Two Runners with Different Acceleration and Holding Phases
Two runners race on a straight 100 m track. Runner A starts from rest at t=0 with constant acceleration 2.0 m/s² for 3.0 s, then runs at the speed reached. Runner B starts 1.0 s later from rest and accelerates at 2.8 m/s² until reaching 8.4 m/s, then holds that speed. A coach claims B will catch A before the finish line because B’s acceleration is larger. Use a consistent coordinate origin at the starting line and treat both runners as particles. Assume no reaction-time delays beyond B’s stated start and no lane differences. The track is long enough for both to finish.
1. Determine whether B catches A before the finish line. If so, find the catch time and position; if not, justify why not.
Read the worked answer
Runner A: For 0 ≤ t ≤ 3.0 s, x_A = 0.5(2.0)t² = t², so at t=3.0 s, v_A = 6.0 m/s and x_A = 9.0 m. For t ≥ 3.0 s, x_A = 9.0 + 6.0(t − 3.0) = 6.0t − 9.0.
Runner B: Starts at t=1.0 s. The acceleration phase lasts 8.4/2.8 = 3.0 s, so it ends at t=4.0 s, with x_B = 0.5(2.8)(3.0)² = 12.6 m. For t ≥ 4.0 s, x_B = 12.6 + 8.4(t − 4.0) = 8.4t − 21.0.
Set positions equal: 6.0t − 9.0 = 8.4t − 21.0, giving 2.4t = 12.0 and t = 5.0 s. Then x = 6.0(5.0) − 9.0 = 21 m. This is well before 100 m, so B does catch A at t=5.0 s and x=21 m.
2. Compare the runners’ speeds at the catch and determine who finishes first, including the time margin.
Read the worked answer
At t=5.0 s, A is in the constant-speed phase with v_A = 6.0 m/s, while B is in the constant-speed phase with v_B = 8.4 m/s. B passes A with the larger speed.
Finish times:
- A: 100 = 6.0t − 9.0 gives t = 109/6.0 = 18.17 s.
- B: 100 = 8.4t − 21.0 gives t = 121/8.4 = 14.40 s.
B finishes first by about 18.17 − 14.40 = 3.77 s. When B finishes, A is at x_A = 6.0(14.40) − 9.0 = 77.4 m, so B wins by about 22.6 m as well as by time.
3. Suppose B’s acceleration phase lasted only 2.0 s before reaching constant speed. Would B still catch A before the finish line? Justify with kinematics rather than a guess.
Read the worked answer
If B accelerates for only 2.0 s, its final speed is (2.8)(2.0) = 5.6 m/s. Starting at t=1.0 s, that phase ends at t=3.0 s, with x_B = 0.5(2.8)(2.0)² = 5.6 m. At t=3.0 s, A is at x_A = 9.0 m, so B is 3.4 m behind.
After t=3.0 s, A moves at 6.0 m/s and B at 5.6 m/s. Since A is already ahead and is faster, the gap increases. B never catches A before the finish line. The larger acceleration alone is not enough; B also needs enough time in the acceleration phase to reach a speed greater than A’s.