For a vehicle moving at 80 , km/hr, how many | vedclass

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(B) The stopping distance $d$ of a vehicle is given by the formula $d = \frac{v_0^2}{2|a|}$, where $v_0$ is the initial velocity and $a$ is the consta

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$4$

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(B) The stopping distance $d$ of a vehicle is given by the formula $d = \frac{v_0^2}{2|a|}$, where $v_0$ is the initial velocity and $a$ is the constant deceleration.Since $d \propto v_0^2$, the ratio of the stopping distances for two vehicles with initial velocities $v_1 = 80 \, km/hr$ and $v_2 = 40 \, km/hr$ is:$\frac{d_1}{d_2} = \left(\frac{v_1}{v_2}\right)^2 = \left(\frac{80}{40}\right)^2 = (2)^2 = 4$.Therefore, the stopping distance for the vehicle moving at $80 \, km/hr$ is $4$ times that of the vehicle moving at $40 \, km/hr$.

For a vehicle moving at $80 \, km/hr$, how many times is the stopping distance compared to a vehicle moving at $40 \, km/hr$, assuming both have the same deceleration (in $\times$)?

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