Question

A car starts from rest at a stop sign. It accelerates at 4.0 m/s2 for 9 seconds, coasts for 2 s, and then slows down at a rate of 3.0 m/s2 for the next stop sign. How far apart are the stop signs?

Answer 1

Acceleration phase:
Vf = Vi + a*t
Vf = 0 + (4.0)(9) = 36 m/s

Vf^2 = Vi^2 + 2a(Xf - Xi)
36^2 = 0 + 2(4.0)(Xf - 0)
1296 = 8 Xf
Xf = 162 m

Coasting phase:
v = 36 m/s
t = 2 s
Xf = v * t = (36)(2) = 72 m

Deceleration phase:
Vf^2 = Vi^2 + 2a(Xf - Xi)
0 = 36^2 + 2(-3.0)(Xf - 0)
-1296 = -6 Xf
Xf = 216 m

Find the total distance:
Xf acceleration = 162 m
Xf coasting = 72 m
Xf deceleration = 216 m

Sum of these: Xtotal = 162 + 72 + 216 = 450 m

Answer 2

thanks so much!