What was Martha’s displacement before turning around?

AI Thread Summary
Martha drives east at 15 m/s for 85 seconds before stopping and making a U-turn. The displacement before turning around can be calculated using the equation d = vt + 1/2 at^2. The initial displacement from her house is 1275 meters, but the calculations for the deceleration and U-turn need clarification. The acceleration during the first 85 seconds is zero, while the acceleration during the next 6.5 seconds is negative as she stops. Accurate values and clear steps in calculations are essential for resolving the problem correctly.
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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