Answer:
a) 6.46 seconds
b) 334 feet
c) Kathy: 103 ft/s; Stan: 89.6 ft/s
Explanation:
At constant acceleration:
x = x₀ + v₀ t + ½ at²
where x is final position,
x₀ is initial position,
v₀ is initial velocity,
a is acceleration,
and t is time.
For Kathy, x₀ = 0 ft, v₀ = 0 ft/s, a = 16 ft/s², and t = t.
For Stan, x₀ = 0 ft, v₀ = 0 ft/s, a = 12 ft/s², and t = t + 1.
Kathy:
x = 0 + (0) t + ½ (16) t²
x = 8t²
Stan:
x = 0 + (0) (t + 1) + ½ (12) (t + 1)²
x = 6 (t + 1)²
a) When Kathy overtakes Stan, they have the same position:
8t² = 6 (t + 1)²
8t² = 6 (t² + 2t + 1)
8t² = 6t² + 12t + 6
2t² − 12t − 6 = 0
t² − 6t − 3 = 0
Solving with quadratic formula (or complete the square):
t = [ -(-6) ± √((-6)² − 4(1)(-3)) ] / 2(1)
t = [ 6 ± √(36 + 12) ] / 2
t = (6 ± √48) / 2
t = (6 ± 4√3) / 2
t = 3 ± 2√3
Since t > 0, t = 3 + 2√3 ≈ 6.46 seconds.
b) Kathy's position when she overtakes Stan is:
x = 8t²
x = 8 (3 + 2√3)²
x ≈ 334 feet
c) To find the final velocities, we use:
v = at + v₀
For Kathy:
v = (16) (3 + 2√3) + 0
v ≈ 103 ft/s
For Stan:
v = (12) (3 + 2√3 + 1) + 0
v ≈ 89.6 ft/s
