Calculating Acceleration at t = 2.20 s for a Particle in 2D Motion

[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.

 

[lostinphys]

... ??

 

[StNowhere]

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.

 

[Pengwuino]

Well the first part, 6.2i + 8.1j is correct. You divide by 2.20 seconds but it looks more like (6.2/2.20)i + (6.10/2.20)j with m/s^2 being the units for the vector. You can treat them separately and think of it like 6.2/2.20 as the dv/dt in the i direction for example.

 

[lostinphys]

The vector representing acceleration would then be (6.2/t)i + (8.1/t)j ?

 

[StNowhere]

That sounds right.

 

[lostinphys]

thanks !