Question

If s^2=at^2 + 2bt+c, prove that the accleration varies as 1/s^3 .

Answer 1

http://www.pinkmonkey.com/studyguides/subjects/calc/chap5/c0505d02.asp

Answer 2

s here is displacement ...
and acceleration is the second derivative of displacement... so differntiate twice to find\[\large{a=\frac{d^2s}{dt^2}}\]do u know how to do it or u want me to show u?

Answer 3

mam please show how to do it.

Answer 4

Lana you can

Answer 5

solve it and show

Answer 6

differentiate implicitly
first time
\[2s \frac{ds}{dt}=2at+2b\] \[s \frac{ds}{dt}=at+b\]\[ss'=at+b\]\[\huge{ss''+s's'=a}\]\[\huge{ss''+(s')^2=a}\]
u know that \[\huge{s'=\frac{at-b}{s}}\]
\[\huge{ss''=a-(\frac{at-b}{s})^2}\]\[\huge{ss''=\frac{as^2-a^2t^2-2abt+b^2}{s^2}}\]and u know that \[\large{s^2=at^2+bt+c}\]so \[\huge{s''=\frac{a(at^2+bt+c)-a^2t^2-2abt+b^2}{s^3}}\]\[\large{s''=\frac{a^2t^2+abt+ac-a^2t^2-2abt+b^2}{s^3}}\]\[\huge{s''=\frac{ac-b^2}{s^3}}\]
ac-b^2 is a constant so
acceleratiom varies inversely with s^3\[acceleration= \frac{K(\text{any constant})}{s^3}\]

Answer 7

Thankyou for simplifying the question for me :)