Kinematics problem with differential equations.

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The problem involves finding the speed of a particle at x=3.0 m given its acceleration function a(x)=(2.0 s^-2)*x and initial conditions. The solution for part (a) shows that the speed at x=3.0 m is 4 m/s. For part (b), the user struggles with integrating to find the time taken to travel from x=1.0 m to x=3.0 m, expressing concern about losing the time variable during integration. Clarification is provided that the integration should be done over the specified x range while keeping track of the corresponding time range.
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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