QUESTION 6a) A particle moves along a horizontal line according to the equation:x = 2t2 + 3t -5,where x is the displacement in m) at time t (in seconds).Find the velocity and acceleration at an instant when time is 5 seconds.(6 marks)b) If distance S travelled by an object in time t is given by:S = e3t(i) Calculate displacement at t = 0 seconds.[2 marks](ii) Prove that the acceleration is nine times distance.(6 marks]c) Find the area enclosed by the curve y = sin 2x, the x-axis and the ordinatesx = 0 and x = T.[6 marks​

(a) The instant velocity and acceleration at a specific time [tex]t_0[/tex] are given by evaluating the first and second derivatives at [tex]t=t_0[/tex]So, we have[tex]\begin{cases}x(t)=2t^2+3t-5\\x'(t)=4t+3\\x''(t)=4\end{cases}[/tex]Which implies[tex]\begin{cases}x'(5)=23\\x''(5)=4\end{cases}[/tex]So, when time is 5 seconds, the object has a velocity of 23 m/s and an acceleration of 4 m/s^2 (which is indeed constant along all motion).(b) The equation for the displacement is already given: [tex]s=e^{3t}[/tex]. So, we simply have to evaluate this function at t=0 to get[tex]s(0)=e^{3\cdot 0}=e^0=1[/tex]So, the displacement at t=0 is 1 unit along the positive side of the line.As for the acceleration, applying the derivation formula[tex]\dfrac{d}{dx} e^{f(x)} = e^{f(x)}\cdot f'(x)[/tex]we have[tex]s'(t) = 3e^{3t},\quad s''(t) = 3\cdot 3e^{3t}=9e^{3t}[/tex]So, we have[tex]s(t) = e^{3t},\quad s''(t) =9e^{3t}[/tex]and thus the acceleration is nine times the distance.(c)We have to evaluate the integral[tex]\displaystyle \int_0^T \sin(2x)[/tex]The antiderivative of sin(2x) is -1/2cos(2x), becase we have[tex]\dfrac{d}{dx} -\dfrac{1}{2}\cos(2x) = \dfrac{1}{2}2\sin(2x)=\sin(2x)[/tex]So, we have[tex]\displaystyle \int_0^T \sin(2x) = \left[-\dfrac{1}{2}\cos(2x)\right]_0^T = -\dfrac{1}{2}\cos(2T)+\dfrac{1}{2}\cos(0)=\dfrac{1-\cos(2T)}{2}[/tex]