10
Question
Consider the following vx=t graph to be parabolic. Plot the acceleration-time graph and analyze the motion of the particle from A to E.
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Solution
Acceleration-time graph is as follows:
We know that the slope of the velocity-time graph is equal to acceleration. The slope at A is maximum, decreases a, linearly, and becomes zero at C and then starts A increasing in the negative side and becomes maximum negative at E.
So at A, acceleration is maximum, at C zero, and at Is, acceleration is negative maximum.
Acceleration is positive and decreasing in magnitude. Also, the magnitude of velocity is decreasing, because velocity and acceleration are in opposite direction.
b.At point B: The velocity is instantaneously zero at this moment So the particle is at rest momentarily. But the particle is still accelerating as the slope has a non-zero value. Acceleration is positive and decreasing in magnitude. From here the particle will start moving along the positive x-direction.
c.Between points B and C: Velocity is positive and increasing. So the acceleration is positive but decreasing in magnitude.
d.At point C: Velocity at point C is maximum as the graph peaks at this point. At this point slope is zero, therefore the acceleration is zero.
e.Between points, C and D: Velocity goes on decreasing between C and D. Due to negative slope acceleration is also negative. Acceleration is increasing in magnitude. Velocity and acceleration are in opposite directions.
f.At point D: Velocity is zero at this point But the slope is still negative, so the acceleration is negative. The particle will start moving in the. negative direction. Acceleration is increasing in magnitude.
g.Between points D and E: Particle is moving in the negative direction. The slope of the graph is negative in this region. It means the acceleration is negative, but the magnitude of the slope is increasing, it means the magnitude of acceleration is increasing. Velocity and acceleration are in the same direction.
We know that the slope of the velocity-time graph is equal to acceleration. The slope at A is maximum, decreases a, linearly, and becomes zero at C and then starts A increasing in the negative side and becomes maximum negative at E.
So at A, acceleration is maximum, at C zero, and at Is, acceleration is negative maximum.
Acceleration is positive and decreasing in magnitude. Also, the magnitude of velocity is decreasing, because velocity and acceleration are in opposite direction.
b.At point B: The velocity is instantaneously zero at this moment So the particle is at rest momentarily. But the particle is still accelerating as the slope has a non-zero value. Acceleration is positive and decreasing in magnitude. From here the particle will start moving along the positive x-direction.
c.Between points B and C: Velocity is positive and increasing. So the acceleration is positive but decreasing in magnitude.
d.At point C: Velocity at point C is maximum as the graph peaks at this point. At this point slope is zero, therefore the acceleration is zero.
e.Between points, C and D: Velocity goes on decreasing between C and D. Due to negative slope acceleration is also negative. Acceleration is increasing in magnitude. Velocity and acceleration are in opposite directions.
f.At point D: Velocity is zero at this point But the slope is still negative, so the acceleration is negative. The particle will start moving in the. negative direction. Acceleration is increasing in magnitude.
g.Between points D and E: Particle is moving in the negative direction. The slope of the graph is negative in this region. It means the acceleration is negative, but the magnitude of the slope is increasing, it means the magnitude of acceleration is increasing. Velocity and acceleration are in the same direction.
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