a car on a straight track is at position (x_i) and has initial speed (v_i) when it begins to slow down with…

a car on a straight track is at position (x_i) and has initial speed (v_i) when it begins to slow down with constant acceleration of magnitude (a). when the car reaches position (x_f) it has speed (v_f). which of the following is a correct expression for (x_f)?\na (\frac{v_f^2 - v_i^2}{2a}+x_i)\nb (\frac{v_i^2 - v_f^2}{2a}+x_i)\nc (\frac{v_f^2 - v_i^2}{2a}-x_i)\nd (\frac{v_i^2 - v_f^2}{2a}-x_i)

a car on a straight track is at position (x_i) and has initial speed (v_i) when it begins to slow down with constant acceleration of magnitude (a). when the car reaches position (x_f) it has speed (v_f). which of the following is a correct expression for (x_f)?\na (\frac{v_f^2 - v_i^2}{2a}+x_i)\nb (\frac{v_i^2 - v_f^2}{2a}+x_i)\nc (\frac{v_f^2 - v_i^2}{2a}-x_i)\nd (\frac{v_i^2 - v_f^2}{2a}-x_i)

Answer

Explanation:

Step1: Recall the kinematic - equation

The kinematic equation $v_{f}^{2}=v_{i}^{2}+2a(x_{f}-x_{i})$ is used for uniformly - accelerated motion.

Step2: Rearrange the equation for $x_{f}$

Starting from $v_{f}^{2}=v_{i}^{2}+2a(x_{f}-x_{i})$, first expand the right - hand side: $v_{f}^{2}=v_{i}^{2}+2ax_{f}-2ax_{i}$. Then, isolate $x_{f}$ terms: $2ax_{f}=v_{f}^{2}-v_{i}^{2}+2ax_{i}$. Finally, solve for $x_{f}$: $x_{f}=\frac{v_{f}^{2}-v_{i}^{2}}{2a}+x_{i}$.

Answer:

A. $\frac{v_{f}^{2}-v_{i}^{2}}{2a}+x_{i}$