Question

Problem 2.19 Item 2 > Review A car starts from rest at a stop sign. It accelerates at 4.6 m/s for 6.8 s, coasts for 2.5 s, and then slows down at a rate of 3.5 m/s for the next stop sign. How far apart are the stop signs? Express your answer to two significant figures and include the appropriate units. Value Units Submit Request Answer

Answer 1

sol" initial velocity of car=om18 its acceleration = 4.6m18² (for t=6688) displacement (8) = ut t I a, ut? &= ox(6-8) + ļ (4.6) (6.872 8 = 106.4 m Now at t = 6.88 its velocity, N= ut aut V=0+ (6-8) (4.6) = 31.28 mis car further moves and decelerate at gate of 3.5m/8ZE for .2.58 then &= ut - I azt² &= (31.28) (2.5)- Į (3.5) (2.572 = 67.26 m W979/5 T RITT STATY

car far away from stop = 8,+82 = 106.4 +67.3 = 173.7 m If we want to write it as two significant sigure then we have to a are then we have to round off » 8a 170m= 17X107m [&= 17810*m/ two significant figure