A swimmer's velocity along a straight path is given by a closed-form function. Students find the times the direction of motion changes, compute acceleration at an instant and decide whether speed is increasing or decreasing, and find displacement between two times and total distance traveled over the full interval.
A swimmer's velocity along a straight path is given by a closed-form function. Students find the times the direction of motion changes, compute acceleration at an instant and decide whether speed is increasing or decreasing, and find displacement between two times and total distance traveled over the full interval.
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Original Tian2 solution. Each part identifies which rubric points are earned and why.
Stephen changes direction exactly when his velocity $v(t)$ **changes sign**, which can only occur where $v(t)=0$. Since $v(t)=2.38\,e^{-0.02t}\sin\!\left(\frac{\pi t}{56}\right)$ and the exponential factor is always positive, set the sine factor to zero: $$\sin\!\left(\frac{\pi t}{56}\right)=0 \;\Rightarrow\; \frac{\pi t}{56}=k\pi \;\Rightarrow\; t=56k.$$ The only such time in $0<t<90$ is $t=56$. Checking signs, $v>0$ just before $t=56$ and $v<0$ just after, so the velocity changes sign there. $$\boxed{t=56 \text{ seconds}}$$
2 points. P1 (considers sign of v): present v(t)=0, or state Stephen changes direction when velocity changes sign. P2 (answer with reason): requires the answer t=56 together with the sign-change reasoning. A bare t=56 with no supporting work earns neither point. Values of t outside (0,90) do not affect scoring.
**Acceleration** is the derivative of velocity, $a(t)=v'(t)$. Using a calculator, $$a(60)=v'(60)\approx -0.0360162.$$ So Stephen's acceleration at $t=60$ is approximately $\boxed{-0.036\ \text{meter per second per second}}$. **Speeding up or slowing down?** Compute the velocity at $t=60$: $$v(60)\approx -0.1595124 < 0.$$ Since $v(60)<0$ and $a(60)<0$, velocity and acceleration have the **same sign**, so the speed $|v|$ is increasing. **Stephen is speeding up at $t=60$ seconds.**
3 points. P1 (a(60) with setup): minimum is v'(60)=-0.036; the connection v'(t)=a(t) (or v'(60)=a(60)) must be shown — a bare a(60)=-0.0360162 is not sufficient. P2 (units): meters per second per second; a value for a(60) must be declared. P3 (speeding up with reason): conclusion must be consistent with v(60)<0 and the declared a(60); stating velocity and acceleration have the same sign (both negative) earns it. Any incorrect sign/value of v(60) forfeits P3. Degree mode does not earn P1.
The (signed) change in position from $t=20$ to $t=80$ is the integral of velocity. Because $v$ does not change sign on this whole interval in a way that matters here — the question asks for the distance between the two positions, i.e. the magnitude of the displacement — we compute $$\int_{20}^{80} v(t)\,dt = 23.383997.$$ Since this value is positive, the distance between Stephen's positions at $t=20$ and $t=80$ is $$\boxed{23.384 \text{ meters}}$$ (a value of $23.383$ is also accepted).
2 points. P1 (integral): present ∫_20^80 v(t)dt (with or without dt). P2 (answer): 23.384 or 23.383, earned regardless of whether P1 was earned. Degree mode gives 2.407982 and does not earn the answer point.
**Total distance** swum is the integral of *speed* $|v(t)|$ over $0\le t\le 90$: $$\int_{0}^{90} |v(t)|\,dt.$$ From part (a), $v>0$ on $(0,56)$ and $v<0$ on $(56,90)$, so $$\int_{0}^{90} |v(t)|\,dt = \int_{0}^{56} v(t)\,dt - \int_{56}^{90} v(t)\,dt = 62.164216.$$ The total distance Stephen swims over $0\le t\le 90$ seconds is $$\boxed{62.164 \text{ meters}}.$$
2 points. P1 (integral): present ∫_0^90 |v(t)| dt, or the equivalent split ∫_0^56 v dt − ∫_56^90 v dt (with or without differentials). P2 (answer): 62.164, earned regardless of whether P1 was earned. Degree mode gives 3.127892 and does not earn the answer point.
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