Thread: 1-D Kinematics Problem

I'm having trouble with this problem.

A car starts from rest at a stop sign. It accelerates at 4.4 for 6.2 , coasts for 2.5 , and then slows down at a rate of 3.4 for the next stop sign. How far apart are the stop signs?



Any help is appreciated.

A car starts from rest at a stop sign. It accelerates at 4.4 for 6.2 , coasts for 2.5 , and then slows down at a rate of 3.4 for the next stop sign. How far apart are the stop signs?


whole distance can be divided into 3 parts
dis. coverd during first 6.2 sec=S1 = 1/2 at^2 = 1/2 X 4.4 X 6.2 X 6.2 =84.56 m ......... second eq. of kinematics

velocityafter 6.2 sec.=V= 0+at =6.2X4.4=27.28m/sec. ............. first eq. of kinematics

now the car moves at this speed for next 2.5 sec.
dist coverd= S2=Vt=(6.2X4.4)X2.5 = 68.2m ........dist= speedXtime

now the car slows down fro V to zero at the rate 3.4m/sec sq

distance covered S3=√ [(V^2-U^2)/2a] ............ third eq. of kinematics
=[6.2X4.4]/√6.8 = 10.46m


S1+ S2+ S3

Thanks alot! I got an answer of approximately 2.6 x 10^2 meters . Appreciate it!