$A$ motor car moving at a speed of $72\, km/h$ cannot come to a stop in less than $3.0\, s$ while for a truck this time interval is $5.0\, s$. On a highway,the car is behind the truck,both moving at $72\, km/h$. The truck driver signals an emergency stop. At what minimum distance should the car be from the truck so that it does not collide with the truck? The human response time is $0.5\, s$.

(10 M) Initial speed of both vehicles,$u = 72 \times \frac{5}{18} = 20\, m/s$.
For the truck,final velocity $v = 0$ at $t_t = 5.0\, s$. Acceleration $a_t = \frac{v - u}{t_t} = \frac{0 - 20}{5} = -4\, m/s^2$.
For the car,final velocity $v = 0$ at $t_c = 3.0\, s$. Acceleration $a_c = \frac{v - u}{t_c} = \frac{0 - 20}{3} = -6.67\, m/s^2$.
Let $d$ be the initial distance. The car travels at constant speed for $0.5\, s$ (reaction time),covering $d_1 = 20 \times 0.5 = 10\, m$.
After $0.5\, s$,the car starts decelerating. Let $t$ be the time taken for the car to stop after the reaction time. $v = u + a_c t \implies 0 = 20 - 6.67t \implies t = 3.0\, s$.
Distance covered by car during deceleration: $s_c = ut + \frac{1}{2} a_c t^2 = 20(3) + 0.5(-6.67)(3^2) = 60 - 30 = 30\, m$.
Total distance covered by car: $D_c = 10 + 30 = 40\, m$.
Distance covered by truck until it stops: $s_t = u t_t + \frac{1}{2} a_t t_t^2 = 20(5) + 0.5(-4)(5^2) = 100 - 50 = 50\, m$.
To avoid collision,$d + s_t = D_c \implies d + 50 = 40$. Since $d$ cannot be negative,we re-evaluate: the car must stop at the same position as the truck. $d = s_t - D_c = 50 - 40 = 10\, m$.