Answer to Question #274842 in Calculus for kong

Question

A particle is moving along a straight line according to the given equation of motion, where s ft is the directed distance of the particle from the origin at t sec. Find the time when the instantaneous acceleration is zero, and then find the directed distance of the particle from the

a. 𝒔 = (𝟏𝟐𝟓)/ (𝟏𝟔𝒕+𝟑𝟐) − 2/5𝒕𝟓 , t≥𝟎

b. 𝒔= 𝟗𝒕^𝟐 + 𝟐 √𝟐𝒕+𝟏, t≥𝟎

Answer 1

a)

$s=\frac{125}{16t+32}-\frac{2}{5t^5},t\ge0$

Acceleration =$\frac{d^2s}{dt^2}$

$\frac{ds}{dt}(s)=\frac{2}{t^6}-\frac{2000}{(16t+32)^2}$

$\frac{d^2s}{dt^2}(s)=\frac{64000}{(16t+32)^3}-\frac{12}{t^7}$

Time when Instantaneous acceleration is zero,

$\frac{64000}{(16t+32)^3}-\frac{12}{t^7}=0$

$64000t^7-12(16t+32)^3=0$

$64000t^7-12(4096t^3+24576t^2+49152t+32768)=0$

Solving to get $t\approx1.6839$

Time=1.68 seconds

$s=\frac{125}{16t+32}-\frac{2}{5t^5}$

$=\frac{125}{16(1.68)+32}-\frac{2}{5(1.68)^5}$

=0.055 ft

b)

$s=9t^2+2\sqrt{2t}+1,t\ge0$

$\frac{ds}{dt}=18t+\frac{\sqrt{2}}{\sqrt{t}}$

$\frac{d^2s}{dt^2}=18-\frac{1}{\sqrt{2}t^{\frac{3}{2}}}$

$18-\frac{1}{\sqrt{2}t^{\frac{3}{2}}}=0$

$18({\sqrt{2}t^{\frac{3}{2}}})-1=0$

$t^{\frac{3}{2}}=\frac{1}{25.46}$

$t=0.12seconds$

$s=9t^2+2\sqrt{2t}+1$

$=9(0.12)^2+2\sqrt{2(0.12)}+1$

=0.1296+1.9798

=2.1094 ft