Respuesta :
Well, Break the problem up into parts.
For speeding up:
The car accelerates at 4 m/s^2 for 3 seconds. Multiplying these values tells you the car reaches a speed of 12 m/s.
Vf^2 = Vi^2 + 2a(Xf - Xi)
12^2 = 0 + 2(4)(Xf - 0)
144 = 8 Xf
Xf = 18 m
For coasting:
The car is at the 12 m/s and does this for 2 seconds.
x = vt = (12)(2) = 24 m
For slowing down:
The car decelerates at 3 m/s^2 to come to a stop at the next sign. From 12 m/s, this would take 12/3 seconds or 4 s.
Vf^2 = Vi^2 + 2a(Xf - Xi)
0 = 12^2 + (2)(-3)(Xf - 0)
-144 = -6 Xf
Xf = 24 m
Summing the distances: Xtotal = 18 + 24 + 24 = 66 m
