Saved Bookmarks
| 1. |
The velocity vecv of a particle moving in the xy plane is given by vecv=(6.0t-4.0t^(2))hati+8.0hatj, with vecv in meters per second and t(gt0) in seconds. (a) What is the acceleration when t = 2.5 s? (b) When (if ever) is the acceleration zero ? (c) When (if ever) is the velocity zero? (d) When (if ever) does the speed equal 10 m/s? |
| Answer» Solution :(a) `(-14m//s^(2))hati,` (B) 0.75 s, ( C) it is never ZERO, (d) 2.2 s | |