Question

Use the position function to find the velocity and acceleration at time t = 2. S(t) = sqrt(t^2 + 8).

Answer 1

@zepdrix can u help?

Answer 2

The position function is given by \(\large s(t)=\sqrt{t^2+8}\).

The derivative function of position is velocity. \(\large s'(t)=v(t)\).
The second dervative of position, which is the first derivative of velocity, is the acceleration function. \(\large s''(t)=v'(t)=a(t)\).

Answer 3

So I'll help you along with the first part, I'll bet you can figure out the acceleration based on this.

\(\large s'(t)=\dfrac{t}{\sqrt{t^2+8}}\)
This is our velocity function, so let's write it like this,
\(\large v(t)=\dfrac{t}{\sqrt{t^2+8}}\)

They want us to find the velocity at time t=2.
\(\large v(\color{royalblue}{2})=\dfrac{\color{royalblue}{2}}{\sqrt{\color{royalblue}{2}^2+8}}\)

Answer 4

Understand the process?
Now take the derivative of your velocity function to find your acceleration function. \(\large v'(t)=a(t)\) and then again, plug t=2 into it to find the acceleration at that time.

Answer 5

okay

Answer 6

so i'm taking the derivative of sqrt(t^2 + 8) right? then i plus in 2 ? @zepdrix

Answer 7

You/I took the derivative of sqrt(t^2 + 8) to get velocity.
Now take the derivative of velocity to get acceleration.

So you're now taking the derivative of \[\large v(t)=\dfrac{t}{\sqrt{t^2+8}}\]

Answer 8

okay so after i take the derivative of that i subtitute 2 in it right?

Answer 9

yes c:

Answer 10

thanks :)

Answer 11

@mathsmind here it is

Answer 12

so i found the derivative of t^2/sqrt(t^2 + 8) and i got 8/t^2 + 8)^3/2 so i substitute the 2 in the t and i got 8/(2^2 + 8)^3/2 which i'm trying to solve now

Answer 13

so whats the problem?

Answer 14

i need to solve 8/(2^2 + 8)^3/2 or should i just leave it like that ?

Answer 15

well either u work this out with ur head or use the calculator to get the final answer...

Answer 16

I did use the calculation but i'm not getting a right answer

Answer 17

@zepdrix i'm stuck on this one can u help?

Answer 18

are u there?

Answer 19

@zepdrix

Answer 20

\[s(t) = \sqrt{t^2 + 8}\]
\[v(t)=\frac{t}{\sqrt{t^2+8}}\]
\[a(t) = \frac{1}{(t^2+8)^{\frac{1}{2}}}-\frac{t^2}{(t^2+8)^{\frac{3}{2}}}\]

Answer 21

ok

Answer 22

now solve for v(2), and a(2) you'll get the answer

Answer 23

so i should subtitute 2 into t?

Answer 24

yes

Answer 25

okay

Answer 26

i got 0

Answer 27

\[v(2)=\frac{(2)}{\sqrt{(2)^2+8}}\]

Answer 28

\[v(2)=\frac{2}{2\sqrt{1+2}}=\frac{1}{\sqrt{3}}\]

Answer 29

\[m.s^{-1}\]

Answer 30

what does m.s -1 mean?

Answer 31

it means meter per a second, u have to put units

Answer 32

okay

Answer 33

so bow i have to solve the problem above?

Answer 34

\[a(2) = \frac{1}{((2)^2+8)^{\frac{1}{2}}}-\frac{(2)^2}{((2)^2+8)^{\frac{3}{2}}}\]

Answer 35

okay so i need to solve that

Answer 36

a(2)=0.192 m/s^2

Answer 37

okay

Answer 38

but u really need to know how to differentiate those simple equations ok

Answer 39

okay