Question

The position of a particle moving along the x axis is given by x= 6.0.-1.0p, where x is in meters and ria seconds. What is the position of the particle when it achieves its maximum speed in the positive x direction? a. 32m b. 12m c. 16 m d. 24 m c. 2.0m 12. 13. The position of a particle moving along the x axis is given by x (21 +221-6.0)m, where t is in s. What is the average velocity during the time interval t = 1.0 s to t = 3.0 s? a. -4.0 m/s b. -2.0 m/s c. -6.0 m's d. 8.0 m/'s e. -8.0 m/s A rock is projected from the edge of the top of a building with an initial velocity of 12.2 m/s at an angle of 53 above the horizontal. The rock strikes the ground a horizontal distance of 25 m from the base of the building Assume that the ground is level and that the side of the building is vertical. How tall is the building? a 29.6 m b. 18.9 m c. 27.4 m d. 23.6 m e. 25.3 m 14. 15. Two identical balls are at rest and side by side at the top of a hill. You let one ball, A, start rolling down the hill. A little later you start the second ball, B, down the hill by giving it a shove. The second ball rolls down the hill along a line parallel to the path of the first ball and passes it. At the instant ball B passes ball A: a. it has the same displacement and the same velocity as A. b. it has the same position and the same velocity as A. c. it has the same velocity and the same acceleration as A. d. it has the same position and the same acceleration as A. e. it has the same position, displacement and velocity as A.

Answer 1

x = 6t 2 - t^3 v = speed = dx/dt = d/dt (6t^2 - t^3) = d/dt (6t^2) - d/dt (t^3) v = speed = 6 times 2t - 3t^2 = 12t - 3t^2 for speed to be maximum dv/dt = 0 Rightarrow d/dt (12t - 3t^2) = 0 12 - 6t = 0 12 = 6t Rightarrow t = 2 So, position of particle at t = 2 sec x(t = 2) = 6 times 2^2 - (2)^3 (from (1)) = 6 times 4 - 8 = 24 - 8 = 16 m