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A particle travels along the x-axis such that its acceleration is given by a(t)=t^(0.3)+3cos(t-5). If the velocity of the particle is v=-5 when t=0, what is the velocity of the particle when t=5 ? You may use a calculator and round your answer to the nearest thousandth. Answer:

A particle travels along the x x -axis such that its acceleration is given by a(t)=t0.3+3cos⁡(t−5) a(t)=t^{0.3}+3 \cos (t-5) . If the velocity of the particle is v=−5 v=-5 when t=0 t=0 , what is the velocity of the particle when t=5 t=5 ? You may use a calculator and round your answer to the nearest thousandth.\newlineAnswer:

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Q. A particle travels along the x x -axis such that its acceleration is given by a(t)=t0.3+3cos⁡(t−5) a(t)=t^{0.3}+3 \cos (t-5) . If the velocity of the particle is v=−5 v=-5 when t=0 t=0 , what is the velocity of the particle when t=5 t=5 ? You may use a calculator and round your answer to the nearest thousandth.\newlineAnswer:
  1. Integrate acceleration function: To find the velocity of the particle at t=5t=5, we need to integrate the acceleration function a(t)a(t) from t=0t=0 to t=5t=5. The velocity function v(t)v(t) is the integral of the acceleration function a(t)a(t) plus the constant of integration, which is the initial velocity v(0)=−5v(0) = -5.
  2. Write integral of acceleration function: First, we write down the integral of the acceleration function from t=0t=0 to t=5t=5: v(t)=∫05(t0.3+3cos⁡(t−5)) dt+v(0)v(t) = \int_{0}^{5}(t^{0.3} + 3\cos(t-5)) \, dt + v(0)
  3. Evaluate integrals and initial velocity: We integrate the function term by term. The integral of t0.3t^{0.3} with respect to tt is (t1.3)/(1.3)(t^{1.3})/(1.3) and the integral of 3cos⁡(t−5)3\cos(t-5) with respect to tt is 3sin⁡(t−5)3\sin(t-5). We will evaluate these integrals from 00 to 55 and add the initial velocity −5-5.
  4. Calculate velocity function: The velocity function v(t)v(t) from t=0t=0 to t=5t=5 is:\newlinev(t) = \left[\frac{t^{\(1\).\(3\)}}{\(1\).\(3\)}\right]_{\(0\)}^{\(5\)} + \left[\(3\sin(t−5-5)\right]_{00}^{55} - 55
  5. Evaluate velocity at t=5t=5: We evaluate the integrals at the upper and lower limits and subtract the lower limit result from the upper limit result for each term: v(\(5) = \left[\frac{(55^{11.33})}{11.33} - \frac{(00^{11.33})}{11.33}\right] + \left[33\sin(55−5-5) - 33\sin(00−5-5)\right] - 55
  6. Perform final calculations: Now we calculate each term using a calculator:\newlinev(5)=[51.31.3−0]+[3sin⁡(0)−3sin⁡(−5)]−5v(5) = \left[\frac{5^{1.3}}{1.3} - 0\right] + \left[3\sin(0) - 3\sin(-5)\right] - 5\newlinev(5)=[51.31.3]+[0−3sin⁡(−5)]−5v(5) = \left[\frac{5^{1.3}}{1.3}\right] + \left[0 - 3\sin(-5)\right] - 5
  7. Round the answer: We know that sin⁡(0)=0\sin(0) = 0 and we use a calculator to find sin⁡(−5)\sin(-5) and 51.35^{1.3}. We then perform the calculations:\newlinev(5)=51.31.3−3sin⁡(−5)−5v(5) = \frac{5^{1.3}}{1.3} - 3\sin(-5) - 5\newlinev(5)≈51.31.3−3×(−0.958924)−5v(5) \approx \frac{5^{1.3}}{1.3} - 3\times(-0.958924) - 5\newlinev(5)≈9.848857−2.876772−5v(5) \approx 9.848857 - 2.876772 - 5\newlinev(5)≈9.848857−2.876772−5v(5) \approx 9.848857 - 2.876772 - 5\newlinev(5)≈6.972085v(5) \approx 6.972085
  8. Round the answer: We know that sin⁡(0)=0\sin(0) = 0 and we use a calculator to find sin⁡(−5)\sin(-5) and 51.35^{1.3}. We then perform the calculations:\newlinev(5)=51.31.3−3sin⁡(−5)−5v(5) = \frac{5^{1.3}}{1.3} - 3\sin(-5) - 5\newlinev(5)≈51.31.3−3∗(−0.958924)−5v(5) \approx \frac{5^{1.3}}{1.3} - 3*(-0.958924) - 5\newlinev(5)≈9.848857−2.876772−5v(5) \approx 9.848857 - 2.876772 - 5\newlinev(5)≈9.848857−2.876772−5v(5) \approx 9.848857 - 2.876772 - 5\newlinev(5)≈6.972085v(5) \approx 6.972085Rounding the answer to the nearest thousandth, we get:\newlinev(5)≈6.972v(5) \approx 6.972

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