A particle's velocity along the x-axis is given by a closed-form function. Students find a time the particle is at rest and its direction of motion, compute acceleration and decide whether speed is increasing or decreasing, and recover the particle's position and total distance traveled by integrating from a known initial position.
A particle's velocity along the x-axis is given by a closed-form function. Students find a time the particle is at rest and its direction of motion, compute acceleration and decide whether speed is increasing or decreasing, and recover the particle's position and total distance traveled by integrating from a known initial position.
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Original Tian2 solution. Each part identifies which rubric points are earned and why.
The particle is at rest when $v(t)=0$. On $0<t<2$ solve $$\ln\!\left(t^2-4t+5\right)-0.2t=0.$$ Using a graphing calculator, the only solution in $0<t<2$ is $$t_R\approx 1.426\ \ (\text{more precisely } 1.425610).$$ To determine direction for $0<t<t_R$, check the sign of $v$ there. For example $v(1)=\ln(1-4+5)-0.2=\ln 2-0.2\approx 0.493>0$. Since $v(t)>0$ on the whole interval $0<t<t_R$, the **particle is moving to the right** on $0<t<t_R$ (position is increasing because velocity is positive).
1 point for solving v(t_R)=0 to get t_R approx 1.426 (1.425 also accepted) on 0<t<2; 1 point for the direction 'to the right' with the reason that v(t)>0 on 0<t<t_R. A bare value of t_R with no equation, or a direction stated without referencing the sign of v, does not earn the corresponding point.
Acceleration is the derivative of velocity, $a(t)=v'(t)$. By calculator, $$a(1.5)=v'(1.5)=-1\ \ (\text{or }-0.999).$$ To decide whether the **speed** is increasing or decreasing, compare the signs of velocity and acceleration. At $t=1.5$, $$v(1.5)=\ln\!\left(1.5^2-4(1.5)+5\right)-0.2(1.5)=\ln(2.25)-0.3\approx -0.077<0,$$ and $a(1.5)=-1<0$. Since $v(1.5)$ and $a(1.5)$ have the **same sign** (both negative), the **speed of the particle is increasing** at $t=1.5$.
1 point for a(1.5)=v'(1.5)=-1 (or -0.999) with a derivative setup shown; 1 point for the conclusion that speed is increasing, justified by v(1.5) and a(1.5) having the same sign. Concluding 'increasing' without comparing the signs of v and a, or comparing the wrong quantities, does not earn the second point.
Position is recovered from velocity by the Fundamental Theorem of Calculus. With $x(1)=-3$, $$x(4)=x(1)+\int_1^4 v(t)\,dt.$$ Evaluating the integral by calculator, $$\int_1^4 v(t)\,dt\approx 0.197117,$$ so $$x(4)=-3+0.197117=-2.802883\approx -2.803.$$ The position of the particle at $t=4$ is approximately $-2.803$.
1 point for the FTC integral setup x(4)=x(1)+int_1^4 v(t)dt; 1 point for using the initial condition x(1)=-3; 1 point for the answer -2.803 (or -2.802). Omitting the initial condition (e.g. reporting just 0.197) forfeits both the initial-condition and answer points.
Total distance traveled is the integral of speed, $$\int_1^4 |v(t)|\,dt.$$ The velocity changes sign inside $[1,4]$ at $t=t_R\approx 1.4256$ (computed in part (a)) and again at $t\approx 2.8832$, so one may split the integral at those points; either way, evaluating the integral of $|v|$ by calculator gives $$\int_1^4 |v(t)|\,dt\approx 0.958.$$ The total distance traveled by the particle over $1\le t\le 4$ is approximately $0.958$.
1 point for the total-distance setup int_1^4 |v(t)| dt (or the equivalent sum of signed integrals split at the sign-change times approx 1.426 and 2.883); 1 point for the answer 0.958 (0.959 and 0.96 also accepted). Integrating v rather than |v| (which gives 0.197) does not earn the answer point.
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