19. a young woman named kathy kool buys a superde-
luxe sports car that can accelerate at the rate of
16 ft/s2. she decides to test the car by dragging with
another speedster, stan speedy. both start from rest,
but experienced stan leaves 1 s before kathy. if stan
moves with a constant acceleration of 12 ft/s2 and
kathy maintains an acceleration of 16 ft/s2, find
(a) the time it takes kathy to overtake stan, (b) the
distance she travels before she catches him, and
(c) the velocities of both cars at the instant she over-
takes him.

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

moving ball hits still ball. mover stops, hit ball moves. momentum transfer.