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A car is traveling at $ 100\;km/h $ when the driver sees an accident $ 80\;m $ ahead and slams on the brakes. What constant deceleration is required to stop the car in time to avoid a pileup?

at least $-62,500 \mathrm{km} / \mathrm{h}^{2}$

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Campbell University

Baylor University

University of Michigan - Ann Arbor

The starboard. This question that saying that a car is traveling 100 km/h when the driver sees an accident 80 m ahead and slams on the brakes. What concerned deceleration is required to stop the car in time to avoid a pile up. All right. So what is given is that We we have a we have a car traveling at 100 km/h And a constant deceleration has been applied toward and it it's supposed to be stopped after traveling another 18 m and if it stops, then definitely the speed That the final speed must be zero. So using this information, we need to find the value of a. We can use the second equation of motion that we square, there could be non square plus two is so this is the first of all. We need to convert this and meet up a second for that as a unit. So 100 kilometers per hour, 100 kilometers per hour is going to be equal 200 times 1000 Over 36 doubles here. These two zeros of cancer, we have 1000 Over 36. This is meet up a second. So we square is zero square because the finance speed is zero. Uh, initiative speed as 1000 over 36 Whole Square Plus two and displacement as 80. So if you re arrange this a little bit, we have -168 and this has taken a word of the left Single, over 36 whole square. So from this will get the value of a as negative 1/1 60 times 1000 or 36 square. So the value of A comes out as let me just grab my calculator here. It's 1000 over 36 square Divided by 1 60. So it's coming as negative 4.82 m/s squared up to two decimal places. So this is the required, it's coming in negative because definitely it's deceleration. So that's why it's coming in negative. So this is the required constant deceleration, which the car must apply. Thank you.