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Why is the answer half of my answer?by jrjack
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#1
Oct1011, 09:05 AM

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  20t^{2}, 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=71t20t^{2} 0=71t20t^{2} 0=7120t 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? 


#3
Oct1011, 09:35 AM

P: 107

Well, because the position of the car is represented by the function s=71t20t^{2}, and since the driver is applying the brakes and coming to a stop, his postion when stopped should be s=0.



#4
Oct1011, 09:37 AM

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Why is the answer half of my answer?



#5
Oct1011, 09:43 AM

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.



#6
Oct1011, 09:53 AM

Mentor
P: 21,215

You need to find a formula for the velocity as a function of t, and determine when the velocity reaches 0. 


#7
Oct1011, 09:54 AM

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? 


#8
Oct1011, 10:02 AM

P: 107

Th velocity would be v=7120t.
if v=0, then 0=7120t, 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 ??? 


#9
Oct1011, 10:03 AM

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#10
Oct1011, 10:03 AM

P: 107

Nevermind, I see my problem.



#11
Oct1011, 10:07 AM

P: 107

Thanks, v=7140t, set equal to 0, gives me t=1.775 or 1.8.
Thanks for all your help. 


#12
Oct1011, 01:50 PM

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P: 39,339

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. 


#13
Oct1011, 02:55 PM

P: 107

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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