Record travel along a straight path

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SUMMARY

The discussion focuses on calculating total displacement and average speeds for a trip involving three distinct legs of motion: constant acceleration, constant velocity, and negative acceleration. The first leg features a constant acceleration of 2.20 m/s² for 15.0 seconds, resulting in a displacement of 247.50 meters. The second leg maintains a constant velocity for 1.65 minutes, while the third leg applies a negative acceleration of -9.92 m/s² for 3.33 seconds. Participants are encouraged to apply the kinematic equations provided to solve for average speeds and total displacement.

PREREQUISITES
  • Understanding of kinematic equations, specifically delta X = Vi t + 1/2 a t² and Vf² = Vi² + 2 a (delta X)
  • Basic knowledge of acceleration and velocity concepts
  • Ability to convert units, such as minutes to seconds
  • Familiarity with the concept of displacement in physics
NEXT STEPS
  • Calculate average speed for leg 2 using the constant velocity formula
  • Determine total displacement by summing displacements from all three legs
  • Explore the effects of varying acceleration on displacement and velocity
  • Review additional kinematic problems to reinforce understanding of motion equations
USEFUL FOR

Students studying physics, particularly those focusing on kinematics, as well as educators looking for examples of motion analysis in real-world scenarios.

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Homework Statement



A record of travel along a straight path is as follows:
1. Start from rest with constant acceleration of 2.20 m/s2 for 15.0 s.
2. Maintain a constant velocity for the next 1.65 min.
3. Apply a constant negative acceleration of −9.92 m/s2 for 3.33 s.
(a) What was the total displacement for the trip? m
(b) What were the average speeds for legs 1, 2, and 3 of the trip, as well as for the complete trip?
leg 1 m/s
leg 2 m/s
leg 3 m/s
complete trip m/s

Homework Equations



delta X = Vi t + 1/2 a t^2
Vf^2 = Vi^2 + 2 a (delta X)

The Attempt at a Solution



I'm sure how to tackle this, but here was my attempt.
delta X = (0)(15) + 1/2(2.20)(15^2) = 247.50
I'm kind of stuck at this point, I don't know what to do?!
 
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Can't you do the same for leg 3 as you did for leg 1? As they are pretty much the same, only the acceleration is now the other way around.
 

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