AP Physics C: Mechanics FRQ Practice

⚙️AP Physics C: Mechanics

Practice FRQ 1 of 141/14

4. A particle moves along a straight horizontal track. The position of the particle as a function of time is given by $x(t) = 4t^2 - 2t^3$ , where $x$ is in meters and $t$ is in seconds, as shown in Figure 1.

Figure 1: Position x versus time t for a particle moving on a straight track, for 0 ≤ t ≤ 2 s, where x(t) = 4t^2 − 2t^3.

The particle's velocity is zero at two different times during the interval from $t = 0$ s to $t = 2$ s. One of these times is $t_1 = 0$ s and the other is $t_2$ .

The magnitude of the acceleration of the particle at $t_1$ is $a_1$ and the magnitude of the acceleration at $t_2$ is $a_2$ .

Indicate whether $a_1$ is greater than, less than, or equal to $a_2$ by writing one of the following.

$a_1 > a_2$
$a_1 < a_2$
$a_1 = a_2$

Justify your answer using qualitative reasoning beyond referencing equations.

Derive an expression for the time $t_2$ when the particle's velocity is zero (other than at $t = 0$ s). Express your answer in terms of physical constants, as appropriate. Begin your derivation by writing a fundamental physics principle or an equation from the reference information.

Indicate whether the speed of the second particle at $t = 2.0$ s is greater than, less than, or equal to the speed of the second particle at $t = 0$ s. In a different scenario, a second particle moves in two dimensions. At time $t = 0$ s, the particle is at the origin with velocity components $v_{0x} = 3.0$ m/s and $v_{0y} = 4.0$ m/s. The particle experiences a constant acceleration with components $a_x = -1.5$ m/s $^2$ and $a_y = 0$ m/s $^2$ .

Briefly justify your answer.

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