Question

A particle is moving along a straight line with an initial velocity of 6 m/s when it is subjected to a deceleration of a = (-1.5012) m/s², where vis in m/s. Determine how far it travels before it stops. How much time does this take?

Answer 1

Sol: 1.5(0)2 acceleration as we have time, rate of of du dt du 0 at du -1.5 dt V? ½ -I/2av v -1.5 dt 1/2 av -1.5 t + C at t=0, u= 6 C Velocity is the rate of change of displacement Given : Initial velocity (W) 6 m/s acceleration (a) Change -1.5(0)², du Zxf6 2 su - 4t +256 Lir 56 (ro -0.957) particle strops when (vo. t = 3·26 sec 0.75+)? (56 0.757) ds 6 +0.5625 +² - 356 t It . 1. tt 2 O 0.75+)' => va (ro 2 ds dt ll 6t +0.5625 +² 2 356 t + c 2 at t = 0, $ . C o 8 3 0.1675t # 1.837t²tot → The particle stops cat t 3.26 sec .: S = 0.1875-3.263 -1.837.3.28? 3.26² +6+3.26 6-53 m Time taken to stop 3.26 seconds. Distance travelled before stopping- 653 m