Question

A particle moves in a straight line and its position X at time T is given by x squared is equal to 2 + t its acceleration is given by

Answer 1

x^2 = 2 + t
d(x^2) / dv = d(2 + t)/dv
2dx/dv = dt/dv
2xdx = dt
dx/dt = 1/(2x)
v = 1/(2x)

Differentiating above equation
dv/dx = -1/(2x^2)

Acceleration is given by
a = vdv/dx
= 1/(2x) * -1/(2x^2)
= -1/(4x^3)

It’s acceleration is given by -1/(4x^3)

Answer 2

Answer:

Explanation:

x^2 = 2 + t

d(x^2) / dv = d(2 + t)/dv

2dx/dv = dt/dv

2xdx = dt

dx/dt = 1/(2x)

v = 1/(2x)

Differentiating above equation

dv/dx = -1/(2x^2)

Acceleration is given by

a = vdv/dx

= 1/(2x) * -1/(2x^2)

= -1/(4x^2)