Calculating Velocity w/ Acceleration & Initial Velocity

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farukh
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1. Given acceleration of an object is given by

a(t) = 6t

and the intial velocity is 9 m/s





2. the general form we learned is

v(t) = -at + v0




3. so i guess

v(t) = -6t^2 + 9

is that right ? can i substitute 6t for a ... or a(t) doesn't equal a
 
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What are you trying to achieve? There is no question.
 
In order to obtain a formula for velocity, you would need to integrate the expression for acceleration.
 
ok basicaly my question is does

a(t) = a

?

becuase the equation is v(t) = -at + v0
and if a(t)=6t then is this right v(t) = -6tt + v0
 
Well, firstly the equation is v=v0+at. Note that this equation only holds for constant acceleration. Here, your acceleration is 6t (i.e. not constant), so in order to derive the equation for velocity, you must integrate the expression for acceleration, using the initial condition v0=9

So, no, your equation is not right!
 
ok i just figured it out how to do the solutions with integration

v(t) = integral 6t
v(t) = 3t^2 + c
9 = 3(0)^2 + c
9 = c

v(t) = 3t^2 + 9

but is there any way to do this without using integration because we haven't learned integration yet..
 
Last edited:
Your equation is still wrong. Acceleration is 6t, so you want to integrate this. Your method is correct, you just need to use the correct expression for a.

I can't think of a way to find the velocity without using integration, but I may be overlooking something!
 
:) now i think its ok
 
Are you asking what is the acceleration if the initial velocity is 9m/s?
 
  • #10
farukh said:
:) now i think its ok

Yup, that looks fine now.

Kurdt said:
Are you asking what is the acceleration if the initial velocity is 9m/s?

No, he wanted a general formula for the velocity, given that the acceleration is a=6t, and subject to the initial condition v0=9
 
  • #11
How to do this problem without integration? Hmm I don't see how that would work...actually I vaugely remember some really fat long painful way that even takes into account the Earth changing gravity at different altitudes, but even that need differentiation.
 
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