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: 𝑣!! = 0.0 m/s, 𝑎! = 2.4 m/s! [right],
Δ𝑡! = Δ𝑡! = Δ𝑡 R: (a) Δ𝑡 (b) Δ𝑑! and Δ𝑑! 𝑣! = 5.4 m/s [left], 𝑎! = 0.0 m/s !
Δ𝑑 ! = 42 m (c) 𝑣!! using time from part (a) A: (a) and (b) Δ𝑑! = 𝑣!!Δ𝑡 + !! 𝑎! Δ𝑡 ! !
= ! 𝑎! Δ𝑡 ! Δ𝑑 ! = Δ𝑑! + Δ𝑑! !
Δ𝑑 ! = ! 𝑎! Δ𝑡 ! + 𝑣!! Δ𝑡 Δ𝑑! = 𝑣!! Δ𝑡 + ! 𝑎! Δ𝑡 ! = 𝑣!! Δ𝑡 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.
(c) 𝑣!! = 𝑎! Δ𝑡 + 𝑣!! = 𝑎! Δ𝑡 S: (a) 42 m = !! (2.4 m/s!)Δ𝑡 ! + (5.4 m/s )Δ𝑡 !
1.2Δ𝑡 ! + 5.4Δ𝑡 − 42 = 0 Allen and Bob collided 8.6 seconds
before they started running ?!?!?
Using the quadratic formula, Δ𝑡 = 4.08 seconds or Δ𝑡 = −8.58 seconds !
(b) Δ𝑑! = ! 2.4 m s ! 4.08 s ! Δ𝑑! = 5.4 m s 4.08 s = 19.98 m = 22.03 m (c) 𝑣!! = 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.