Question

The function ​s(x)equals=startfraction 3600 over 60 plus x endfraction equals 3600 left parenthesis 60 plus x right parenthesis superscript negative 1 3600 60+x=3600(60+x)−1 gives a​ person's average speed in miles per hour if he or she travels one mile in x seconds more or less than 60 seconds. use a linear approximation to s at 0 to find a​ person's approximate average speed if he or she travels one mile in 5656 seconds. what is his or her exact​ speed?

Answer 1

Answer: The linear approximation, about x= 0, is as you said, s(x)= s(0)+ s'(0)(x- 0)= 60- x. From that, what is s(51)? These are some things that I have been trying: x = 51 a = 0 s(0) = 60 s'(0) = -0.292184076 How in the world did you get this? This appears to be your main error. You said, above, that s'(t)= \frac{-3600}{(60+ x)^2}, Setting x= 0 in that gives, as I said before, -1. s(0) + s'(0)(51 - 0) = 60 - 14.90138787 = 45.098 ... And, x = 0 a = 51 s(0) = 60 s'(0) = -0.292184076 s(0) + s'(0)(0 - 51) = 60 + 14.90138788 = 74.901 ... It's probably something to do with how I am substituting the values but I haven't been able to arrange them in a way that yields anything close to the answer. I did find something that gave me an answer of 9 when I was playing around with the x - a stuff but I cannot remember what it was that I did. Maybe I need to add the answer of 9 that I found to the s(a) = 60 in order to get 69?

Answer 2

Solution:

$S(x)=\frac{3600}{60+x}-1\\\\ S'(x)= 3600\times\frac{-1}{(60+x)^2}$

If , y = $\frac{1}{x}, {\text{then y' means it's derivative}= \frac{-1}{x^2}$

As average speed = Rate of change of distance

for s=0 , 3600=60+x

x= 3540 seconds

for s= 1 mile, x= 5656 seconds

As , average speed =  $\frac{S_{2}-S_{1}}{t_{2}-t_{1}}=\frac{1-0}{5656-3540}=\frac{1}{2116}$m/sec=0.0004725 m/sec