A particle confined to motion along the x axis moves with constant acceleration from x = 2.0 m to x = 8.0 m during a 2.5-s time interval. The velocity of the particle at x = 8.0 m is 2.8 m/s. What is the acceleration during this time interval? A) 0.48 m/s² B) 0.32 m/s² C) 0.64 m/s² D) 0.80 m/s² E) 0.57 m/s²

Question

Quizplus · Accepted Answer

To find the acceleration, we can use the formula $v_f = v_i + at$, where $v_f$ is the final velocity, $v_i$ is the initial velocity, $a$ is the acceleration, and $t$ is the time. We are not directly given $v_i$, but we can find it using the displacement formula $x = x_0 + v_i t + \frac{1}{2}at^2$, where $x$ is the final position, $x_0$ is the initial position, and the other symbols have their usual meanings. Substituting the given values into the displacement formula, we can solve for $v_i$, and then use the first formula to solve for $a$. However, without the initial velocity, we cannot directly solve for acceleration using the given information alone. The correct approach involves using both the given displacement and the final velocity to find acceleration, but the provided information and method in the explanation are insufficient to directly arrive at the correct answer without the initial velocity or a more detailed calculation. Therefore, a mistake was made in the explanation process. To correctly solve for acceleration, one would typically use the kinematic equations, taking into account the total displacement and the time taken, but the direct calculation from the initial setup was not provided.

A Particle Confined to Motion Along the X Axis Moves