Question

The position equation for a particle is s(t)= sqrt{t ^{3}+1} where s is measured in feet and t is measured in seconds. Find the acceleration of the particle at 2 seconds

the correct answer is 2/3 ft/sec2 but i do not know how you get to this answer

Answer 1

@3mar

Answer 2

recall for position s, velocity v, and acceleration a, at a time t...

s(t)
\[v(t)=s'(t)\]
\[a(t)=v'(t)=s''(t)\]

Answer 3

instantaneous velocity is derivative of position, and instantaneous acceleration is derivative of velocity

Answer 4

Oh so just derive it twice and plug in 2? @DanJS

Answer 5

Right?

Answer 6

yes, to get the acceleration function, you take second derivative of position..

Answer 7

but how do you derive with the square root

Answer 8

\[\large \sqrt{u}=u^{1/2}\]

\[\large s(t)=(t^3+1)^{\frac{ 1 }{ 2 }}\]

Answer 9

chain rule
and
\[\sqrt{x} = x^{1/2}\]
\[\frac{ d}{dx} x^n = n x^{n-1}\]

Answer 10

use the chain rule, for a quantity u,
\[\large [u]^a = a*[u^{a-1}]*u'\]

Answer 11

so 1/2(t^3+1^-.5)(2t^2 ??

Answer 12

@3mar could you check if i derived that correctly? for the first derivative?

Answer 13

\[\large v(t)=s'(t)=\frac{ 3t^2 }{ 2\sqrt{t^3+1} }\]

Answer 14

Oh, thats what i meant. So then how would you do the second derivative

Answer 15

because we have another square root on the bottom

Answer 16

@3mar

Answer 17

the quotient rule works..

\[\frac{ d }{ dt }[\frac{ u }{ v }]=\frac{ v*u'-u*v' }{ v^2 }\]

Answer 18

ok thank you