\(\begin{aligned} 2a(x-x_{0})x&=(2a)v_{0}\left(\frac{v-v_{0}}{a} \right)+(2a)\frac{1}{2}a\left(\frac{v-v_{0}}{a} \right)^{2} \\ 2a(x-x_{0})&=(2v_{0})a\left(\frac{v-v_{0}}{a} \right)+a^{2}\left( \frac{...2a(x−x0)x=(2a)v0(v−v0a)+(2a)12a(v−v0a)22a(x−x0)=(2v0)a(v−v0a)+a2(v−v0a)22a(x−x0)=2v0(v−v0)+(v−v0)2 So, we are interested in the value of t when xR=xV, where xR is the position of Rob, and xV is the position of the velociraptor.