Calculate the time at which the object changes direction

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SUMMARY

The discussion centers on calculating the time at which an object changes direction after undergoing acceleration and constant velocity phases. The object accelerates to 20.0 m/s in 5.0 minutes, travels at that speed for 10 minutes, and then accelerates to 15.0 m/s in the opposite direction over 3.0 minutes. A key point of confusion arises from the calculation of acceleration, where a precise value of -0.194444 m/s² is critical for accurate results, as rounding to -0.19 m/s² leads to discrepancies in the final time calculation.

PREREQUISITES
  • Understanding of kinematics, specifically acceleration and velocity calculations.
  • Familiarity with graphing techniques, particularly velocity-time graphs.
  • Knowledge of linear equations, specifically the equation of a line y=mx+b.
  • Ability to interpret changes in direction and displacement in motion problems.
NEXT STEPS
  • Review kinematic equations for uniformly accelerated motion.
  • Learn how to accurately plot and analyze velocity-time graphs.
  • Study the implications of negative acceleration and its effect on direction change.
  • Practice problems involving multiple phases of motion to solidify understanding.
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Students studying physics, particularly those focusing on kinematics and motion analysis, as well as educators looking for examples of direction change calculations in real-world scenarios.

Snipes

Homework Statement


An object starts from rest and uniformly acquires a velocity of 20.0 m/s in 5.0 minutes. For the next 10. minutes the object moves with a constant speed. After 10. minutes moving at 20. m/s the object then again accelerates and reaches a speed of 15 m/s in the opposite direction in 3.0 minutes time.

Calculate the time at which the object changes direction

Homework Equations


y=mx+b
v=d/t

The Attempt at a Solution


made velocity time graph. tried to find the equation y=mx + b. already knew that slope + -.19 unknown what b is and i found that velocity = 0 when time is at 1005.275 seconds however it said this was wrong. I am very confused and jumbled. Unsure how to find displacement since it crosses the x-axis and moves into negative.
 
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Snipes said:

Homework Statement


An object starts from rest and uniformly acquires a velocity of 20.0 m/s in 5.0 minutes. For the next 10. minutes the object moves with a constant speed. After 10. minutes moving at 20. m/s the object then again accelerates and reaches a speed of 15 m/s in the opposite direction in 3.0 minutes time.

Calculate the time at which the object changes direction

Homework Equations


y=mx+b
v=d/t

The Attempt at a Solution


made velocity time graph. tried to find the equation y=mx + b. already knew that slope + -.19 unknown what b is and i found that velocity = 0 when time is at 1005.275 seconds however it said this was wrong. I am very confused and jumbled. Unsure how to find displacement since it crosses the x-axis and moves into negative.

Could you post image of your graph ?

I am getting value of time which is slightly lesser than what you are getting .
 
Last edited:
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Likes   Reactions: Merlin3189
When you calculated the acceleration did you get a figure of -0.194444m/s^2. If so don't round it down to -0.19m/s^2 for the next step as that seems to account for your slightly different answer to the one I get.
 

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