Question

a car traveling along a road begins accelerating with a constant acceleration of 1.5 m/s² in the direction of motion. after traveling 392 m at this acceleration, its speed is 35 m/s. determine the speed of the car when it began accelerating. 1.5 m/s 49 m/s 34 m/s 2.3 m/s 7.0 m/s

Explanation:

Step1: Identify the kinematic - equation

We use the equation $v^{2}=v_{0}^{2}+2ax$, where $v$ is the final velocity, $v_{0}$ is the initial velocity, $a$ is the acceleration, and $x$ is the displacement.

Step2: Rearrange the equation for $v_{0}$

$v_{0}^{2}=v^{2}-2ax$. Then $v_{0}=\sqrt{v^{2}-2ax}$.

Step3: Substitute the given values

Given $v = 35\ m/s$, $a=1.5\ m/s^{2}$, and $x = 392\ m$.
$v_{0}=\sqrt{35^{2}-2\times1.5\times392}$
$=\sqrt{1225 - 1176}$
$=\sqrt{49}$
$ = 7\ m/s$

Answer:

E. 7.0 m/s