Question

Show the qualitatively correct position versus time graph and the matching velocity versus time graph for an object that passes through the point X0 = -1 m at t0 = 0 s, and speeds up in the negative X-direction until it reaches a constant speed at tA. Between tA and tB, the object maintains this speed. At tB the object slows down and makes a momentary stop at tC, and immediately resumes the motion in the opposite direction (show a small fragment of the graphs after tC.

Answer 1

Important notes for plotting :-

1) if acceleration is constant and not equal to zero then the graph of x-t graph will be a quadratic curve because

s = ut + 1/2at², Now for the graph to know whether it will go upward or downward you can basically see whether if it is along or opposite to the direction of velocity, which from the first part you can see that it is in the direction of velocity towards negative x-direction. Also for the bump or concavity of the curve you can check the value of a, if it is negative then the concavity will be downwards else upwards. Now from the question as it is increasing the velocity in negative direction this means that the acceleration is negative w.r.t the x-direction. So the concavity will be downwards.

2) If acceleration is zero then the x-t graph will be a straight line with slope equal to the velocity of the initial point or the slope of the final point of the last curve.

Using this the plot may look like-