expressing acceleration and distance in dependance of time

buk1337

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Hello, i am studying for my math exam and i have problems solving this task, i would be very thankful if anyone could help me. (excuse me if im not writing correctly, english is not my native language).

Task is: express acceleration and distance in dependance of time (alpha and beta are positive constant).

v(t)= alpha * 3sqrt(t) + beta * t^2 (at 3sqrt(t) here is not regular sqrt but sqrt is on 3,i mean 3rd root of t)

i start :

v = ds / dt
ds = v * dt // integrate
integrate ds = integrate v * dt
integrate ds = integrate ( alpha * t^1/3 + beta * t^2) * dt
s integrated = alpha * t^4/3 / 4/3 + beta * t^3 / 3 + c integrated

I think this is the expression for distance s, i dont know how to express acceleration further, if anyone can help im very thankful.
 
Hello, i am studying for my math exam and i have problems solving this task, i would be very thankful if anyone could help me. (excuse me if im not writing correctly, english is not my native language).

Task is: express acceleration and distance in dependance of time (alpha and beta are positive constant).

v(t)= alpha * 3sqrt(t) + beta * t^2 (at 3sqrt(t) here is not regular sqrt but sqrt is on 3,i mean 3rd root of t)

i start :

v = ds / dt
ds = v * dt // integrate
integrate ds = integrate v * dt
integrate ds = integrate ( alpha * t^1/3 + beta * t^2) * dt
s integrated = alpha * t^4/3 / 4/3 + beta * t^3 / 3 + c integrated

I think this is the expression for distance s, i dont know how to express acceleration further, if anyone can help im very thankful.

hint: a = dv/dt
 
hey ty for the hint and fast reply.

so if i understand correctly i diffirentiate the equation?

a = dv/dt
a = (alpha * t^1/3 + beta * t^2) / dt
a = alpha * 1/3 * t^-2/3 + beta * 2t
this is it?

something is telling me im not doing this right? help. What about the first part the distance s is that correct?
 
[Hello,] [thank you] for the hint and fast reply.

[So,] if understand correctly[, I] [differentiate] the equation?

Which "equation"? Do you perhaps mean to ask if you should differentiate the velocity function in order to determine the acceleration function?

And, if so, does the following:

a = dv/dt
a = (alpha * t^1/3 + beta * t^2) / dt
a = alpha * 1/3 * t^-2/3 + beta * 2t
...actually mean something like the below?

. . .\(\displaystyle \mbox{Given that }\, a\, =\, \dfrac{dv}{dt}\, \mbox{ and that }\, v(t)\, =\, \alpha\, \sqrt[3]{\strut t\,}\, + \, \beta\, t^2,\, \mbox{ then:}\)

. . . . .\(\displaystyle a(t)\, =\, v'(t)\, =\, \dfrac{dv}{dt}\, =\, \dfrac{d\left(\,\alpha\,\sqrt[3]{\strut t\,}\, + \, \beta\, t^2 \,\right)}{dt}\)

. . . . . . .=(13)αt23+2βt,\displaystyle =\, \left(\dfrac{1}{3}\right)\, \alpha\, t^{-\frac{2}{3}}\, +\, 2\, \beta\, t,

. . .\(\displaystyle \mbox{so }\, a(t)\, =\, \dfrac{\alpha}{3\, \sqrt[3]{\strut t^2\,}}\, +\, 2\, \beta\, t\)


something is telling me im not doing this right?
What is the "something"? Please clarify. Did you plug in the given value for t, and end up with a value for a(t) which did not match the answer you'd been given (maybe in the back of the book)? Thank you! ;)
 
ty for reply stapel. Both answers to your questions are yes, i meant do i differentiate "base" equation v(t)= alpha * 3sqrt(t) + beta * t^2 to get the acceleration function, ok i understand this now, but i wasnt sure if i differentiate correctly it was just a hinch that was telling me that im doing this wrong, and i dont have the results for this task so i dont know the correct result. And second question yes that differentiation what i wrote is excatly what u wrote whit that u went 1 line/step further. So this is the answer right...

I just dont understand sometimes when do i have to integrate and when differentiate, which is pretty sad i guess, anyways ty for the answer
 
ty for reply stapel. Both answers to your questions are yes, i meant do i differentiate "base" equation v(t)= alpha * 3sqrt(t) + beta * t^2 to get the acceleration function, ok i understand this now, but i wasnt sure if i differentiate correctly it was just a hinch that was telling me that im doing this wrong, and i dont have the results for this task so i dont know the correct result. And second question yes that differentiation what i wrote is excatly what u wrote whit that u went 1 line/step further. So this is the answer right...

I just dont understand sometimes when do i have to integrate and when differentiate, which is pretty sad i guess, anyways ty for the answer

if i might,

You differentiate if you have velocity and need acceleration. Integrate when you have acceleration and need velocity. In kindergarden terms: if you differentiate SOMETHING you get Something1. to get SOMETHING then you integrate Something1. Try thinking of it like this: Integrating is reversing a differentiation.

hope that makes sense and helps you
 
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