## Related questions with answers

Suppose that you were called upon to give some advice to a lawyer concerning the physics involved in one of her cases. The question is whether a driver was exceeding a $30$-mi/h speed limit before he made an emergency stop, brakes locked and wheels sliding. The length of skid marks on the road was $19.2$ ft. The police officer made the assumption that the maxi- mum deceleration of the car would not exceed the accelera- tion of a freely falling body $(=32$ ft/s$^2)$ and did not give the driver a ticket. Was the driver speeding? Explain.

Solution

VerifiedCalculate the velocity of the car and determine if the driver was exceeding the speed limit. First, we identify which kinematic equation of motion is to be used to solve for the unknown. Hence, our useful equation is given by

$\begin{align} v_x^2 = v_{0x}^2 +2a(x-x_0)\end{align}$

We assume that the car has initial velocity when it starts to slide and brought to rest after traveling the length of skid marks on the road which was 19.2 feet from the point he made an emergency stop. Hence, $v_x = 0, x= 19.2\text{ ft}, \text{ and } a = \text{-}32\mathrm{~\dfrac{ft}{s^2}}$.

Solving equation(1) for $v_{0x}$ gives

$\begin{align*} v_{0x} &=\sqrt{v_{x}^2 -2a(x-x_0)}\\ & = \sqrt{0^2 -2\left( \text{-}32\mathrm{~\dfrac{ft}{s^2}}\right)(19.2\text{ ft}-0\text{ ft})}\\ & = \sqrt{1228.8\mathrm{~\dfrac{ft^2}{s^2}}}\\ & =35.05\mathrm{~\dfrac{ft}{s}} \end{align*}$

From Appendix G, $1\mathrm{~\dfrac{ft}{s}} = 0.6818\mathrm{~\dfrac{mi}{h}}$.

Converting the value to desired unit yields

$\begin{align*} v_{0x} &= 35.05\mathrm{~\dfrac{ft}{s}} \left(\dfrac{0.6818\mathrm{~\dfrac{mi}{h}}}{1\mathrm{~\dfrac{ft}{s}}}\right)\\ & = 24 \mathrm{~\dfrac{mi}{h}} \end{align*}$

$\boxed{\therefore v_{0x} =24 \mathrm{~\dfrac{mi}{h}}}$

Thus, the driver was not over speeding because his speed was less than the $30\mathrm{~\dfrac{mi}{h}}$ speed limit.

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