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(II) A car traveling 85 $\mathrm{km} / \mathrm{h}$ slows down at a constant0.50 $\mathrm{m} / \mathrm{s}^{2}$ just by "letting up on the gas" Calculate $(a)$ thedistance the car coasts before it stops, $(b)$ the time it takesto stop, and $(c)$ the distance it travels during the first andfifth seconds.

a) $557 \mathrm{m} \approx 560 \mathrm{m}$b) 47 sc) 23 m 21 m

Physics 101 Mechanics

Chapter 2

Describing Motion: Kinematics in One Dimension

Physics Basics

Motion Along a Straight Line

Motion in 2d or 3d

Newton's Laws of Motion

University of Michigan - Ann Arbor

University of Washington

Simon Fraser University

Hope College

Lectures

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So for part A, we want to find the total distance it takes for the car to stop. So we know that the final velocity of the car zero s so we can say that velocity final squared equals lost the initial squared plus two times a day out X we're solving for Delta X. So Delta X, the change in displacement in the ex direction would be equal to velocity Final squared minus lost the initial squared divided by two times the acceleration. We know that the car is coming to a stop, so this could be considered negative together, this could be considered zero. Um and so we can say that this is gonna be equal to negative times 85 kilometers per hour. We're going to use a conversion so one meters per second for every 3.6 kilometers per hour and then this quantity er this quantity would be squared. And then this would be divided by two times negative 20.50 meters per second squared, and we find that Delta X is going to be equal to approximately 557 meters for part B. We want to find the time to stop so we can say that philosophy final equals velocity initial plus 80. We know that the final velocity is zero. So we can say that the time taken would simply be equal to the negative lost negative one times the velocity initial divided by the acceleration. So this would be equal Thio negative again. 85 kilometers per hour, uh multiplied by one meters per site, one meter per second for every 3.6 kilometers per hour And this is going to be divided by a So this would be a negative 0.50 meters per second squared and we find that tea is going to be equal to 47.2 seconds. Yes, So this would be your answer for B the time it takes to stop At the time, the told this since traveled when it's trying to stop and then for part c, they want the distance traveled in the 1st 2nd of the 5th 2nd So for part c, we can say that ex initial would be equal to ex at T equals zero seconds which, of course, that the zero meters and we can say that again A is gonna equal negative 0.50 meters per second squared and we're going again the initial velocity of 85 kilometers. So we can say that for X in the 1st 2nd this would be equal to zero plus 85 kilometers per hour multiplied by one meters, one meter per second, divided by 3.6 kilometers per hour, uh, multiplied by one second plus 1/2 times negative 0.50 meters per second squared multiplied by one second quantity squared and we find that traveled in the 1st 2nd would be equal to 23.4 meters. Now we need to find the total dis untroubled in the fifth seconds. So we can say that we need to find X to the X two x At T equals four seconds and ex at T equals five seconds and subtract the two values. So let's get a new workbook again. Ex at T equals four seconds. Aah! This would be equal to again 85 kilometers per hour. We know that the initial position is zero, so we have one meter per second, divided by 3.6 kilometers per hour, multi of other multiplied by four seconds and then plus 1/2 times negative 0.50 meters per second squared multiplied by four seconds. Quantity squared. This is going to be equal to 90.44 meters. Don't round until the very end and then x. So my apologies accepted. T equals five seconds would be equal to 85 kilometers for our multiplied at one meters per second. Divided by 3.6 kilometers per hour, uh, multiplied by five seconds and then plus 1/2 times native 0.50 meters per second. Squared times five seconds quantity squared and this is going to equal 111.81 meters. And we find that X to the fight except five seconds minus except fourths at four seconds is gonna equal 111 top a one minus 90.44 and the distance traveled in that 5th 2nd would be equal to 21.4 meters. That is the end of the solution. Thank you for watching

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