At t = 0, a particle moving in the xy plane with constant acceleration has a velocity of vi = (3.00 i - 2.00 j) m/s and is at the origin. At t = 2.20 s, the particle's velocity is v = (9.20 i + 6.10 j) m/s.

i am asked to find the acceleration at time t.
i know that the acceleration is the derivative of the velocity, meaning the change in velocity divided by the time. so i took the difference between the vectors (v - vi) and got (6.2i + 8.1j)/(2.20 - 0). But i don't think it is expressed correctly in vectors. any help would be appreciate. thanks.
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 Quote by lostinphys At t = 0, a particle moving in the xy plane with constant acceleration has a velocity of vi = (3.00 i - 2.00 j) m/s and is at the origin. At t = 2.20 s, the particle's velocity is v = (9.20 i + 6.10 j) m/s. i am asked to find the acceleration at time t. i know that the acceleration is the derivative of the velocity, meaning the change in velocity divided by the time. so i took the difference between the vectors (v - vi) and got (6.2i + 8.1j)/(2.20 - 0). But i don't think it is expressed correctly in vectors. any help would be appreciate. thanks.

I think you're on the right track, but it looks like you're trying to use time as a vector here. remember, it divides both components of your $$\Delta v$$ vector.

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