multiple - concept example 8 illustrates the concepts that are pertinent to this problem. a cab driver picks…

Explanation:

Step1: Let acceleration be $a$ and deceleration be $a_d$, distance of acceleration phase be $x_1$ and deceleration phase be $x_2$. Given $a_d = 3.91a$. Also, let the maximum - speed be $v$. Using the kinematic equation $v^{2}=v_0^{2}+2ax$ (where $v_0 = 0$ for the acceleration phase and $v_f = 0$ for the deceleration phase). For the acceleration phase: $v^{2}=2ax_1$. For the deceleration phase: $0 = v^{2}-2a_dx_2$, so $v^{2}=2a_dx_2$. Then $2ax_1=2a_dx_2$.

Since $a_d = 3.91a$, we have $x_1 = 3.91x_2$.

Step2: The total distance $x=x_1 + x_2$. Given $x=1.66\ km = 1660\ m$. Substitute $x_1 = 3.91x_2$ into $x=x_1 + x_2$. We get $1660=3.91x_2+x_2=(3.91 + 1)x_2=4.91x_2$.

Solve for $x_2$: $x_2=\frac{1660}{4.91}\ m\approx338.1\ m$.

Step3: Solve for $x_1$. Since $x_1 = 3.91x_2$, then $x_1=3.91\times338.1\ m\approx1321.9\ m$.

Answer:

(a) Number: $1321.9$; Units: $m$ (b) Number: $338.1$; Units: $m$