Kinematics - Finding position and acceleration

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SUMMARY

The discussion focuses on solving kinematic equations for a test car's position and acceleration, defined by the polynomial function x(t) = 2.17 m + 4.80 m/s²t² - 0.100 m/s⁶t⁶. Participants addressed finding the car's position and acceleration at the first and second instances when the car's velocity is zero. The correct approach involves taking the first derivative to obtain velocity v(t) and the second derivative for acceleration a(t). The user encountered difficulties in determining the correct time values for zero velocity, despite successfully solving for position and acceleration in other parts.

PREREQUISITES
  • Understanding of kinematic equations and polynomial functions
  • Knowledge of derivatives and their applications in physics
  • Familiarity with the concepts of velocity and acceleration
  • Ability to solve equations for specific variables
NEXT STEPS
  • Study the application of derivatives in kinematics, focusing on velocity and acceleration
  • Learn how to solve polynomial equations for specific roots
  • Explore the implications of higher-order derivatives in motion analysis
  • Practice problems involving zero velocity conditions in kinematic equations
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Students studying physics, particularly those focusing on kinematics, as well as educators seeking to clarify the application of derivatives in motion problems.

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Homework Statement



The position of the front bumper of a test car under microprocessor control is given by x(t) = 2.17 m + 4.80 m/s2t2 - 0.100 m/s6t6.
a. Find its position at the first instant when the car has zero velocity.
b. Find its acceleration at the first instant when the car has zero velocity.
c. Find its position at the second instant when the car has zero velocity.
d. Find its acceleration at the second instant when the car has zero velocity.

Homework Equations


i took the derivative and to get v(t) for parts a,b,c of this
then the second derivative to get a(t)


The Attempt at a Solution


a(t)= 9.6-3t4 solve for t we get 1.337 when i plug that into the v(t) equation i get ~10 but when i plugg this in it's wrong. I have to idea what is going wrong, this is how my professor set it up... i got a,b,c right though...
 
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Are you not suppose to solve for t at v(t) = 0?
 


yes for part a and c
 

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