sladerThe position s(t) of a particle moving along the x-axis at time t is given by s(t) = −t3 + 3t2 + 5 4 , where s is measured in meters and t is measured in seconds. At what time is the particle's instantaneous velocity equal to its average velocity on the interval [0, 6]?

Answer: t = 3.646 secondsStep-by-step explanation:We have given instantaneous velocity [tex]s(t) = -t^3 + 3t^2 + 5 4[/tex]Now,[tex]\dfrac {ds}{dt} = -3t^2 + 6t ...(1)[/tex]Now we will find the average velocity[tex]\text {Average velocity} =\dfrac {s(t_2) - s(t_1)}{t_2 - t_1}[/tex][tex]\text {Average velocity} = \dfrac {s(6) - s(0)}{6 - 0} = \dfrac{((-6)^3 + 3\times 6^2 + 54)- (0 + 3\times 0 + 54)}{6} \\\\\text {Average velocity} = \dfrac {-216 + 108 + 54 - 54}6 = \dfrac {-108}6 =-18 ... (2)\\\\ \text {Average velocity} = -18[/tex]It is also given that particle's instantaneous velocity equal to its average velocity. From equation (1) and (2)[tex]-3t^2 + 6t = -18\\3t^2 - 6t - 18 = 0\\t^2 -2t - 6 = 0\\[/tex]Now we factorize the equation by the rule of Shri dharacharya [tex]\dfrac {-b \pm \sqrt{b^2 - 4ac} }{2a}[/tex][tex]t=\dfrac {-2 \pm\sqrt{(-2)^2 - 4 \times 1 \times -6} }{2 \times 1} \\\\t=\dfrac {-2 \pm\sqrt{ 4 + 24} }{2 } \\\\t= \dfrac {-2 \pm\sqrt{ 28} }{2 }\\\\t= \dfrac {-2 \pm 2\sqrt{ 7} }{2 }\\t= 1 \pm \sqrt 7[/tex]If we put the value of [tex]\sqrt{7}[/tex] If we take positive value than t = 1 + 2.646 = 3.646If we take negative value thant = 1 - 2.646 =  -1.646We can not take negative value of time So, t = 3.646 seconds