# If a block moves on a straight line (no friction) for time $T$ and then along an incline plane, what will the displacement vs. time graph look like?

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?

• Is your incline up or down, thats important to know if the later part of the graph May 8 '16 at 9:40

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?