Answer
Distance between them when t=4 s:
152ft
The total distance each has traveled in t = 4 s:
Xa= 40ft
Xb=192ft
Work Step by Step
\[
aA = 6t - 3 \quad aB = 12t^2 - 8
\]
\[
\frac{dv}{dt} = 6t - 3 \quad \frac{dv}{dt} = 12t^2 - 8
\]
\[
\int dv = \int 6t - 3 \, dt \quad \int dv = \int 12t^2 - 8 \, dt
\]
\[
VA = 3t^2 - 3t + 4 \quad VB = 4t^3 - 8t
\]
\[
\int dx = \int 3t^2 - 3t \, dt \quad X_B = t^4 - 4t^2
\]
\[
X_A = \frac{3}{p}t^3 - \frac{3}{2}t^2 + 4
\]
\[
X_{A|t=4} = 40 \, ft \quad X_{B|t=4} = 192 \, ft
\]
Distance between them =
\[
192 - 40 = 152 \, dt
\]
