WALL-E Problem Set 1 - Solutions

1. 2.362 x 10⁴ in/min
2.
WALL-E: v0 = 0, delta_x = 0.5*a*(delta_t)^2. Rearranging, delta_t = 11.55 seconds.
EVE: delta_x = v0*delta_t so delta_t = 100 m / (10 m/s) = 10 seconds
3.
WALL-E: Travels 75 meters during the first 10 seconds (from constant acceleration equation used in #2). His average velocity is 75 m / 10 s = 7.5 m/s
EVE: average velocity is the same as initial velocity since her velocity is constant, so average = 10 m/s.
4. Using either kinematics equation and the answer to #2 for WALL-E, the final speed is 17.325 m/s.
5.At time t = 11.55 seconds, EVE has traveled 115.5 meters, and WALL-E has traveled 100 meters. This means that EVE is still 15.5 meters ahead.

WALL-E position function = EVE position function at time t₂ after they have BOTH traveled 100 meters.
100 m + (17.321 m/s)*t₂ = 115.5 m + (10 m/s)*t₂

Solving, t₂ = 2.117 seconds.

Since this occurs after the 11.55 seconds it takes for WALL-E to finish accelerating, WALL-E catches up to EVE after 11.55 + 2.117 = 13.68 seconds.

6.
EVE: 15 seconds.
WALL-E: Must travel 50 meters past the 1st 100 meters. v0 = 17.321 m/s, delta_x = 50 m, so t = 2.887 seconds. Total time of 14.44 seconds.

7. WALL-E will observe EVE moving to the left at 7.321 m/s.