20 m/s t = 0 x (m) -30-20-10 0 10 PHYS 120-Module 1 Name: 1-3) A car initially located at x = 10 m is traveling in the -x direction with a speed of 20 m/s. The car slows down and comes to a stop 5 seconds later after traveling a distance of 40 meters. (Note: The acceleration is not constant.) 1) What is the direction of the acceleration? (3 points) A) B) C) D) The acceleration is zero. E) 2) What is the displacement vector? (3 points) A) Ax = +40 m B) Ax = -40 m C) Ax-30 m D) Ax = +10 m E) Ax+30 m 3) What is the average velocity vector? (3 points) A) Vavx +4 m/s = B) Vavx +10 m/s C) Vavx=-10 m/s D) Vavx-8 m/s = E) Vavx-4 m/s = t=5s