1. **Problem statement:** Car A starts with a velocity of $26.5\ \frac{km}{h}$ and accelerates at $1.7\ \frac{m}{s^{2}}$. Car B moves at a constant velocity of $53.1\ \frac{km}{h}$. We need to find the time $t$ in seconds when car A catches car B. 2. **Convert all units to meters per second (m/s):** $$26.5\ \frac{km}{h} = 26.5 \times \frac{1000}{3600} = 7.3611\ \frac{m}{s}$$ $$53.1\ \frac{km}{h} = 53.1 \times \frac{1000}{3600} = 14.75\ \frac{m}{s}$$ 3. **Set up the equations for positions:** Let $t$ be the time in seconds when A catches B. Position of A at time $t$: $$x_A = v_{A0} t + \frac{1}{2} a t^2 = 7.3611 t + \frac{1}{2} \times 1.7 \times t^2 = 7.3611 t + 0.85 t^2$$ Position of B at time $t$: $$x_B = v_B t = 14.75 t$$ 4. **Find $t$ when $x_A = x_B$:** $$7.3611 t + 0.85 t^2 = 14.75 t$$ Bring all terms to one side: $$0.85 t^2 + 7.3611 t - 14.75 t = 0$$ $$0.85 t^2 - 7.3889 t = 0$$ 5. **Factor out $t$:** $$t (0.85 t - 7.3889) = 0$$ 6. **Solve for $t$:** Either $t=0$ (initial time) or $$0.85 t - 7.3889 = 0$$ $$0.85 t = 7.3889$$ $$t = \frac{7.3889}{0.85}$$ 7. **Calculate $t$:** $$t = 8.69\ s$$ **Answer:** Car A catches car B after approximately $8.69$ seconds.