! + (5.4 m/s )Δ 1.2Δ ! + 5.4Δ − 42 = 0 Using the quadratic formula, Δ = 4 .08 seconds or Δ = −8.58 seconds (b) Δ ! = ! ! 2.4 m s ! 4.08 s ! Δ ! = 5.4 m s 4.08 s = 19 .98 m = 22 .03 m (c) !! = 2.4 m s ! right 4.08 s = 9 .79 m/s right S: Therefore, the football players collide after 4.1 seconds. At this time, Allen moved 20 m and Bob moved 22 m. Allen was moving at 9.8 m/s when they collided. Even though the two vectors are in opposite directions, the magnitudes of the two displacements is 42 m. So we just add the magnitudes and we don’t need to use the vectors. Allen and Bob collided 8.6 seconds before they started running ?!?!?
