Try Magic Notes and save time.Try it free
Try Magic Notes and save timeCrush your year with the magic of personalized studying.Try it free

Related questions with answers

Question

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 3030-mi/h speed limit before he made an emergency stop, brakes locked and wheels sliding. The length of skid marks on the road was 19.219.2 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(=32 ft/s2)^2) and did not give the driver a ticket. Was the driver speeding? Explain.

Solution

Verified
Step 1
1 of 2

Calculate 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

vx2=v0x2+2a(xx0)\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, vx=0,x=19.2 ft, and a=-32 fts2v_x = 0, x= 19.2\text{ ft}, \text{ and } a = \text{-}32\mathrm{~\dfrac{ft}{s^2}}.

Solving equation(1) for v0xv_{0x} gives

v0x=vx22a(xx0)=022(-32 fts2)(19.2 ft0 ft)=1228.8 ft2s2=35.05 fts\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 fts=0.6818 mih1\mathrm{~\dfrac{ft}{s}} = 0.6818\mathrm{~\dfrac{mi}{h}}.

Converting the value to desired unit yields

v0x=35.05 fts(0.6818 mih1 fts)=24 mih\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*}

v0x=24 mih\boxed{\therefore v_{0x} =24 \mathrm{~\dfrac{mi}{h}}}

Thus, the driver was not over speeding because his speed was less than the 30 mih30\mathrm{~\dfrac{mi}{h}} speed limit.

Create an account to view solutions

Create an account to view solutions

Recommended textbook solutions

Physics for Scientists and Engineers: A Strategic Approach with Modern Physics 4th Edition by Randall D. Knight

Physics for Scientists and Engineers: A Strategic Approach with Modern Physics

4th EditionISBN: 9780133942651 (4 more)Randall D. Knight
3,508 solutions
Mathematical Methods in the Physical Sciences 3rd Edition by Mary L. Boas

Mathematical Methods in the Physical Sciences

3rd EditionISBN: 9780471198260Mary L. Boas
3,355 solutions
Physics, Volume 1 5th Edition by David Halliday, Kenneth S. Krane, Robert Resnick

Physics, Volume 1

5th EditionISBN: 9780471320579 (2 more)David Halliday, Kenneth S. Krane, Robert Resnick
2,456 solutions
Fundamentals of Physics 10th Edition by David Halliday, Jearl Walker, Robert Resnick

Fundamentals of Physics

10th EditionISBN: 9781118230718David Halliday, Jearl Walker, Robert Resnick
8,971 solutions

More related questions

1/4

1/7