A motor car moving at a speed of $72\, km/h$ cannot come to a stop in less than $3.0\,\sec $ whilefor a truck this time interval is $5.0\,\sec $. On a highway, the car is behind the truck  both moving at $72\, km/h$. The truck gives a signal that it is going to stop at emergency. At what distance the car should be from the truck so that it does not bump onto (collide with) the truck. Human response time is $0.5\,\sec $.

Given, speed of car $=$ speed of truck $=72 \mathrm{~km} / \mathrm{h}$
$=72 \times \frac{5}{18} \mathrm{~m} / \mathrm{s}=20 \mathrm{~m} / \mathrm{s}$
Now, $v=u+a_{t} t$
$0=20+a_{t} \times 5$
$a_{t}=-4 \mathrm{~m} / \mathrm{s}^{2}$
For car, $v=u+a_{c} t$
$0=20+a_{c} \times 3$
$a_{c}=-\frac{20}{3} \mathrm{~m} / \mathrm{s}^{2}$
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 $0.5 \mathrm{~s}$, therefore, time of retarded motion of car is $(t-0.5) \mathrm{s}$. 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 collision,
$v_{c} =v_{t}$
$20 -\frac{20}{3}(t-0.5)=20-4 t$
$4 t =\frac{20}{3}(t-0.5)$
$t =\frac{5}{3}(t-0.5)$
$3 t =5 t-2.5$
$\therefore t =\frac{2.5}{2}=\frac{5}{4} \mathrm{~s}$