What was Martha’s displacement before turning around?

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Homework Help Overview

The problem involves calculating Martha's displacement before she turns around after driving east at a constant velocity. The context includes her initial velocity, the time driven, and subsequent deceleration and acceleration phases.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the use of the displacement equation and question the appropriateness of the graphs needed for the problem. There are inquiries about specific values used in calculations and the interpretation of terms in the displacement equation.

Discussion Status

Some participants are seeking clarification on the relevant equations and the calculations presented. There is an ongoing exploration of the values used in the equations and the understanding of acceleration during different phases of the trip.

Contextual Notes

Participants are encouraged to show their work and clarify their reasoning, indicating a focus on understanding the problem rather than providing direct solutions.

Snipes

Homework Statement


Use the following information to construct any necessary graphs to help you complete the next four questions.

Martha leaves her house and drives with a constant velocity due east at 15 m/s for 85 seconds. She realizes she forgot something at home and stops the car in 6.5 seconds, makes a u-turn and then accelerates at 0.25 m/ss until she arrives back at home.

What was Martha’s displacement before turning around?

Homework Equations


d= vt+1/2 at^2

The Attempt at a Solution


tried using the displacement equation initial velocity acceleration but didnt get it right
Not sure what graph to use position vs time or what?

(15 m/s)(91.5 s)+1/2(.1639 m/s^2)(8.37 s^2) =d
2058.60589 =d
 
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Please list the "relevant equations" in section 2. Show your work in section 3 so that we can see where you are making a mistake.
 
Snipes said:
(15 m/s)(91.5 s)+1/2(.1639 m/s^2)(8.37 s^2) =d
Can you elaborate on how you got the expression on the left? What does the first term represent? How did you get the values .1639 m/s2 and 8.37 s2 in the second term?
 
used d=vt+1/2 at^2
 
What is the value of the acceleration during the first 85 seconds of the trip? Please explain your answer.

What is the value of the acceleration during the next 6.5 seconds of the trip? Please explain your answer.
 

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