A particle starts from rest at x_i = 0 and moves for 10 s with an acceleration of +2.0 cm/s². For the next 20 s, the acceleration of the particle is −1.0 cm/s². What is the position of the particle at the end of this motion? A) zero B) +3.0 m C) −1.0 m D) +2.0 m E) −3.0 m

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Quizplus · Accepted Answer

For the first 10 seconds, the particle accelerates at +2.0 cm/s^2. Using the equation $x = \frac{1}{2}at^2$, where $x$ is the displacement, $a$ is the acceleration, and $t$ is the time, the displacement is $x_1 = \frac{1}{2}(2)(10)^2 = 100$ cm or 1.0 m. For the next 20 seconds, the particle accelerates at -1.0 cm/s^2. The displacement in this period is $x_2 = \frac{1}{2}(-1)(20)^2 = -200$ cm or -2.0 m. The total displacement is $x_{total} = x_1 + x_2 = 1.0 m - 2.0 m = -1.0 m$. However, the correct calculation should consider the velocity at the end of the first phase as the initial velocity for the second phase. The velocity at the end of the first 10 seconds is $v = at = (2)(10) = 20$ cm/s. For the next 20 seconds, the displacement using $x = vt + \frac{1}{2}at^2$ is $x_2 = (20)(20) + \frac{1}{2}(-1)(20)^2 = 400 - 200 = 200$ cm or 2.0 m. Thus, the total displacement is $1.0 m + 2.0 m = 3.0 m$, indicating an error in the initial explanation.

A Particle Starts from Rest at Xi = 0 and Moves

A Particle Starts from Rest at X_i = 0 and Moves