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A car travelling at a speed of $v_{0}$ applies its breaks, skidding to a stop over a distance of $x$ m. Assuming that the deceleration due to the breaks is constant, what would be the skidding distance of the same car if the breaking were twice as effective (doubling the magnitude of deceleration)?(A) 0.25$x \mathrm{m}$(B) 0.5$x \mathrm{m}$(C) $x \mathrm{m}$(D) 2$x \mathrm{m}$

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Chapter 13

Practice Test 3

Section 1

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high in the given problem. Initially speed of the car is we i. Is equal to we're not. It is given as we not and finally it is coming to rest so it's finally speed. We f Will be taken as zero total distance traveled before coming to rest is X. No. Using third equation of motion which says the Squire of finally speed, we have his square equals to a square of initially speed plus two times of the product of acceleration with their displacement. So far we have this zero for V. I. This is the note square plus two times of a acceleration with the displacement cover means X. So from here there's two X comes out to be minus the notice where means acceleration or we can see retardation Produced in this motion of the car is minus we not required by two X. No it is said that if the retardation is made twice means if a dash is twice of eight means this is twice off minus we notice required by two X cancelling this too. Yes, this is minus we not square by X. So using The 3rd equation of motion again we left. The square is equal to B i square plus two eight dash X dash. We have zero again, V I S. We note again plus two for a dash, this is minus. We not be squared by X into X dash. So rearranging the terms we get to wise off. Me not is squared by X into X. Dash is equal to be noted square. So canceling this we know T square. We kept An expression for the distance covered by the car before coming to rest. Now will be given by X by two Or this is 0.5 times of X means here we can see our option B is correct. Thank you.

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