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The motion of a set of particles moving along the x-axis is governed by the differential equation dx dt = t 3 - x3, where x1t2 denotes the position at time t of the particle. (a) If a particle is located at x = 1 when t = 2, what is its velocity at this time? (b) Show that the acceleration of a particle is

Respuesta :

Answer:

a)V=7 m/s

b)a=3t²-3x² t³ +3 x ⁵

Explanation:

Given that

[tex]\dfrac{dx}{dt}=t^3-x^3[/tex]

a)

We know that velocity V is given as follows

[tex]V=\dfrac{dx}{dt}[/tex]

[tex]V=t^3-x^3[/tex]

At t= 2 s and x= 1 m

[tex]V=2^3-1^3=7 m/s[/tex]

V=7 m/s

b)

Acceleration a is given as follows

[tex]a=\dfrac{dV}{dt}[/tex]

[tex]a=3t^2-3x^2\dfrac{dx}{dt}[/tex]

Now by putting the values

[tex]a=3t^2-3x^2\times (t^3-x^3)[/tex]

a=3t²-3x² t³ +3 x ⁵

Therefore the acceleration of a particle will be 3t²-3x² t³ +3 x ⁵.

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