To determine the minimum stopping distance for a car moving at a different speed, we need to understand the relationship between speed and stopping distance.
Given:
- Initial speed ([tex]\( v_1 \)[/tex]) = 36 km/hr
- Initial stopping distance ([tex]\( d_1 \)[/tex]) = 2 meters
- New speed ([tex]\( v_2 \)[/tex]) = 72 km/hr
First, let's convert the speeds from kilometers per hour (km/hr) to meters per second (m/s) because we want to be consistent with the units for distance (meters) and time (seconds).
1. Convert the initial speed:
[tex]\[ v_1 = 36 \, \text{km/hr} \][/tex]
We know that:
[tex]\[ 1 \, \text{km/hr} = \frac{1000 \, \text{m}}{3600 \, \text{s}} = \frac{5}{18} \, \text{m/s} \][/tex]
So:
[tex]\[ v_1 = 36 \times \frac{5}{18} \, \text{m/s} = 10 \, \text{m/s} \][/tex]
2. Convert the new speed:
[tex]\[ v_2 = 72 \, \text{km/hr} \][/tex]
[tex]\[ v_2 = 72 \times \frac{5}{18} \, \text{m/s} = 20 \, \text{m/s} \][/tex]
Next, we use the fact that the stopping distance is proportional to the square of the speed. This can be expressed as:
[tex]\[ d \propto v^2 \][/tex]
So, we set up a proportion:
[tex]\[ \frac{d_2}{d_1} = \left( \frac{v_2}{v_1} \right)^2 \][/tex]
We know:
[tex]\[ d_1 = 2 \, \text{m} \][/tex]
[tex]\[ v_1 = 10 \, \text{m/s} \][/tex]
[tex]\[ v_2 = 20 \, \text{m/s} \][/tex]
Plug these values into the proportion:
[tex]\[ \frac{d_2}{2} = \left( \frac{20}{10} \right)^2 \][/tex]
Simplify the right side:
[tex]\[ \frac{d_2}{2} = 2^2 \][/tex]
[tex]\[ \frac{d_2}{2} = 4 \][/tex]
Now, solve for [tex]\(d_2\)[/tex]:
[tex]\[ d_2 = 4 \times 2 \][/tex]
[tex]\[ d_2 = 8 \, \text{m} \][/tex]
Therefore, the minimum stopping distance when the car is moving at 72 km/hr is:
[tex]\[ \boxed{8 \, \text{meters}} \][/tex]
