The acceleration of a particle at time t is given by

a = cos ti + sin tj .

At time t = 0, its position relative to the origin is – i and its velocity is – j. Prove that the particle always lies on the circle x² + y² = 1 .

Question

The acceleration of a particle at time t is given by

a = cos ti + sin tj .

At time t = 0, its position relative to the origin is - i and its velocity is - j. Prove that the particle always lies on the circle x² + y² = 1 .

Accepted Answer

The velocity and then the position of the body are obtained by successive integration of

a = cos ti + sin tj .

The velocity is therefore

v = (sin t + [latex]c_{1}[/latex])i + (- cos t + [latex]c_{2}[/latex])j

where [latex]c_{1}[/latex] and [latex]c_{2}[/latex] are the constants of integration.

When t = 0, v = [latex]c_{1}[/latex]i + (-1 + [latex]c_{2}[/latex])j. But the velocity is -j when t = 0, so [latex]c_{1}[/latex] and [latex]c_{2}[/latex] are both zero. Hence

v = sin ti - cos tj .

Integrating again gives

r = (- cos t + [latex]k_{1}[/latex])i + (- sin t + [latex]k_{2}[/latex])j.

Again, the constants of integration [latex]k_{1}[/latex] and [latex]k_{2}[/latex] are zero, since r is - i when t is zero. So

r = - cos ti - sin tj .

The co-ordinates of the particle at time t are given by x = - cos t, y = - sin t . This always lies on the circle x² + y² = 1 (since sin² t + cos² t = 1).

Question 17.3: The acceleration of a particle at time t is given by a = co...

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