A car starts from rest at a stop sign. It accelerates at 2.0 m/s^2 for 6.0 seconds, coasts for 2.0s, and then slows down at a rate of 1.5 m/s^2 for the next stop sign.How far apart are the stop signs?

### 2 Answers

breakdown of trip between signs:

pt 1:

a = 2.0 m/s²

t = 6.0 s

d1 = 1/2at² = (0.5)(2)(6)² = 36 m

pt 2:

V = at = (2)(6) = 12 m/s {from data of pt 1}

d2 = Vt = (12)(2) = 24 m

pt 3:

a = – 1.5 m/s²

(Vf)² = (Vi)² + 2a(d3) {where Vf = 0, Vi = 12, a = -1.5}

0 = (12)² – (2)(1.5)d3

144 = 3d3

d3 = 144/3 = 48 m

Total distance between signs = d1+d2+d3 = 36+24+48 = 108 m ANS

“starts off from relax and speeds up at a velocity of three m/s^2 for 20 seconds” exchange in v = at. Use the given values of a and t to ascertain the excellent velocity. “decelerates on the value of 5 m/s^2” exchange in v = at. Use the given values of (exchange in v) and a to ascertain t.