Question

A particle moves in a straight line and it's position x at time t is given by square of x =2+t. It's acceleration is given by

Answer 1

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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)

Answer 2

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⏬⏬ANSWER⏬⏬


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)



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