Kinematics problem with differential equations.

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Homework Help Overview

The problem involves kinematics and differential equations, specifically focusing on the acceleration of a particle as a function of its position. The original poster presents a scenario where the acceleration is defined as a(x) = (2.0 s-2) * x, and seeks to determine the speed of the particle at a specific position and the time taken to travel between two points.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to derive the velocity function using integration and applies initial conditions. Some participants question how to combine the velocity equation with the differential form and express time in terms of position. Others express uncertainty about integrating the equations without losing variables.

Discussion Status

The discussion is ongoing, with participants exploring different mathematical approaches to solve for time and velocity. Some guidance has been offered regarding the integration process, but there is no clear consensus on how to proceed with the calculations.

Contextual Notes

Participants note the importance of maintaining the relationship between position and time during integration, as well as the initial conditions provided in the problem statement. There is an acknowledgment of the constraints posed by the problem's setup and the need for clarity in variable definitions.

Spoti112
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Homework Statement


Suppose the acceleration of a particle is a function of x, where a(x)=(2.0 s-2)*x.
(a) If the velocity is zero when x= 1.0 m, what is the speed when x=3.0 m?
(b) How long does it take the particle to travel from x=1.0 m to x=3.0 m. a(x)=(2.0 s-2) * x
(a) V(x=3) = ? , V(x=1) = 0
(b) t=? (from x=1 to x=3)

Homework Equations


a=dV/dt
V=dx/dt

The Attempt at a Solution


(a) a = dV/dt = (dV/dx) * (dx/dt)= (dV/dx) * V ⇒
⇒ a = 2x = (dV/dt) * V ⇒ VdV = 2xdx ⇒ ∫VdV = ∫2xdx ⇒
⇒ V2/2 = x2 + c
x=1↔V=0 ⇒ 0=12 + c ⇒ c = -1
⇒ V=√( 2 * (x2 - 1) ) ⇒ V(x=3) = 4 m/s

(b) i have no idea... ;(
 
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Can you combine these:
Spoti112 said:
V=dx/dt

Spoti112 said:
V=√( 2 * (x2 - 1) )
 
dx/dt = √( 2 * (x2 - 1)) → dt = dx/√(2(x2 - 1)) ... and i`m stuck with the math
if i integrate here in an interval of 1 to 3 (both sides ) , won't i lose the t in the equation ?
 
Spoti112 said:
an interval of 1 to 3 (both sides )
The 1 to 3 is the x range. The t range is the duration, from tx=1m=0 to t=tx=3m.
 
thank you very much for the help!
 

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