Find initial velocity with accelration as a funt. of t

AI Thread Summary
The particle's acceleration is defined as a_{x}(t) = -1.94 m/s² + (3.05 m/s³)t. To find the initial velocity v_{0x} for the particle to have the same x-coordinate at t = 4.00 s as at t = 0, the acceleration must be integrated twice to derive the position function x(t). The initial attempt yielded an incorrect velocity of 16.64 m/s, but after integrating correctly, the final result for v_{0x} was determined to be -4.25 m/s. This solution aligns with the requirement that x(4 sec) equals x(0). The integration process is crucial for accurately solving the problem.
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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