QuestionJuly 17, 2025

2. A particle is released as part of an experiment. Its speed t seconds after release is given by v(t)=-0.5t^2+10t where v(t) is in meters per second. a) How far does the particle travel during the first five seconds? b) How far does it travel during the second five seconds?

2. A particle is released as part of an experiment. Its speed t seconds after release is given by v(t)=-0.5t^2+10t where v(t) is in meters per second. a) How far does the particle travel during the first five seconds? b) How far does it travel during the second five seconds?
2. A particle is released as part of an experiment. Its speed t seconds after release is given by v(t)=-0.5t^2+10t where v(t) is in meters per second. a) How far does the particle travel during the first five seconds? b) How far does it travel during the second five seconds?

Solution
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Answer

a) \frac{625}{6} meters ### b) \frac{375}{6} meters Explanation 1. Find the position function Integrate v(t) = -0.5t^2 + 10t to find the position function s(t). The integral of velocity gives position: s(t) = \int v(t) \, dt = \int (-0.5t^2 + 10t) \, dt = -\frac{0.5}{3}t^3 + 5t^2 + C. 2. Determine constant of integration Assume initial position s(0) = 0, so C = 0. Thus, s(t) = -\frac{1}{6}t^3 + 5t^2. 3. Calculate distance for first five seconds Evaluate s(t) from t=0 to t=5: s(5) - s(0) = \left(-\frac{1}{6}(5)^3 + 5(5)^2\right) - 0 = -\frac{125}{6} + 125 = \frac{625}{6} meters. 4. Calculate distance for second five seconds Evaluate s(t) from t=5 to t=10: s(10) - s(5) = \left(-\frac{1}{6}(10)^3 + 5(10)^2\right) - \left(-\frac{1}{6}(5)^3 + 5(5)^2\right) = \left(-\frac{1000}{6} + 500\right) - \frac{625}{6} = \frac{375}{6} meters.

Explanation

1. Find the position function<br /> Integrate $v(t) = -0.5t^2 + 10t$ to find the position function $s(t)$. The integral of velocity gives position: $s(t) = \int v(t) \, dt = \int (-0.5t^2 + 10t) \, dt = -\frac{0.5}{3}t^3 + 5t^2 + C$.<br /><br />2. Determine constant of integration<br /> Assume initial position $s(0) = 0$, so $C = 0$. Thus, $s(t) = -\frac{1}{6}t^3 + 5t^2$.<br /><br />3. Calculate distance for first five seconds<br /> Evaluate $s(t)$ from $t=0$ to $t=5$: $s(5) - s(0) = \left(-\frac{1}{6}(5)^3 + 5(5)^2\right) - 0 = -\frac{125}{6} + 125 = \frac{625}{6}$ meters.<br /><br />4. Calculate distance for second five seconds<br /> Evaluate $s(t)$ from $t=5$ to $t=10$: $s(10) - s(5) = \left(-\frac{1}{6}(10)^3 + 5(10)^2\right) - \left(-\frac{1}{6}(5)^3 + 5(5)^2\right) = \left(-\frac{1000}{6} + 500\right) - \frac{625}{6} = \frac{375}{6}$ meters.
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