Question

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.)
a. find the velocity and acceleration as functions of ( t ).
( v(t)=6t^{2}-20t - 7 )
( a(t)=12t - 20 )
b. find the acceleration after 1 second.
( -8.00\frac{m}{s^{2}} )
c. find the acceleration at the instant when the velocity is 0.
( \frac{m}{s^{2}} )

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:

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