Question

What are the steps in finding the velocity and acceleration functions when x(t)=6

Then how do I find the velocity and acceleration at 2.5s?

Answer 1

x(t)=6 is your position function?
velocity is the derivative of position and acceleration is the derivative of velocity, so
v(t)=x'(t)=0
a(t)=v'(t)=x''(t)=0
so a(2.5)=0... how boring...
in other words the object in question is just sitting at point x=6, so it is neither accelerating or moving at all. If your position function depended on time t then it might be more interesting. Like if it was
x(t)=6t^3+3t+7
v(t)=18t^2+3
a(t)=36t
then a(2.5)=36(2.5)=90m/s^2
does that help?

Answer 2

yes! so that would mean if
x(t)=5t^2+17t then
v(t)=10t^4+17
a(t)=40t
so then a(2.5)=40(2.5)=100m/s^2

Answer 3

you got the derivatives a wrong, but you have the right idea
x(t)=5t^2+17t
v(t)=10t+17
a(t)=10 so in this case a(2.5)=10 (t doen't matter if it's not in the equation, right?)
remember differentiation: Der(x^n)=nx^n-1
don't know how you got t^4 for your velocity equation

Answer 4

5-1

Answer 5

but clearly that was wrong

Answer 6

subtract 1 from the EXPONENT when you take the derivative. Leave the coefficient alone.
Derivative of\[ax^n\]\[=anx ^{n-1}\]
I think you need a bit of practice with derivatives.
http://tutorial.math.lamar.edu/

Answer 7

i really do

Answer 8

no calculus yet?

Answer 9

nope in calc 2 but i travled last year and skipped pre-calc. so trying to catch up. also in pysics 2 which is going fast than my calc class

Answer 10

just calc not calc 2...opps

Answer 11

you'll have derivatives like those down in no time, but it sucks when the physics teachers aren't in sync with the math you're working on. keep at it. :)

Answer 12

thanks!