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I'm new to Kinematics.

I have the velocity vector as v $ = (2i - 4j) \ m/s $ , and the acceleration vector as a $ = (-2i + 4j) \ m/s^2 $ .

Now I need to tell what type of motion this is in terms of a whether it is uniformly accelerated or non-uniformly accelerated. And the second parameter is - whether it is two dimensional motion or one dimensional motion.

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The coefficients of your unit vectors are just numbers, so your acceleration is constant -- meaning uniformly accelerated. In this case you will get two possible trajectories. The path is either a piece of a parabola or is straight-line motion with acceleration. The straight line case will occur if the velocity and acceleration are co-linear, and the parabolic case if they are not. Compute the cross-product of the velocity and the acceleration. If you get zero, then the vectors are co-linear.

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a = -(2i-4j) a = -v

This means acceleration is in opposite direction of velocity and independent of time so it is uniform retarding acceleration.

And as you can see a is defined by i and j ( unit vector for x and y axis respectively) hence the the motion takes place in x-y plane and therefore it is a 2-D motion.

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  • $\begingroup$ Answering homework questions may be well intentioned, I do appreciate that, but it is against the policy of this site. In this case, the OP has learnt less than they might have by following the links. $\endgroup$
    – user163104
    Commented Jul 23, 2017 at 5:25
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    $\begingroup$ @Countto10 ok thanks i will keep that in mind $\endgroup$
    – Rohan
    Commented Jul 23, 2017 at 5:28
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    $\begingroup$ Rohan, this is not 2D. If a motion is along only one path - in other words, of you could have described it with just one axis - then it is 1D. It just happens to be written in a 2D coordinate system. You could also have written it in a 3D coordinate system, such as $$\vec v=2i-4j+0k$$ And if you tilted the system and choice a different orientation, none of the three terms would be zero. A 1D motion in a 2D space is still a 1D motion. $\endgroup$
    – Steeven
    Commented Jul 23, 2017 at 8:00
  • $\begingroup$ This may seem like nitpicking, but it's not: $a \neq -v$ because they describe different physical quantities. Always treat the physical units like they were algebra symbols of their own. $\endgroup$
    – Apollonius
    Commented Aug 6, 2017 at 8:14
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  • Is it uniform acceleration?
    Is the acceleration of constant magnitude?
    Is the direction of the acceleration constant?
    If the answer to both questions is "Yes" then the acceleration is uniform.

  • Is the motion in one dimension or in two dimensions?
    Can the velocity $\vec v$ and the acceleration $\vec a$ be written in terms of the same (unit) vector $\vec w$ such that $\vec v = v_{\rm w} \vec w$ and $\vec a = a_{\rm w} \vec w$ where $v_{\rm w}$ and $a_{\rm w}$ are the components of the velocity and acceleration of the vector $\vec w$?
    If the answer is "Yes" then the motion is one dimensional.

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