# Kinematics problem with differential equations.

Spoti112

## 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)

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... ;(

Homework Helper
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Can you combine these:
V=dx/dt

V=√( 2 * (x2 - 1) )

Spoti112
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 ?