A particle starts from the origin with velocity 2i m/s at t = 0 and moves in the xy plane with a varying acceleration given by 2 (sq root t) where is in meters per second squared and t is in seconds.

(a) Determine the velocity of the particle as a function of time.

So I calculated and got
v(t) = 2i + (4 (sq root t^3)/3) j m/s

(b) Determine the position of the particle as a function of time.
x(t) = ? m

now for part b I got x(t)=2ti + 8 (sq root t^5)j but it says this answer is incorrect. I'm not sure what I'm doing wrong? I thought I did it correctly???

To determine the position of the particle as a function of time, you need to integrate the velocity function with respect to time. Let's go through the process step by step.

Given: Velocity function v(t) = 2i + (4√(t^3)/3)j m/s

To find the position function x(t), integrate the x-component of the velocity function with respect to t:

∫(2i) dt = 2ti + C

where C is the constant of integration.

Next, integrate the y-component of the velocity function with respect to t:

∫((4√(t^3)/3)j) dt

To integrate this, let's make a substitution:
Let u = t^(3/2), then du = (3/2)t^(1/2)dt

∫((4√(t^3)/3)j) dt = (4/3) ∫(u) j du = (4/3)(uj) + C

Now, substitute back u = t^(3/2):

(4/3)(t^(3/2)j) + C

Therefore, the position function x(t) = 2ti + (4/3)(t^(3/2)j) + C

It seems like you missed the 4/3 coefficient in front of the second term. Make sure to double-check your calculations.