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 find the position of the particle as a function of time, you need to integrate the velocity function with respect to time. Let's go step by step.
Given:
v(t) = 2i + (4 sqrt(t^3))/3 j m/s
To find x(t), integrate the x-component and y-component of the velocity separately with respect to time:
∫(2) dt = 2t (integration with respect to t yields t)
∫((4 sqrt(t^3))/3) dt = (4/3) * (2/5) * t^(5/2) = (8/15) * t^(5/2) (integration with respect to t yields t^(5/2))
Therefore, we have:
x(t) = 2t i + (8/15) * t^(5/2) j m
This should give you the correct position-time function.