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In my opinion, the graph for the first part of the motion will be a straight line passing thorugh the origin (at the beiginning of motion) and then it will be a falling parabolic curve ( $x$ is proportional $ -t^2$). The red part corresponds to the linear motion when the velocity is constant whereas the green part corresponds to the constant retardation $a=g \sin(\theta)$ experienced by the block. Is this correct?

enter image description here

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  • $\begingroup$ Is your incline up or down, thats important to know if the later part of the graph $\endgroup$
    – Shubham
    May 8 '16 at 9:40
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The straight line should be followed by a continuous, downward parabolic curve. There is a gradual change between the straight line and the parabola. Because while the block is on the plane, the x coordinate is still increasing with time but now at a decreasing rate because of the acceleration $gsin(\theta)$ along the plane, opposite to the velocity of the block. In your figure the green part shows a decrease in x with time after reaching a maximum point. Is this correct? Does 'x' really start decreasing when the block climbs up the incline? Or, is it the velocity?

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  • $\begingroup$ I wanted to draw a parabolic curve but I don't know how would it look like. $\endgroup$
    – nls
    May 8 '16 at 8:56
  • $\begingroup$ Your red curve which has correctly been drawn as a straight line would gradually change into the green parabola while x keeps increasing with time, eventually reaching a maximum point. Think a little more on it. You have almost got it. Hint: both curves are individually depicting the correct variation of x coordinate. You just have to change the position on one curve. $\endgroup$ May 8 '16 at 9:05
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The displacement in x direction will be a straight line all the time including the travel on the incline plane. There is no reason it will change its value.

The displacement in y direction will be zero on the horizontal plane and change in parabolic because of gravity.

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