Solving 1D Motion Question with x>0, v<0, a>0

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SUMMARY

The discussion focuses on providing examples of one-dimensional motion where position (x) is greater than zero, velocity (v) is less than zero, and acceleration (a) is greater than zero. The equation x(t) = A + Be-γt is presented as a valid model, where A and B are positive constants, and γ is a positive decay constant. Another example given is x(t) = 3t2 - 20t + 100, evaluated at t=2, demonstrating the required conditions through calculus by differentiating to find velocity and acceleration.

PREREQUISITES
  • Understanding of calculus, specifically differentiation and its applications in motion analysis.
  • Familiarity with the concepts of position, velocity, and acceleration in physics.
  • Knowledge of exponential decay functions and their properties.
  • Ability to solve quadratic equations and analyze their graphs.
NEXT STEPS
  • Study the implications of exponential decay in motion using the equation x(t) = A + Be-γt.
  • Explore the relationship between position, velocity, and acceleration through calculus.
  • Investigate other forms of motion equations, such as polynomial motion equations.
  • Learn how to graph motion equations to visualize the relationships between x, v, and a.
USEFUL FOR

Students studying physics, particularly those focusing on kinematics, as well as educators looking for examples of motion with specific conditions.

mttal24
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Homework Statement


Give example of a motion where x>0, v<0, a>0 at a particular instant.
x-->Position
v-->Velocity
a-->Acceleration


Homework Equations


I thought I had to give an example such as a car, ball etc.
But the answer says:
x(t) ie; position for time t; given by
x(t)=A+Be[itex]^{-\gamma t}[/itex]
where A>B, [itex]\gamma[/itex]>0 are chosen +ve constants.

The Attempt at a Solution


I don't know how to do it. Please help.
 
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the given example wasnt the only solution

another example would be
x(t) = 3t^2 -20t+100 at t=2

for your example use calculus, for at t=1, x>0 as u can check
differentiate dx/dt = -γBe^(-γt), which is less than zero at t=1
differentiating again gives a=(γ^2)(B)(e^-yt) which is positive at t=1
 

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