Two particles move along the x-axis, one given by a position function and one by a velocity function. Students find a velocity, determine the intervals where the particles move in opposite directions, analyze whether one particle's speed is increasing or decreasing, and recover a position by integrating from a known initial position.
Two particles move along the x-axis, one given by a position function and one by a velocity function. Students find a velocity, determine the intervals where the particles move in opposite directions, analyze whether one particle's speed is increasing or decreasing, and recover a position 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 velocity of particle $H$ is the derivative of its position. With $x_H(t)=e^{t^2-4t}$, the chain rule gives $$v_H(t)=x_H'(t)=(2t-4)\,e^{t^2-4t}.$$ At $t=1$: $$v_H(1)=(2(1)-4)\,e^{1-4}=(-2)\,e^{-3}=-2e^{-3}.$$ So $v_H(1)=-2e^{-3}\approx -0.100$ (moving in the negative direction).
2 points. P1: considers x_H'(t) — presenting x_H'(t), x_H'(1), the expression (2t-4)e^(t^2-4t), or (2*1-4)e^(1-4) all qualify. P2: the answer -2e^(-3). An unsupported -2e^(-3) earns P2 but not P1.
Particle $H$: from part (a), $v_H(t)=(2t-4)e^{t^2-4t}$. Since $e^{t^2-4t}>0$ always, the sign of $v_H$ is the sign of $2t-4$, which is $0$ at $t=2$. So $v_H<0$ on $0<t<2$ (moving left) and $v_H>0$ on $2<t<5$ (moving right). Particle $J$: $v_J(t)=2t(t^2-1)^3$. On $0<t<5$ the factor $2t>0$, so the sign is that of $(t^2-1)^3$, i.e. of $t^2-1$, which is $0$ at $t=1$. So $v_J<0$ on $0<t<1$ (moving left) and $v_J>0$ on $1<t<5$ (moving right). The particles move in **opposite** directions where their velocity signs differ. Comparing: - $0<t<1$: both negative (same direction). - $1<t<2$: $v_J>0$ (right) while $v_H<0$ (left) — **opposite**. - $2<t<5$: both positive (same direction). Therefore $H$ and $J$ move in opposite directions on $\boxed{1<t<2}$.
3 points. P3: considers the sign of v_H or v_J — e.g. sets x_H'(t)=0 (t=2), sets v_J(t)=0 (t=1), or identifies the interval 1<t<2. P4: a correct sign/direction analysis on (0,5) for one of the two particles. P5: the answer 1<t<2 with a reason; eligibility for P5 requires correct sign analyses for BOTH particles. Only behavior within 0<t<5 is scored.
Speed is increasing when velocity and acceleration have the **same sign**. At $t=2$, $v_J(2)=2(2)(2^2-1)^3=4\cdot 27=108>0$, and we are given $v_J'(2)=486>0$. Since $v_J(2)$ and $v_J'(2)$ are both positive (same sign), the speed of particle $J$ is **increasing** at $t=2$.
1 point (P6): conclude the speed is increasing because v_J(2) and v_J'(2) have the same sign. Evaluating the numbers is not required, but any stated values must be correct (v_J(2)=108, v_J'(2)=486). The sign of v_J(2) may be imported from part (b).
Position is the initial position plus the accumulated displacement. With $x_J(0)=7$, $$x_J(2)=x_J(0)+\int_0^2 v_J(t)\,dt = 7+\int_0^2 2t(t^2-1)^3\,dt.$$ Let $u=t^2-1$, $du=2t\,dt$ (so $u=-1$ at $t=0$ and $u=3$ at $t=2$): $$\int_0^2 2t(t^2-1)^3\,dt = \Big[\tfrac{1}{4}(t^2-1)^4\Big]_0^2 = \tfrac{1}{4}\big((3)^4-(-1)^4\big) = \tfrac{1}{4}(81-1)=\tfrac{1}{4}(80)=20.$$ Therefore $x_J(2)=7+20=27$.
3 points. P7: a definite or indefinite integral with integrand v_J(t)=2t(t^2-1)^3. P8: an antiderivative of the form k(t^2-1)^4 with k>0; if k != 1/4 the response cannot earn P9. P9: the answer 27. Writing 7 + (1/4)((3)^4-(-1)^4) banks P9 (later simplification errors not penalized), but an ambiguous form must resolve to 27.
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