A particle is moving with a constant acceleration of 4.0 m/s². Its speed at t = 1.0 s is 4.0 m/s and at t = 3.0 s it is 12.0 m/s. What is the area under the position-time graph for the interval from t = 1.0 s to t = 3.0 s? A) 8.0 m/s B) 8.0 m C) 12 m D) 16 m E) 16 m/s²

Question

Quizplus · Accepted Answer

The area under the position-time graph represents the displacement of the particle. From the information given, we can use the following kinematic equation:

$v_f = v_i + a\Delta t$

where $v_i$ is the initial velocity, $v_f$ is the final velocity, $a$ is the acceleration, and $\Delta t$ is the time interval.

Substituting the given values, we have:

$12.0	ext{ m/s} = 4.0	ext{ m/s} + 4.0	ext{ m/s}^2	imes\Delta t$

Simplifying, we get:

$\Delta t = 2.0	ext{ s}$

Therefore, the displacement of the particle from $t=1.0	ext{ s}$ to $t=3.0	ext{ s}$ is:

$\Delta x = v_i\Delta t + \frac{1}{2}a(\Delta t)^2 = 4.0	ext{ m/s}	imes 2.0	ext{ s} + \frac{1}{2}(4.0	ext{ m/s}^2)	imes(2.0	ext{ s})^2 = 16.0	ext{ m}$

Hence, the area under the position-time graph for the interval from $t=1.0	ext{ s}$ to $t=3.0	ext{ s}$ is 16.0 m$^2$, which is answer choice D.

A Particle Is Moving with a Constant Acceleration of 4