by jrjack
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 P: 107 1. The problem statement, all variables and given/known data The driver of a car traveling at 71 ft/sec suddenly applies the brakes. The position of the car is s = 71t - 20t2, t seconds after the driver applies the brakes. After how many seconds does the car come to a stop? Round your answer to the nearest tenth. 2. Relevant equations 3. The attempt at a solution s=71t-20t2 0=71t-20t2 0=71-20t t=71/20 t=3.55 = 3.6 However the correct answer is 1.8, which is half my answer, but I do not understand why you would divide by 2? Or where I went wrong?
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P: 9,772
 Quote by jrjack 0=71t-20t2
 P: 107 Well, because the position of the car is represented by the function s=71t-20t2, and since the driver is applying the brakes and coming to a stop, his postion when stopped should be s=0.
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PF Gold
P: 9,772

 Quote by jrjack Well, because the position of the car is represented by the function s=71t-20t2, and since the driver is applying the brakes and coming to a stop, his postion when stopped should be s=0.
Just because the car is stopped doesn't mean the displacement (s) is zero.
 P: 107 I don't fully understand displacement, the problems where we had to solve for displacement included a range for t, 0<= t<= 3 or something like that.
Mentor
P: 21,284
 Quote by jrjack The position of the car is s = 71t - 20t2, t seconds after the driver applies the brakes.
According to this formula, at the moment the driver steps on the brakes (t = 0), s = 0.
You need to find a formula for the velocity as a function of t, and determine when the velocity reaches 0.
 P: 156 That function S describes the cars position as a function of time. You're interested in knowing when the car comes to a rest, hence you want to know how the cars velocity is changing in time. How can you figure out velocity if you know the cars position?
 P: 107 Th velocity would be v=71-20t. if v=0, then 0=71-20t, so t=3.55. The time when the cars velocity reaches 0 is 3.55 sec. But if I plug t=3.55 into my origonal equation, I get s=0 ???? Or if I set s=3.55, I get t=.0507, 3.499 ???
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P: 9,772
 Quote by jrjack Th velocity would be v=71-20t.
Not quite, be careful with your differentiation.
 P: 107 Nevermind, I see my problem.
 P: 107 Thanks, v=71-40t, set equal to 0, gives me t=1.775 or 1.8. Thanks for all your help.
 Math Emeritus Sci Advisor Thanks PF Gold P: 39,552 By the way, the reason the correct answer is exactly half of your answer is that with a constant acceleration, the average speed over a time interval is just the numeric average of the speed at the beginning and end of the interval. If the initial speed is v and the final speed is 0 (the car is stopped means the speed is 0, not the position!) the average speed is (v+ 0)/2= v/2.
P: 107
 Quote by HallsofIvy By the way, the reason the correct answer is exactly half of your answer is that with a constant acceleration, the average speed over a time interval is just the numeric average of the speed at the beginning and end of the interval. If the initial speed is v and the final speed is 0 (the car is stopped means the speed is 0, not the position!) the average speed is (v+ 0)/2= v/2.
That is kinda where I was getting confused, I could get the average speed, but this question wasn't asking for that. And with my bad differentiation I was getting the wrong answers no matter which way I tried to solve it.

Thanks for the explanation, this is an online class and there is very little explanation, I most use the Kahn Academy, You Tube, and of course the experts of PF.

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