Acceleration Vector from Position Functions
The position of a particle moving in the xy-plane is given by the functions $$x(t) = 3t^2$$ and $$y(t) = 4t - 5t^2$$. What is the direction of the particle’s acceleration vector at $$t = 2$$ s, measured counterclockwise from the positive $$x$$-axis?
A
Directly opposite to the velocity vector
B
Perpendicular to the velocity vector
C
At an angle of 301° measured counterclockwise from +x
D
Directly toward the origin
Question Leaderboard
| Rank | |||||
|---|---|---|---|---|---|
| #1 | dxie26808 | 2 | 2 | 0m 00s | 200 |
| #2 | skopjeajvar091 | 1 | 3 | 0m 46s | 34 |
| #3 | coolboy19521 | 0 | 1 | 0m 00s | -10 |
| #4 | aibekn386 | 1 | 2 | 2m 39s | -69 |
| #5 | kietngiuss14042021 | 0 | 1 | 1m 04s | -74 |
Items per page:
10
1 – 5 of 5