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 $x$ of the particle as a function of time $t$ is given by $x(t) = 2t^3 - 15t^2 + 24t$ , where $x$ is in meters and $t$ is in seconds. At time $t = 0$ , the particle is at position $x = 0$ .

Figure 1: Position x versus time t for the particle, shown from t = 0 s to t = 5 s, for x(t) = 2t^3 − 15t^2 + 24t.

The magnitude of the particle's average velocity during the time interval from $t = 0$ s to $t = 1$ s is $v_{avg,1}$ . The magnitude of the particle's average velocity during the time interval from $t = 1$ s to $t = 4$ s is $v_{avg,2}$ .

Indicate whether $v_{avg,1}$ is greater than, less than, or equal to $v_{avg,2}$ by writing one of the following.

$v_{avg,1} > v_{avg,2}$
$v_{avg,1} < v_{avg,2}$
$v_{avg,1} = v_{avg,2}$

Justify your answer using qualitative reasoning based on the graph in Figure 1, beyond referencing equations.

Derive an expression for the time $t$ at which the particle first changes direction after $t = 0$ . Express your answer in terms of the coefficients in the position function and physical constants, as appropriate. Begin your derivation by writing a fundamental physics principle or an equation from the reference information.

Indicate whether, as measured by the moving observer, the particle first changes direction before, at the same time as, or after $t = 1$ s. An observer moves along the same track with a constant velocity of $+3$ m/s in the positive $x$ -direction, starting from position $x = 0$ at time $t = 0$ .

Briefly justify your answer by describing how the particle's motion appears in the moving observer's reference frame.

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