A particle travels along the x-axis such that its velocity is given by v(t)=t^(0.7)sin(2t+3). What is the acceleration of the particle at time t=4 ? You may use a calculator and round your answer to the nearest thousandth. Answer:

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A particle travels along the  x-axis such that its velocity is given by  v(t)=t^(0.7)sin(2t+3). What is the acceleration of the particle at time  t=4 ? You may use a calculator and round your answer to the nearest thousandth. Answer:

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Identify Given Function: Identify the given function and what is asked. We are given the velocity function v(t) = t^(0.7)sin(2t+3) and we need to find the acceleration at time t=4. Acceleration is the derivative of velocity with respect to time.Differentiate Velocity Function: Differentiate the velocity function to find the acceleration function. To find the acceleration, we need to take the derivative of the velocity function with respect to time t. The velocity function is a product of two functions of t, so we will use the product rule for differentiation.Apply Product Rule: Apply the product rule for differentiation. The product rule states that the derivative of a product of two functions is the derivative of the first function times the second function plus the first function times the derivative of the second function. Let's denote the first function as f(t) = t^(0.7) and the second function as g(t) = sin(2t+3).Differentiate f(t) and g(t): Differentiate f(t) and g(t). The derivative of f(t) with respect to t is f'(t) = 0.7t^(-0.3). The derivative of g(t) with respect to t is g'(t) = cos(2t+3) * derivative of (2t+3) with respect to t, which is 2. So, g'(t) = 2cos(2t+3).Combine Derivatives: Combine the derivatives using the product rule. Using the product rule, the acceleration function a(t) is: a(t) = f'(t)g(t) + f(t)g'(t) a(t) = (0.7t^(-0.3))sin(2t+3) + (t^(0.7))(2cos(2t+3))Evaluate at t=4: Evaluate the acceleration function at t=4. Now we substitute t=4 into the acceleration function to find the acceleration at that time. a(4) = (0.7(4)^(-0.3))sin(2(4)+3) + (4^(0.7))(2cos(2(4)+3))Calculate Numerical Value: Calculate the numerical value using a calculator. Using a calculator, we find: a(4) = (0.7(4)^(-0.3))sin(11) + (4^(0.7))(2cos(11)) a(4) ≈ (0.7 * 0.316)sin(11) + (2.297)(2cos(11)) a(4) ≈ (0.2212)sin(11) + (4.594)(2cos(11)) a(4) ≈ (0.2212 * 0.1908) + (4.594 * 1.9645) a(4) ≈ 0.0422 + 9.0277 a(4) ≈ 9.0699

A particle travels along the $x$ -axis such that its velocity is given by $v(t)=t^{0.7} \sin (2 t+3)$ . What is the acceleration of the particle at time $t=4$ ? You may use a calculator and round your answer to the nearest thousandth. $\newline$ Answer:

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A particle travels along the x x x-axis such that its velocity is given by v(t)=t0.7sin⁡(2t+3) v(t)=t^{0.7} \sin (2 t+3) v(t)=t0.7sin(2t+3). What is the acceleration of the particle at time t=4 t=4 t=4 ? You may use a calculator and round your answer to the nearest thousandth.\newlineAnswer:

A particle travels along the $x$ -axis such that its velocity is given by $v(t)=t^{0.7} \sin (2 t+3)$ . What is the acceleration of the particle at time $t=4$ ? You may use a calculator and round your answer to the nearest thousandth. $\newline$ Answer: