Question

problem displacement an automobile starts at rest and accelerates at 5.7 m/s² after a traffic light turns green. how far will it have gone when it is traveling at 15 m/s? 1. analyze and sketch the problem knowns unknown x_i = =? v_i = v_f = overline{a}=a = 2. solve for the unknown use the relationship among velocity, acceleration, and displacement to find x_f.

Explanation:

Step1: Identify known values

The automobile starts at rest, so the initial position $x_i = 0$ m, the initial velocity $v_i=0$ m/s, the final velocity $v_f = 15$ m/s, and the acceleration $a = 5.7$ m/s². The unknown is the final - position $x_f$.

Step2: Select the appropriate kinematic equation

The kinematic equation that relates initial velocity ($v_i$), final velocity ($v_f$), acceleration ($a$), and displacement ($\Delta x=x_f - x_i$) is $v_f^{2}=v_i^{2}+2a(x_f - x_i)$. Since $x_i = 0$ and $v_i = 0$, the equation simplifies to $v_f^{2}=2ax_f$.

Step3: Solve for $x_f$

We can re - arrange the equation $v_f^{2}=2ax_f$ to solve for $x_f$. So, $x_f=\frac{v_f^{2}}{2a}$.
Substitute $v_f = 15$ m/s and $a = 5.7$ m/s² into the formula:
$x_f=\frac{15^{2}}{2\times5.7}=\frac{225}{11.4}\approx19.7$ m

Answer:

$x_f\approx19.7$ m