91Ó°ÊÓ

To find the position function for object A, we need to integrate its velocity function with respect to time, \(v(t)=2at\). The position at time \(t=0\) is \(s=0\), so we will use that as our initial condition: $$ s_A(t) = \int 2at\,dt \\ s_A(t) = at^2 + C $$ Since the initial position is \(s=0\) at \(t=0\), we have \(0 = a(0)^2 + C \Rightarrow C=0\). Therefore, the position function for object A is: $$ s_A(t) = at^2 $$

The velocity of object B is given by a constant function, \(V(t)=b\). To find the position function for object B, we can simply integrate its velocity function with respect to time, using the initial condition that its position is \(s=c\) at \(t=0\): $$ s_B(t) = \int b\,dt \\ s_B(t) = bt + D $$ With the initial condition, \(c = b(0) + D \Rightarrow D=c\). Therefore, the position function for object B is: $$ s_B(t) = bt + c $$

To find the time at which object A overtakes object B, we need to find when their positions are equal. That is, when \(s_A(t) = s_B(t)\): $$ at^2 = bt + c $$ Rearranging the equation and solving for \(t\), we get: $$ at^2 - bt - c = 0 \\ t^2 - \frac{b}{a}t - \frac{c}{a} = 0 $$ Now we have a quadratic equation in the form of \(t^2 - pt - q = 0\), where \(p=\frac{b}{a}\) and \(q=\frac{c}{a}\). Using the quadratic formula to solve for \(t\), we get: $$ t = \frac{-(-p) \pm \sqrt{(-p)^2 - 4(-q)}}{2} \\ t = \frac{p \pm \sqrt{p^2 + 4q}}{2} $$ Substituting back our values for \(p\) and \(q\), we get: $$ t = \frac{\frac{b}{a} \pm \sqrt{\left(\frac{b}{a}\right)^2 + 4\frac{c}{a}}}{2} \\ t = \frac{b \pm \sqrt{b^2 + 4ac}}{2a} $$ Since we are looking for the time when object A overtakes object B (that is, catches up to it), we only consider the positive value of the square root. Therefore, the time when object A overtakes object B is: $$ t = \frac{b+\sqrt{b^{2}+4 a c}}{2 a} $$