Acceleration Vector from Position Functions

Question

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?

Accepted Answer

At an angle of 301° measured counterclockwise from +x

Answer

Directly opposite to the velocity vector

Answer

Perpendicular to the velocity vector

Answer

Directly toward the origin

Rank	User	Correct Count	Attempt Count	Total Time	Score
#1	dxie26808	2	2	0m 00s	200
#2	skopjeajvar091	1	3	0m 46s	34
#3	nikhilmahankali56	1	1	1m 40s	0
#4	coolboy19521	0	1	0m 00s	-10
#5	129.keilana	0	1	0m 46s	-56
#6	aibekn386	1	2	2m 39s	-69
#7	kietngiuss14042021	0	1	1m 04s	-74

Acceleration Vector from Position Functions

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