## onegirl Group Title Use the position function to find the velocity and acceleration at time t = 2. S(t) = sqrt(t^2 + 8). one year ago one year ago

1. onegirl Group Title

@zepdrix can u help?

2. zepdrix Group Title

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. $$\large s''(t)=v'(t)=a(t)$$.

3. zepdrix Group Title

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. $$\large v(\color{royalblue}{2})=\dfrac{\color{royalblue}{2}}{\sqrt{\color{royalblue}{2}^2+8}}$$

4. zepdrix Group Title

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.

5. onegirl Group Title

okay

6. onegirl Group Title

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

7. zepdrix Group Title

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 $\large v(t)=\dfrac{t}{\sqrt{t^2+8}}$

8. onegirl Group Title

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

9. zepdrix Group Title

yes c:

10. onegirl Group Title

thanks :)

11. onegirl Group Title

@mathsmind here it is

12. onegirl Group Title

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

13. mathsmind Group Title

so whats the problem?

14. onegirl Group Title

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

15. mathsmind Group Title

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

16. onegirl Group Title

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

17. onegirl Group Title

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

18. onegirl Group Title

are u there?

19. onegirl Group Title

@zepdrix

20. mathsmind Group Title

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

21. onegirl Group Title

ok

22. mathsmind Group Title

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

23. onegirl Group Title

so i should subtitute 2 into t?

24. mathsmind Group Title

yes

25. onegirl Group Title

okay

26. onegirl Group Title

i got 0

27. mathsmind Group Title

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

28. mathsmind Group Title

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

29. mathsmind Group Title

$m.s^{-1}$

30. onegirl Group Title

what does m.s -1 mean?

31. mathsmind Group Title

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

32. onegirl Group Title

okay

33. onegirl Group Title

so bow i have to solve the problem above?

34. mathsmind Group Title

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

35. onegirl Group Title

okay so i need to solve that

36. mathsmind Group Title

a(2)=0.192 m/s^2

37. onegirl Group Title

okay

38. mathsmind Group Title

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

39. onegirl Group Title

okay