Did I make the velocity vs time graph correctly?

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SUMMARY

The discussion centers on creating accurate velocity vs time and acceleration vs time graphs based on a distance-time graph analysis. The slopes for each segment were calculated, revealing a zero slope for segments A, C, and G, a constant slope of +1 for segment B, a sharp change to -3 for segment D, and a constant slope of -2/5 for segment F. It is confirmed that the velocity graph consists of horizontal segments reflecting the calculated slopes, while the acceleration graph will display instantaneous changes in acceleration due to sudden velocity changes, rather than being composed solely of horizontal lines.

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  • Understanding of slope calculation using the formula slope = deltaY/deltaX
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  • Ability to interpret instantaneous changes in motion
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  • Explore the concept of instantaneous acceleration and its implications
  • Practice creating distance-time, velocity-time, and acceleration-time graphs with various motion scenarios
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Students studying physics, particularly those focusing on kinematics, as well as educators looking to enhance their teaching methods for graphing motion concepts.

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Homework Statement


a) Calculate the slope for each segment
b) Describe the motion the object experiences in each segment
c) Make a velocity vs time graph using the slopes
d) Make a acceleration time graph

(for the distance-time graph)
Segment A slope of zero
Segment B constant slope of +1
Segment C slope of zero
Segment D sharp change to slope of -3
Segment E sudden change to slope of zero
Segment F constant slope of -2/5
Segment G slope of zero


Homework Equations



slope = deltaY/deltaX
acceleration = deltaV/deltaT



The Attempt at a Solution



I'm fine on a and b, but for c my graph just has horizontal segments at varying y values that correspond to the earlier velocities (slopes). I think this may be wrong. Anyway if it isn't does that mean that my acceleration-time graph will be a load of horizontal lines since all the velocity changes were sudden and constant?
 
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Last edited:
ziltoid said:

I'm fine on a and b, but for c my graph just has horizontal segments at varying y values that correspond to the earlier velocities (slopes).


That is correct. The Velocity in each section is the slope of the distance/time graph in that section.

...does that mean that my acceleration-time graph will be a load of horizontal lines since all the velocity changes were sudden and constant?

No. The acceleration at any instant is equal to the slope of the velocity/time graph at that instant. Sudden or rapid change in velocity mean large acceleration. The Velocity graph has instantaneous changes in velocity followed by periods of constant velocity. This means infinite acceleration for very short periods of time followed by periods of zero acceleration.
 
Last edited:

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