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