Question

A motor car moving at a speed of $72km/h$ can not come to a stop in less than $3.0s$ while for a truck this time interval is $5.0s$. On a higway the car is behind the truck both moving at $72km/h.$ 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 $0.5s:$

• A. $1.25m$
• B. $1.5m$
• C. $1m$
• D. $0.5m$

Answer 1

The correct option is A $1.25m$

Given that;

Speed of both car and truck, $v=72km/h$

or, $v=20m/s$

Time taken to stop the car after applying the break,$t_c=3.0s$

Time taken to stop the truck after applying the break$=t_t=5.0s$

Therefore,

Retardation of the car,$a_c=\frac{0-20}{3}=-\frac{20}{3}m/s^2$

Retardation of the truck, $a_t=\frac{0-20}{5}=-4m/s^2$

Let at time $t=0$, the truck signals to stop and applies the break.

At time $t=0$, let the truck and the car be at a distance $sm$ from each other.

Time taken by the car to travel $sm=t$ seconds.

To avoide the bump, the velocity of the car and the truck after time $t$ should be equal.

Thus,

Velocity of the truck after time t :

$v_t=u-at$

$v_t=20-4t$

The car will apply the break after 0.5 seconds of signal.

Thus;

Velocity of car after time t will be:

$v_c=u-a(t-0.5)$

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

Therefore,

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

$t=\frac{5}{4}sec$

Thus,

Distance travelled by truck in time $t$

$s_t=ut+\frac{1}{2}a{t}^2$

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

$s_t=21.875$

Distance travelled by the car in $t$ seconds, $s_s$;

Distance travelled by the car in 0.5 s + distance travelled by the car in $(t-0.5)sec$

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

$s_s=23.125$

Therefore, minimum distance required for the car to be distant from the truck will be $23.125-21.875=1.25m$