The position equation for a particle is s of t equals the square root of the quantity 2 times t plus 1 where s is measured in feet and t is measured in seconds. Find the acceleration of the particle at 4 seconds?

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The position equation for a particle is s of t equals the square root of the quantity 2 times t plus 1 where s is measured in feet and t is measured in seconds. Find the acceleration of the particle at 4 seconds?

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Answer 1 · 2018-01-23T05:49:01

-1/27 ft/s^2

Explanation:

The position function is $s (t) = \sqrt{2 t + 1}$ $s(t)=sqrt(2t+1)$ .
Find the first derivative to find the velocity function: $v (t) = \frac{1}{\sqrt{2 t + 1}}$ $v(t)=1/sqrt(2t+1)$ .
The second derivative will result in the acceleration function:
$a (t) = - \frac{1}{{(2 t + 1)}^{\frac{3}{2}}}$ $a(t)=-1/(2t+1)^(3/2)$ .
Therefore, $a (4) = - \frac{1}{27}$ $a(4)=-1/27$ .

Answer 2 · 2018-01-23T06:04:23

Please see below.

Explanation:

.

It is not clear what your position equation is from the wording of your problem. If you meant:

$s (t) = \sqrt{2} (t + 1)$ $s(t)=sqrt2(t+1)$

The velocity equation is the derivative of the position equation:

$v (t) = \sqrt{2}$ $v(t)=sqrt2$ feet per second

The acceleration equation would be the derivative of the velocity equation:

$a (t) = 0$ $a(t)=0$ , i.e. no acceleration (constant speed)

If you meant:

$s (t) = \sqrt{2 t} + 1 = \sqrt{2} {(t)}^{\frac{1}{2}} + 1$ $s(t)=sqrt(2t)+1=sqrt2(t)^(1/2)+1$

$v (t) = \frac{\sqrt{2}}{2} {(t)}^{- \frac{1}{2}} = \frac{\sqrt{2}}{2 t^{\frac{1}{2}}} = \frac{\sqrt{2}}{2 \sqrt{t}} = \frac{\sqrt{2 t}}{2 t}$ $v(t)=sqrt2/2(t)^(-1/2)=sqrt2/(2t^(1/2))=sqrt2/(2sqrtt)=sqrt(2t)/(2t)$ feet per second

$a (t) = (- \frac{1}{2}) (\frac{\sqrt{2}}{2}) {(t)}^{- \frac{3}{2}} = \frac{\sqrt{2}}{4 t^{\frac{3}{2}}} = \frac{\sqrt{2}}{4 t \sqrt{t}} = \frac{\sqrt{2 t}}{4 t^{2}}$ $a(t)=(-1/2)(sqrt2/2)(t)^(-3/2)=sqrt2/(4t^(3/2))=sqrt2/(4tsqrtt)=sqrt(2t)/(4t^2)$ feet per $sec o n d^{2}$ $second^2$

$a (4) = \frac{\sqrt{2 (4)}}{4 {(4)}^{2}} = \frac{\sqrt{2}}{32}$ $a(4)=sqrt(2(4))/(4(4)^2)=sqrt2/32$ feet per $sec o n d^{2}$ $second^2$

If you meant:

$s (t) = \sqrt{2 (t + 1)} = \sqrt{2} {(t + 1)}^{\frac{1}{2}}$ $s(t)=sqrt(2(t+1))=sqrt2(t+1)^(1/2)$

$v (t) = \frac{\sqrt{2}}{2} {(t + 1)}^{- \frac{1}{2}} = \frac{\sqrt{2}}{2 \sqrt{(t + 1)}} = \frac{\sqrt{2 (t + 1)}}{2 (t + 1)}$ $v(t)=sqrt2/2(t+1)^(-1/2)=sqrt2/(2sqrt((t+1)))=sqrt(2(t+1))/(2(t+1))$ feet per second

$a (t) = (- \frac{1}{2}) (\frac{\sqrt{2}}{2}) {(t + 1)}^{- \frac{3}{2}} = - \frac{\sqrt{2}}{4 (t + 1) \sqrt{t + 1}} = - \frac{\sqrt{2 (t + 1)}}{4 {(t + 1)}^{2}}$ $a(t)=(-1/2)(sqrt2/2)(t+1)^(-3/2)=-sqrt2/(4(t+1)sqrt(t+1))=-sqrt(2(t+1))/(4(t+1)^2)$ feet per $sec o n d^{2}$ $second^2$

$a (4) = - \frac{\sqrt{2 (4 + 1)}}{4 {(4 + 1)}^{2}} = - \frac{\sqrt{10}}{100}$ $a(4)=-sqrt(2(4+1))/(4(4+1)^2)=-sqrt10/100$ feet per $sec o n d^{2}$ $second^2$

If you meant

$s (t) = \sqrt{2} t + 1$ $s(t)=sqrt2t+1$

$v (t) = \sqrt{2}$ $v(t)=sqrt2$ feet per second

$a (t) = 0$ $a(t)=0$

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The position equation for a particle is s of t equals the square root of the quantity 2 times t plus 1 where s is measured in feet and t is measured in seconds. Find the acceleration of the particle at 4 seconds?

2 Answers

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