Find initial velocity with accelration as a funt. of t

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SUMMARY

The discussion focuses on calculating the initial velocity \( v_{0x} \) of a particle given its acceleration function \( a_{x}(t) = -1.94 \, \text{m/s}^2 + (3.05 \, \text{m/s}^3)t \). The correct approach involves integrating the acceleration function twice to derive the position function \( x(t) \) and applying the condition that \( x(4 \, \text{s}) = x(0) \). The final result for the initial velocity is determined to be \( -4.25 \, \text{m/s} \).

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mnafetsc
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Homework Statement



The acceleration of a particle is given by a_{x}(t)= - 1.94 m/s2 +( 3.05 m/s3 )t.

Find the initial velocity v_{0x} such that the particle will have the same x-coordinate at time t= 4.00 s as it had at t=0.

Homework Equations



vx= v0x + the integral ax dt

The Attempt at a Solution



What I did was set velocity to 0 moved initial velocity over then integrated acceleration to give me this:

-v0x= (-1.94 m/s2)t + ((3.05 m/s3)t2)/2

I then plugged in t=4 and t=0 into the equation which gave me 16.64 -wrong

I feel I am close, I just don't know how else to approach this problem.
 
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Hello mnafetsc,

Welcome to Physics Forums!

I'll give you a hint. You need to integrate twice. Ask yourself this. If you know what a(t) is, how do get an expression for x(t)? After you know that, then consider that the problem statement tells your that x(4 sec) = x0. :wink: You should be able to take it from there.
 
Last edited:
Perfect, integrated twice set x to 0 gives me -4.25 m/s

Thanks a bunch
 

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