Applications of the Equations of Kinematics for Constant Acceleration

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SUMMARY

The discussion focuses on calculating the total displacement of a speedboat undergoing three distinct acceleration phases: +2.01 m/s² for 6.55 seconds, +0.518 m/s² for 6.45 seconds, and -1.49 m/s² for 8.80 seconds. The initial displacement after the first phase is confirmed as 43.117 meters. However, there is a discrepancy in the calculation of the initial velocity for the second phase, with a correction to 13.1655 m/s instead of 16.5066 m/s. The conversation emphasizes the importance of accurately interpreting the problem statement regarding the duration of the second phase.

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  • Understanding of kinematic equations for constant acceleration
  • Familiarity with basic physics concepts such as displacement, velocity, and acceleration
  • Ability to perform calculations involving time, acceleration, and initial velocity
  • Knowledge of units of measurement in physics (meters, seconds)
NEXT STEPS
  • Review the kinematic equations for constant acceleration, specifically the equations of motion
  • Practice problems involving multiple phases of acceleration to solidify understanding
  • Explore the concept of instantaneous velocity and its calculation during varying acceleration
  • Investigate common mistakes in kinematic calculations and how to avoid them
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Students studying physics, particularly those focusing on kinematics, as well as educators looking for examples of common calculation errors in motion problems.

jacksonpeeble
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Homework Statement


A speedboat starts from rest and accelerates at +2.01 m/s2 for 6.55 s. At the end of this time, the boat continues for an additional 6.45 s with an acceleration of +0.518 m/s2. Following this, the boat accelerates at -1.49 m/s2 for 8.80 s.

Find the total displacement of the boat.

Homework Equations


http://www.webassign.net/ebooks/cj6/pc/c02/read/main/c02x2_4.htm
(Hopefully this is public access, if not, the equations can be determined from the work below, assuming that I am using the correct equations.)


The Attempt at a Solution


v0=0
d1=0*6.55+.5*2.01*6.55^2=43.117
v1=2.01*6.55+.518*6.45=16.5066
d2=16.5066*6.45+.518*6.45^2=128.018
v2=16.5066+-1.49*8.8=3.3946
d3=3.3946*8.8+-1.49*8.8^2=-85.5131

Any help is, as always, greatly appreciated! Thank you in advance!
 
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I agree with your first distance (43.117) after time 6.55 s.
However, at this time I get
velocity = Vi*t + .5*a*t^2
= 0 + .5*2.01*6.55^2 = 13.1655 m/s
This is the initial velocity for the second interval, not 16.5066.
 
Delphi51 said:
I agree with your first distance (43.117) after time 6.55 s.
However, at this time I get
velocity = Vi*t + .5*a*t^2
= 0 + .5*2.01*6.55^2 = 13.1655 m/s
This is the initial velocity for the second interval, not 16.5066.

I see that you changed 6.45 to 6.55. Is this correct, since the problem states "...the boat continues for an additional 6.45 s..."?
 

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