Chapter 2 - One-Dimensional Kinematics - Problems and Conceptual Exercises - Page 48: 11

$Distance=3.5km$ $Elapsed$ $time=0.0038h=14s$

(a) We know that $s=vt$ We plug in the known values to obtain: $s=(18.056m/s)(192s)=3466.752m=3.5Km$ (b) We know that $t=\frac{s}{v}$ We plug in the known values to obtain: $t=\frac{0.25\times 10^3m}{18.056m/s}=14s=0.0038h$