The position of an object as a function of time is given as \(x=A t^{3}+B t^{2}+C t+D .\) The constants are \(A=2.1 \mathrm{~m} / \mathrm{s}^{3}\) \(B=1.0 \mathrm{~m} / \mathrm{s}^{2}, C=-4.1 \mathrm{~m} / \mathrm{s},\) and \(D=3 \mathrm{~m}\) a) What is the velocity of the object at \(t=10.0 \mathrm{~s}\) ? b) At what time(s) is the object at rest? c) What is the acceleration of the object at \(t=0.50 \mathrm{~s} ?\) d) Plot the acceleration as a function of time for the time interval from \(t=-10.0 \mathrm{~s}\) to \(t=10.0 \mathrm{~s}\).

The velocity of an object is the derivative of its position function with respect to time. Given the position function: \(x(t) = 2.1t^3 + 1.0t^2 - 4.1t + 3\) Now, differentiate the position function with respect to time: \(v(t) = \dfrac{dx}{dt} = 6.3t^2 + 2.0t - 4.1\) To find the velocity at t=10.0 s, substitute this value of t into the velocity function: \(v(10.0) = 6.3(10.0)^2 + 2.0(10.0) - 4.1 = 6.3(100) + 20 - 4.1 = 630 + 15.9 = 645.9 m/s\) The object's velocity at t=10.0 s is 645.9 m/s.

An object is at rest when its velocity is zero. Using the velocity function we obtained previously, we set it to zero and solve for t. \(v(t) = 6.3t^2 + 2.0t - 4.1 = 0\) This is a quadratic equation, and we can use the quadratic formula to find the roots: \(t = \dfrac{-b\pm\sqrt{b^2-4ac}}{2a}\) In this case, a = 6.3, b = 2.0, and c = -4.1. Substituting these values, we get: \(t = \dfrac{-2\pm\sqrt{2^2 - 4(6.3)(-4.1)}}{2(6.3)}\) Now solve for t: \(t = \dfrac{-2\pm\sqrt{4+102.52}}{12.6} = \dfrac{-2\pm\sqrt{106.52}}{12.6}\) Calculating the two possible values of t: \(t_1 = \dfrac{-2+\sqrt{106.52}}{12.6} \approx 0.42 s\) \(t_2 = \dfrac{-2-\sqrt{106.52}}{12.6} \approx -1.52 s\) The object is at rest at t ≈ 0.42 s and t ≈ -1.52 s.

The acceleration of the object is the derivative of its velocity function with respect to time. We differentiate the velocity function we obtained in part (a) with respect to time: \(a(t) = \dfrac{dv}{dt} = 12.6t + 2.0\) Now, we substitute t=0.50 s to find the acceleration at this time: \(a(0.50) = 12.6(0.50) + 2.0 = 6.3 + 2.0 = 8.3 m/s^2\) The object's acceleration at t=0.50 s is 8.3 m/s².

We have obtained the acceleration function in part (c): \(a(t) = 12.6t + 2.0\). Now, we can create a plot of this function in the given time interval from t = -10.0 s to t = 10.0 s. Use your preferred graphing software or graphing calculator to plot the function \(a(t) = 12.6t + 2.0\) in the interval \(-10.0 \leq t \leq 10.0\). The resulting graph is a straight line with a positive slope, starting at \(a(-10.0) = -123.0 m/s^2\) and ending at \(a(10.0) = 127.0 m/s^2\)