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$3.1 \textrm{s}$

Giancoli 6th Edition, Chapter 2, Problem 31

(6:59)

Chapter 2, Problem 31 is solved.

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Transcript for this Giancoli solution

The sprinter has three minutes left which we’ll convert to one hundred and eighty seconds and he has to cover the remaining one thousand one hundred meters of the ten thousand meter race. We’re going to estimate what his initial speed is in order to solve this, so we’re going to assume that the ten thousand meters minus the one thousand one hundred remaining which is covered in twenty seven minutes times sixty seconds per minute gives an average speed which we’ll consider our initial speed as five point four nine three eight meters per second and the sprinter’s acceleration is zero point two zero meters per second squared. Think of this one thousand one hundred meters as consisting of two portions, the first portion is going to be while the sprinter is accelerating and he’ll have some time ‘ta’ spent accelerating and that'll be some distance ‘da’ and then the remaining portion of the one thousand one hundred meter total is going to be at constant speed ‘tc’ and distance ‘dc’ as so we can see that ‘da’ plus ‘dc’ is one thousand one hundred meters and we know that ‘ta’ plus ‘tc’ equals the total one hundred and eighty seconds remaining in the race. The distance equation given initial velocity and knowing the time ‘ta’ and the acceleration ‘a’ is: ‘da’ plus ‘dc’ equals ‘vi’ times ‘ta’ plus one half ‘a’ times ‘ta’ squared, we don’t know ‘ta’ so we’ll solve for it. Substituting for ‘dc’, we’re going at constant speed so it’ll be whatever final speed he reaches after the acceleration during the first portion multiplied the time spent at the constant speed ‘tc’ and that total is one thousand one hundred meters. Substituting for ‘vf’ and ‘tc: ‘vi’ times ‘ta’ plus one half ‘a’ times ‘a’ times ‘ta2’, the ‘vf’ is going to be the initial speed plus acceleration times time spent accelerating and ‘tc’ is one hundred and eighty minus ‘ta’, that equals one thousand one hundred meters. The only variable in the expression we don’t know is ‘ta’ and we need to use the quadratic equation as the last step. Expanding the binomials: ‘vi’ times ‘ta’ plus one half ‘a’ times ‘ta2’ plus one hundred and eighty times ‘vi’ minus ‘vi’ times ‘ta’ plus one hundred and eighty times ‘a’ times ‘ta’ minus ‘a’ times ‘ta2’ which equals one thousand one hundred and cancelling out collecting terms we’ll have ‘ta2’ multiplied by negative one half ‘a’ plus one hundred and eighty ‘a’ times ‘ta’ and after subtracting one thousand one hundred from both sides that will combine with the one hundred and eighty times ‘vi’ to make our constant term in this quadratic equation all of which equals zero. Substitute in some numbers we have: negative one half times zero point two meters per second squared times ‘ta2’ plus one hundred and eighty times zero point two meters per second squared times ‘ta’ plus one hundred ad eighty times five point four nine there eight meters per second minus one thousand one hundred meters equals zero and this works put to negative zero point one times ‘ta2’ plus thirty six ‘ta’ minus one hundred and eleven point one equals zero and this is the point you refer back to the quadratic formula and solving it for ‘ta’ gives tree point one seconds. If you want to know more about the quadratic formula and more about this last step just [write] a comment below this video and I’ll tell you more.

Comments

Hi can you please explain the last step