The coordinates of a particle moving in a plane are given by x(t)= a

1.	The coordinates of a particle moving in a plane are given by x(t)= a cos(pt) and y(t) = b sin(pt) where a, b (<a) and p are positive constants of appropriate dimensions. Then(a) The path of the particle is an ellipse(b) The velocity and acceleration of the particle are normal to each other at t = π(2p)(c) The acceleration of the particle is always directed towards a focus(d) The distance travelled by the particle in time interval t = 0 to t = π/(2p) is a
Answer» x = a cos pt => cos pt = x/a y = b sin pt => sin pt = y/b squaring and adding them we get x²/a² + y²/b² = 1 Therefore path is an ellipse ---------- (option A is correct) dx/dt = v_x = -a p sin pt d²x/ dt² = a_x = - ap² cos pt dy/dt = v_y = bp cos pt d²y/dt² = - bp² sin pt At time t = pi/2p or pt = pi/2 a_x and v_y become 0 only v_x and v_y are left or we can say velocity is along the negative x axis and accelaration along the negative y axis at t = pi/2p , velocity and accelaration are normal to each other , so option B is also correct. At t=t , position of the particle r = x i + y j = a cos pt i + b sin pt j and accelaration of the particle is a(t) = a_x i + a_y j = - p² [ a cos pt i + b sin pt j ] = - p² [ x i + y j ] = - p² r (t) So the accelaration of the particle is always directed towards origin hence option (C) is also correct.

The coordinates of a particle moving in a plane are given by x(t)= a cos(pt) and y(t) = b sin(pt) where a, b (<a) and p are positive constants of appropriate dimensions. Then(a) The path of the particle is an ellipse(b) The velocity and acceleration of the particle are normal to each other at t = π(2p)(c) The acceleration of the particle is always directed towards a focus(d) The distance travelled by the particle in time interval t = 0 to t = π/(2p) is a

Answer»

x = a cos pt => cos pt = x/a

y = b sin pt => sin pt = y/b

squaring and adding them we get

x²/a² + y²/b² = 1

Therefore path is an ellipse ---------- (option A is correct)

dx/dt = v_x = -a p sin pt

d²x/ dt² = a_x = - ap² cos pt

dy/dt = v_y = bp cos pt

d²y/dt² = - bp² sin pt

At time t = pi/2p or pt = pi/2

a_x and v_y become 0

only v_x and v_y are left

or we can say velocity is along the negative x axis and accelaration along the negative y axis

at t = pi/2p , velocity and accelaration are normal to each other , so option B is also correct.

At t=t , position of the particle

r = x i + y j = a cos pt i + b sin pt j and accelaration of the particle is

a(t) = a_x i + a_y j

= - p² [ a cos pt i + b sin pt j ]

= - p² [ x i + y j ]

= - p² r (t)

So the accelaration of the particle is always directed towards origin hence option (C) is also correct.