Question

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

Answer 1

Answer:

a)V=7 m/s

b)a=3t²-3x² t³ +3 x ⁵

Explanation:

Given that

$\dfrac{dx}{dt}=t^3-x^3$

a)

We know that velocity V is given as follows

$V=\dfrac{dx}{dt}$

$V=t^3-x^3$

At t= 2 s and x= 1 m

$V=2^3-1^3=7 m/s$

V=7 m/s

b)

Acceleration a is given as follows

$a=\dfrac{dV}{dt}$

$a=3t^2-3x^2\dfrac{dx}{dt}$

Now by putting the values

$a=3t^2-3x^2\times (t^3-x^3)$

a=3t²-3x² t³ +3 x ⁵

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