istance can be obtained from the work-energy principle:

$$ F \cdot s = \frac{1}{2} m v^2 $$

where:

$ F $ is the force applied by the brakes,

$ s $ is the stopping distance,

$ m $ is the mass of the moving object,

$ v $ is the velocity of the object.

Assuming the brakes apply a constant stopping force in both scenarios, we compare the initial and new stopping scenarios.

Total mass $ m_1 = 80 \, \text{kg (mass of man)} + 100 \, \text{kg (mass of scooter)} = 180 \, \text{kg} $

Stopping distance: $ s_1 $

New total mass $ m_2 = 80 \, \text{kg (mass of man)} + 100 \, \text{kg (mass of scooter)} + 60 \, \text{kg (mass of rice)} = 240 \, \text{kg} $

New stopping distance: $ s_2 $

Using the work-energy principle for both scenarios and assuming velocity $ v $ and force $ F $ remain constant:

Initial condition:

$$ F \cdot s_1 = \frac{1}{2} \times 180 \times v^2 $$

New condition:

$$ F \cdot s_2 = \frac{1}{2} \times 240 \times v^2 $$

Dividing the second equation by the first:

$ \frac{s_2}{s_1} = \frac{\frac{1}{2} \times 240 \times v^2}{\frac{1}{2} \times 180 \times v^2} = \frac{240}{180} = \frac{4}{3} $

Rearranging gives:

$$ s_2 = \frac{4}{3} s_1 $$

or equivalently:

$$ 3 s_2 = 4 s_1 $$

Thus, the correct relationship is given by:

Option D: $ 4 s_1 = 3 s_2 $

Work Energy And Power

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