question 2 (1 point)\n(02 02 mc)\na car crosses the start line with an initial velocity of 7 m/s. after 12…

Explanation:

Step1: Recall the kinematic - equation

The kinematic equation for position is $x = x_0+v_0t+\frac{1}{2}at^{2}$. Assuming $x_0 = 0$ (starting - point as origin), we have $x=v_0t+\frac{1}{2}at^{2}$. Given $v_0 = 7$ m/s, $t = 12$ s, and $x = 105$ m.

Step2: Substitute values into the equation

Substitute the values into $x=v_0t+\frac{1}{2}at^{2}$: $105=7\times12+\frac{1}{2}a\times(12)^{2}$.

Step3: Simplify the right - hand side

First, calculate $7\times12 = 84$. Then the equation becomes $105=84 + 72a$.

Step4: Solve for acceleration $a$

Subtract 84 from both sides: $105 - 84=72a$, so $21 = 72a$. Then $a=\frac{21}{72}=\frac{7}{24}\approx0.29$ m/s². Since $a>0$, the acceleration is positive.

Answer:

A. It is moving with a positive acceleration.