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Question

Consider the following x-t graph to be parabolic. Draw the velocity-time graph and acceleration-time graph and analyze the motion of the particle regarding its velocity and acceleration.
992211_5f1d2c3f3de241a1894524a87d02bcf3.png

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Solution

Velocity-time graph and acceleration-time graph are as shown.

Analysis of motion

a. Between points A and B: Between these points, x has negative value; it means the particle is to the left of the origin. But the value of x is decreasing, so the particle is moving towards the origin. It means the velocity of the particle is positive. Between these points, the slope of the x-t graph is increasing; it means the magnitude of velocity is increasing. It means the velocity and acceleration both are in the same direction. So acceleration is positive. Note that acceleration is constant between A and B.
Also, the graph is concave up, so the acceleration is positive.

b.At point B: Particle is at origin at this point. At this point, the slope of the x-t graph is
maximum, so velocity is maxi-
mum at this point. Before point B, acceleration is positive, but after point B acceleration will be negative as the slope of the x-t graph will start decreasing after this fig.
We cannot define acceleration at point B.
c.Between points B and C: Between these points, x has a positive value; it means the particle is to the right of the origin. But the value of x is increasing, so the particle is moving away from the origin. It means the velocity of the particle is positive.

Between these points, the slope of the x-t graph is decreasing; it means the magnitude of velocity is decreasing. It means the velocity and acceleration are in opposite directions. So the acceleration is negative. Note that acceleration is constant between B and C.
Also, the graph is concave down, so acceleration is negative.
d.At point C: Particle is maximum away from the origin at this point. At this point, the slope of the x-t graph is zero, so velocity is zero. But acceleration is negative as the graph is concave down. At this point, the particle will change its direction of motion.
So after this, the particle will start moving towards the origin.
e.Between points C and D: Between these points, x has a positive value; it means the particle is to the right of the origin. But the value of x is decreasing, so the particle is moving towards the origin. It means the velocity of the particle is negative.
Between these points, the magnitude of the slope of the x-r graph is increasing; it means the magnitude of velocity is increasing but in the negative direction. So acceleration is negative. Note that acceleration is constant between C and D.
Also, the graph is concave down, so acceleration is negative.
f.At point D: Particle is to the right of the origin at this point. At this point, the slope of the x-t graph is negative maximum, so velocity is maximum at this point in the negative direction. Before point D, acceleration is negative, but after point D, the acceleration will be positive (as the graph is concave up after D). We cannot define acceleration at point D.
g. Between points D and E: Between these points, x has a positive value; it means the particle is to .the right of the origin. But the value of x is decreasing, so the particle is moving towards the origin. It means the velocity of the particle is negative.

Between these points, the magnitude of the slope of the x-t graph is decreasing; it means the magnitude of velocity is decreasing but in the negative direction. So, acceleration is positive. Note that acceleration is constant between D and E.
Also, the graph is concave up, so acceleration is positive.

1029333_992211_ans_f20d0f30d9f845c3bd8d38238534ad8e.png

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