To solve the problem, we need to work with the velocity function \( v(t) = t^2 - 2t - 15 \). We’ll perform two tasks: find the displacement and then find the distance traveled by the particle from \( t = 2 \) to \( t = 8 \).
(a) Finding the Displacement
The displacement over the time interval can be found by integrating the velocity function:
\[ \text{Displacement} = \int_{2}^{8} v(t) , dt = \int_{2}^{8} (t^2 - 2t - 15) , dt \]
First, we compute the integral:
\[ \int (t^2 - 2t - 15) , dt = \frac{t^3}{3} - t^2 - 15t + C \]
Now, we evaluate this from \( t = 2 \) to \( t = 8 \):
\[ \left[ \frac{t^3}{3} - t^2 - 15t \right]_{2}^{8} \]
Calculating at \( t = 8 \):
\[ \frac{8^3}{3} - 8^2 - 15(8) = \frac{512}{3} - 64 - 120 \] \[ = \frac{512}{3} - \frac{192}{3} - \frac{360}{3} = \frac{512 - 192 - 360}{3} = \frac{-40}{3} \]
Now calculating at \( t = 2 \):
\[ \frac{2^3}{3} - 2^2 - 15(2) = \frac{8}{3} - 4 - 30 = \frac{8}{3} - \frac{12}{3} - \frac{90}{3} = \frac{8 - 12 - 90}{3} = \frac{-94}{3} \]
Now we find the displacement:
\[ \text{Displacement} = \left( \frac{-40}{3} \right) - \left( \frac{-94}{3} \right) = \frac{-40 + 94}{3} = \frac{54}{3} = 18 \text{ meters} \]
Thus, the displacement is:
\[ \boxed{18} \text{ meters} \]
(b) Finding the Distance Traveled
To find the distance traveled, we need to take into account when the velocity changes sign within the interval \( [2, 8] \). We first find the roots of the velocity function \( v(t) = 0 \):
\[ t^2 - 2t - 15 = 0 \]
Using the quadratic formula:
\[ t = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(1)(-15)}}{2(1)} = \frac{2 \pm \sqrt{4 + 60}}{2} = \frac{2 \pm \sqrt{64}}{2} = \frac{2 \pm 8}{2} \]
Calculating the roots:
\[ t = \frac{10}{2} = 5 \quad \text{and} \quad t = \frac{-6}{2} = -3 \text{ (not in the interval)} \]
The velocity changes sign at \( t = 5 \). We evaluate the distance by integrating the absolute value of the velocity in the intervals \( [2, 5] \) and \( [5, 8] \):
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For \( t \in [2, 5] \): \( v(t) \) is negative since \( v(4) = 4 - 8 - 15 < 0 \).
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For \( t \in [5, 8] \): \( v(t) \) is positive since \( v(6) = 36 - 12 - 15 > 0 \).
Now we compute:
\[ \text{Distance} = \int_{2}^{5} -v(t) , dt + \int_{5}^{8} v(t) , dt \]
Calculating the first integral:
\[ \int_{2}^{5} -v(t) , dt = \int_{2}^{5} -(t^2 - 2t - 15) , dt = \int_{2}^{5} (-t^2 + 2t + 15) , dt \]
Calculating the integral:
\[ \int (-t^2 + 2t + 15) , dt = -\frac{t^3}{3} + t^2 + 15t + C \]
Evaluate from 2 to 5:
\[ \left[-\frac{t^3}{3} + t^2 + 15t \right]_{2}^{5} \]
Calculating at \( t = 5 \):
\[ -\frac{5^3}{3} + 5^2 + 15(5) = -\frac{125}{3} + 25 + 75 = -\frac{125}{3} + 100 = \frac{300 - 125}{3} = \frac{175}{3} \]
Calculating at \( t = 2 \):
\[ -\frac{2^3}{3} + 2^2 + 15(2) = -\frac{8}{3} + 4 + 30 = -\frac{8}{3} + 34 = \frac{102 - 8}{3} = \frac{94}{3} \]
Now evaluate the first integral:
\[ \int_{2}^{5} -v(t) , dt = \left( \frac{175}{3} - \frac{94}{3} \right) = \frac{81}{3} = 27 \]
Now for the second integral from \( t = 5 \) to \( t = 8 \):
\[ \int_{5}^{8} v(t) , dt = \int_{5}^{8} (t^2 - 2t - 15) , dt \]
We already computed the first part up to \( t = 8 \) earlier. So,
\[ \text{Distance} = \int_{2}^{5} -v(t) , dt + \int_{5}^{8} v(t) , dt = 27 + \left( \frac{-40}{3} + \frac{94}{3} \right) = 27 + \frac{54}{3} = 27 + 18 = 45 \]
Thus, the total distance traveled is:
\[ \boxed{45} \text{ meters} \]
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