Question

The position equation for a particle is s of t equals the square root of the quantity t cubed plus 1 where s is measured in feet and t is measured in seconds. Find the acceleration of the particle at 2 seconds.

1 ft/sec2
two thirds ft/sec2
negative 1 over 108 ft/sec2
None of these

Answer 1

$\bf s(t)=\sqrt{t^3+1} \\\\\\ \cfrac{ds}{dt}=\cfrac{1}{2}(t^3+1)^{-\frac{1}{2}}\cdot 3t^2\implies \boxed{\cfrac{ds}{dt}=\cfrac{3t^2}{2\sqrt{t^3+1}}}\leftarrow v(t) \\\\\\ \cfrac{d^2s}{dt^2}=\cfrac{6t(2\sqrt{t^3+1})-3t^2\left( \frac{3t^2}{\sqrt{t^3+1}} \right)}{(2\sqrt{t^3+1})^2}\implies \cfrac{d^2s}{dt^2}=\cfrac{ \frac{6t(2\sqrt{t^3+1})-1}{\sqrt{t^3+1}} }{4(t^3+1)}$

$\bf \cfrac{d^2s}{dt^2}=\cfrac{6t[2(t^3+1)]-1}{4(t^3+1)\sqrt{t^3+1}}\implies \boxed{\cfrac{d^2s}{dt^2}=\cfrac{12t^4+12t-1}{4t^3+4\sqrt{t^3+1}}}\leftarrow a(t)\\\\ -------------------------------\\\\a(2)=\cfrac{215~ft^2}{44~sec}$

Answer 2

Answer:

Acceleration of the particle is two thirds ft/s².

Step-by-step explanation:

It is given that, the position of the particle at time t is given by :

$s(t)=\sqrt{t^3+1}$

Where

s is in feet

t is in seconds

We need to find the acceleration of the particle at 2 seconds. Firstly calculating the velocity of the particle as :

$v=\dfrac{ds(t)}{dt}$

$v=\dfrac{d(\sqrt{t^3+1})}{dt}$    

This gives, $v=\dfrac{3t^2}{2\sqrt{t^3+1}}$

Since, $a=\dfrac{dv}{dt}$

$a=\dfrac{d(\dfrac{3t^2}{2\sqrt{t^3+1}})}{dt}$

$a=\dfrac{3t^4+12t}{4(t^3+1)^{3/2}}$

At t = 2 seconds,

$a=\dfrac{3(2)^4+12(2)}{4((2)^3+1)^{3/2}}$

$a=0.66\ ft/s^2$

or

$a=\dfrac{2}{3}\ ft/s^2$

So, the acceleration of the particle at 2 seconds is 2/3 ft/s². Hence, this is the required solution.