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(II) $(a)$ Show that the minimum stopping distance for an auto-mobile traveling at speed $v$ is equal to $v^{2} / 2 \mu_{\mathrm{s}} g,$ where $\mu_{\mathrm{s}}$ isthe coefficient of static friction between the tires and the road,and $g$ is the acceleration of gravity. $(b)$ What is this distance fora $1200-\mathrm{kg}$ car traveling 95 $\mathrm{km} / \mathrm{h}$ if $\mu_{\mathrm{s}}=0.65 ?(c)$ What wouldit be if the car were on the Moon (the acceleration of gravityon the Moon is about $g / 6$ ) but all else stayed the same?

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a. $\Delta x=\frac{v_{o}^{2}}{2 \cdot u_{s} \cdot g}$b. $\Delta x=54.62 \mathrm{m}$c. $330 \mathrm{m}$

Physics 101 Mechanics

Chapter 5

Using Newton's Laws: Friction, Circular Motion, Drag Forces

Motion Along a Straight Line

Motion in 2d or 3d

Newton's Laws of Motion

Applying Newton's Laws

Rotation of Rigid Bodies

Dynamics of Rotational Motion

Equilibrium and Elasticity

Cornell University

Simon Fraser University

University of Sheffield

Lectures

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In physics, a rigid body is an object that is not deformed by the stress of external forces. The term "rigid body" is used in the context of classical mechanics, where it refers to a body that has no degrees of freedom and is completely described by its position and the forces applied to it. A rigid body is a special case of a solid body, and is one type of spatial body. The term "rigid body" is also used in the context of continuum mechanics, where it refers to a solid body that is deformed by external forces, but does not change in volume. In continuum mechanics, a rigid body is a continuous body that has no internal degrees of freedom. The term "rigid body" is also used in the context of quantum mechanics, where it refers to a body that cannot be squeezed into a smaller volume without changing its shape.

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In physics, rotational dynamics is the study of the kinematics and kinetics of rotational motion, the motion of rigid bodies, and the about axes of the body. It can be divided into the study of torque and the study of angular velocity.

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so if we drop free body diagram of the car, you see that MG is acting straight downwards. Um, then we have the normal force, which is perpendicular. Tow the horizontal plane. We call it s N. Then we have frictional force, which is working against the car. So if the car is moving from left to right, then friction forces from right to left. Call it f off fr If we apply Newton's second Law on this, um, system, we see that in extraction, we have f X, which is, um, negative f off if are alone. Now, the reason we're taking native is because we're choosing from left to right direction as our positive X and from on upwards direction as our positive light. So, um, negative fr is equal to m times, eh? Now, since the car is, um, not sliding, we can assume the friction on the maximum friction force using the static friction condition. So if off fr will be new S s, as in static times, fr friend. And what is, if often we see that since there's no vertical motion, so if often should be, will do mg. So this country right here is us times, mg. Now, once we know that, we can put that back in our FX equation and we see that it's negative. Mu s m G is equal to end things A And from here we see that a is equal. Do us times, g. And then there's a good site. No, uh, for for the minimum stopping distance, we can use the question 12. Sorry, We can use the question to point 12 c and see that the squared minus we all squared is equal toe price a X minus x o where view is the initial velocity of the car. The square of these defining velocity is the acceleration and ah x minus X not is the displacement. So from here we see that displacement should be equal to explain its external, which is B squared, minus video squared, divided by twice a day. And as we mentioned that, um, the final velocity zero. And if we consider Theo as small, we are our initial velocity V. We can replace this quantity as zero minus. He's good. Then we divide that by two way also we saw that is equal to name us time ski so we can put that here and we get negative. Reece squared, divided by negative us trying to G, which is equal to B squared over us time. See? And then there's a two. Yeah, right. So that's the displacement on. That's the algebraic solution for that. No, um given, for with the given values we see that V is equal to, uh, 95 kilometers per hour. We convert that toe meters for a second, and to do so, we multiply one meter per second by 3.6 kilometers for ours, which is 26.38 meetings for a second. Now, um, if we put that value over here and solve for X minus X not we see that X minus ex not is equal to 26.38 meters per second. Put a square divided by place off 0.65 which is mu s times, 9.8 meat over seconds word. And that's equal to 55 meters. No, in part C. We need to find out the distance if the gravity has changed or if the car is on the moon. So we know that if the car is on the moon. The gravity is decreased by ahh factor of six. So if we say GMs in gravity of the moon, that's gonna be gravity of the earth, divided by six. So that means our displacement will be increased by a factor of six. So previously are displacement X minus X not on. All right was 55 centimeters. So the same thing on moon will be six times 55 centimeters, which is 3 30 meters. Thank you.

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