(b)

In this problem equations related to one dimensional motion will be applied for acceleration positive sign will be used and for retardation negative sign will be used.

Given, speed of car as well as truck $=72 \,km / h$

$=72 \times \frac{5}{18}\, m / s =20 \,m / s$

Retarded motion for truck $v=u+a_t t$

$0=20+a_t \times 5 \text { or } a_t=-4\,m / s ^2$

Retarded motion for the car $v=u+a_c t$

$0=20+a_c \times 3 \text { or } a_c=-\frac{20}{3}\,m / s ^2$

Let car be at a distance $x$ from truck, when truck gives the signal and $t$ be the time taken to cover this distance.

As human response time is $0.5 s$, therefore, time of retarded motion of car is $(t-0.5) s$.

Velocity of car after time $t$,

$v_c=u-a t=20-\left(\frac{20}{3}\right)(t-0.5)$

Velocity of truck after time $t$,

$v_t=20-4\,t$

To avoid the car bump onto the truck,

$v_c=v_t$

$20-\frac{20}{3}(t-0.5)=20-4 t$

$4 t=\frac{20}{3}(t-0.5) \Rightarrow t=\frac{5}{3}(t-0.5)$

$3 t=5 t-2.5 \Rightarrow t=\frac{2.5}{2}=\frac{5}{4} s$

Distance travelled by the truck in time $t$,

$s_t =u_t t+\frac{1}{2} a_t t^2$

$=20 \times \frac{5}{4}+\frac{1}{2} \times(-4) \times\left(\frac{5}{4}\right)^2$

$ =21.875\,m$

Distance travelled by car in time $t=$ Distance travelled by car in $0.5 s$ (without retardation) + Distance travelled by car in $(t-0.5) s$ (with retardation)

$s_c =(20 \times 0.5)+20\left(\frac{5}{4}-0.5\right)-\frac{1}{2}\left(\frac{20}{3}\right)\left(\frac{5}{4}-0.5\right)^2$

$=23.125\,m$

$\therefore s_c-s_t=23.125-21.875=1.250\,m$

Therefore, to avoid the bump onto the truck, the car must maintain a distance from the truck more than $1.250\,m$.