Problem 1: Time Interval Between Cars
Step-by-Step Derivation
Let the initial speed of the cars be $v_1$ and the time interval between their departure be $\Delta t_1$.
The distance separation ($d$) between any two consecutive cars while they are moving at speed $v_1$ is given by:
$$ d = v_1 \times \Delta t_1 $$According to the problem, at a specific instant, all cars simultaneously reduce their speed to $v_2$.
Since the change happens simultaneously for all cars, the relative velocity between any two cars at the moment of change is zero. Therefore, the spatial separation $d$ remains unchanged immediately after the speed reduction.
Now, both cars are moving at the new speed $v_2$, maintaining the same separation distance $d$. The new time interval $\Delta t_2$ between the arrival of the cars is the time it takes to cover the distance $d$ at the new speed $v_2$.
$$ \Delta t_2 = \frac{d}{v_2} $$Substituting the expression for $d$ from the first equation:
$$ \Delta t_2 = \frac{v_1 \Delta t_1}{v_2} $$Calculation
Given values:
- Initial speed $v_1 = 60 \text{ km/h}$
- Final speed $v_2 = 40 \text{ km/h}$
- Initial interval $\Delta t_1 = 30 \text{ s}$
Substituting these into our formula:
$$ \Delta t_2 = \left( \frac{60}{40} \right) \times 30 \text{ s} $$ $$ \Delta t_2 = 1.5 \times 30 \text{ s} $$ $$ \Delta t_2 = 45 \text{ s} $$Correct Option: (d)