Why is the answer half of my answer?

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Homework Help Overview

The problem involves a car's motion described by the position function s = 71t - 20t², where t is the time in seconds after the brakes are applied. The original poster is trying to determine the time it takes for the car to come to a stop, noting a discrepancy between their calculated time and the expected answer.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to solve for the time when the car stops by setting the position function to zero. Some participants question the interpretation of displacement and the conditions under which the car is considered to have stopped.

Discussion Status

Participants are exploring different interpretations of the problem, particularly regarding the relationship between position and velocity. Some guidance has been offered on how to derive the velocity from the position function, and there is acknowledgment of confusion around the calculations and concepts involved.

Contextual Notes

There is a discussion about the implications of constant acceleration and how it affects average speed calculations. The original poster expresses a lack of understanding regarding displacement and the conditions for stopping, which may be influencing their approach to the problem.

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Homework Statement


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.

Homework Equations





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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What made you write this?
jrjack said:
0=71t-20t2
 
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.
 
jrjack said:
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.
 
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.
 
jrjack said:
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.
 
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?
 
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 ?
 
jrjack said:
Th velocity would be v=71-20t.
Not quite, be careful with your differentiation.
 
  • #10
Nevermind, I see my problem.
 
  • #11
Thanks, v=71-40t, set equal to 0, gives me t=1.775 or 1.8.

Thanks for all your help.
 
  • #12
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
HallsofIvy said:
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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