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Mechanics help!

FFF6EE7C-6E8F-40C9-A9A4-AA5DFF016352.jpg.jpeg

I’ve had a go At this question and I thought I knew what I was doing. However, I am getting confused if the particle is accelerating or decelerating.
And my initial velocity comes out negative...?

Anyways would really appreciate someone’s method to do it, or tell me where I’ve gone wrong.

What I did:
Added the vectors
Formed triangle with new vector to find resultant force (hypotenuse)
Used equation F=ma
Then used suvat equation - v=u at, to find u (v=0, a= answer to part i), t=4)
(edited 6 years ago)
237FBF12-9647-4DC6-865F-497DB13438CD.jpg.jpeg36.5KB
Original post by littlebitthick
FFF6EE7C-6E8F-40C9-A9A4-AA5DFF016352.jpg.jpeg

I’ve had a go At this question and I thought I knew what I was doing. However, I am getting confused if the particle is accelerating or decelerating.
And my initial velocity comes out negative...?

Anyways would really appreciate someone’s method to do it, or tell me where I’ve gone wrong.

What I did:
Added the vectors
Formed triangle with new vector to find resultant force (hypotenuse)
Used equation F=ma
Then used suvat equation - v=u at, to find u (v=0, a= answer to part i), t=4)


You can't really talk about whether it's acceleration or decelerating because there are two separate components for the acceleration vector.
You should have it as a=(3,−12)\mathbf{a} = (3, -\frac{1}{2}) which means that as time goes on, it is accelerating along the i\mathbf{i} direction, but it's decelerating along the j\mathbf{j} direction.
Original post by RDKGames
You can't really talk about whether it's acceleration or decelerating because there are two separate components for the acceleration vector.
You should have it as a=(3,−12)\mathbf{a} = (3, -\frac{1}{2}) which means that as time goes on, it is accelerating along the i\mathbf{i} direction, but it's decelerating along the j\mathbf{j} direction.


I don’t think we’ve done that in class.. hmm

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