# Why is the answer half of my answer?

1. Oct 10, 2011

### jrjack

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?

2. Oct 10, 2011

### Hootenanny

Staff Emeritus
What made you write this?

3. Oct 10, 2011

### 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.

4. Oct 10, 2011

### Hootenanny

Staff Emeritus
Just because the car is stopped doesn't mean the displacement (s) is zero.

5. Oct 10, 2011

### jrjack

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. Oct 10, 2011

### Staff: Mentor

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.

7. Oct 10, 2011

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. Oct 10, 2011

### jrjack

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 ???

9. Oct 10, 2011

### Hootenanny

Staff Emeritus
Not quite, be careful with your differentiation.

10. Oct 10, 2011

### jrjack

Nevermind, I see my problem.

11. Oct 10, 2011

### jrjack

Thanks, v=71-40t, set equal to 0, gives me t=1.775 or 1.8.

Thanks for all your help.

12. Oct 10, 2011

### HallsofIvy

Staff Emeritus
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. Oct 10, 2011

### jrjack

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.