a particle is moving along a straight line and its position with respect to a reference point is ( s =…

a particle is moving along a straight line and its position with respect to a reference point is ( s = 2t^{3}-10t^{2}-7t + 11 ) (where ( s ) is in meters and ( t ) is in seconds, and assume that ( t ) is non - negative). (round all decimal answers to 2 decimal places.)\na. find the velocity and acceleration as functions of ( t ).\n( v(t)=6t^{2}-20t - 7 )\n( a(t)=12t - 20 )\nb. find the acceleration after 1 second.\n( -8.00\frac{m}{s^{2}} )\nc. find the acceleration at the instant when the velocity is 0.\n( \frac{m}{s^{2}} )

a particle is moving along a straight line and its position with respect to a reference point is ( s = 2t^{3}-10t^{2}-7t + 11 ) (where ( s ) is in meters and ( t ) is in seconds, and assume that ( t ) is non - negative). (round all decimal answers to 2 decimal places.)\na. find the velocity and acceleration as functions of ( t ).\n( v(t)=6t^{2}-20t - 7 )\n( a(t)=12t - 20 )\nb. find the acceleration after 1 second.\n( -8.00\frac{m}{s^{2}} )\nc. find the acceleration at the instant when the velocity is 0.\n( \frac{m}{s^{2}} )

Answer

Explanation:

Step1: Find the time when velocity is 0

Set (v(t)=6t^{2}-20t - 7 = 0). Using the quadratic formula (t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}) for (ax^{2}+bx + c = 0). Here (a = 6), (b=-20), (c=-7). [t=\frac{20\pm\sqrt{(-20)^{2}-4\times6\times(-7)}}{2\times6}=\frac{20\pm\sqrt{400 + 168}}{12}=\frac{20\pm\sqrt{568}}{12}=\frac{20\pm2\sqrt{142}}{12}=\frac{10\pm\sqrt{142}}{6}] (t=\frac{10+\sqrt{142}}{6}\approx\frac{10 + 11.92}{6}=\frac{21.92}{6}\approx3.65) (we take the positive value since (t\geq0))

Step2: Calculate the acceleration at (t = 3.65)

Given (a(t)=12t-20). Substitute (t = 3.65) into (a(t)): (a(3.65)=12\times3.65-20) (a(3.65)=43.8-20) (a(3.65)=23.80)

Answer:

(23.80\frac{m}{s^{2}})