A particle of unit mass is moving on a plane. Its trajectory, in polar coordinates, is given by r(t) = t2, q (t)= t, where t is time. The kinetic energy of the particle at time t = 2 isa)4b)12c)16d)2Correct answer is option 'C'. Can you explain this answer?

Question

Answer

Given:
The trajectory of the particle is given by the polar coordinates r(t) = t^2 and θ(t) = t, where t is the time.

To find:
The kinetic energy of the particle at time t = 2.

Solution:
The kinetic energy (KE) of a particle is given by the formula:
KE = (1/2) m v^2

Step 1:
Find the velocity vector of the particle.

The position vector of the particle in polar coordinates is given by:
r(t) = t^2 (cosθ(t), sinθ(t))

Differentiating r(t) with respect to time t, we get:
v(t) = (d/dt) r(t) = (d/dt) t^2 (cosθ(t), sinθ(t))

Using the chain rule, we can differentiate each component of r(t):
v(t) = (2t)(cosθ(t), sinθ(t)) + t^2 (-sinθ(t), cosθ(t))

Simplifying, we get:
v(t) = 2t (cosθ(t), sinθ(t)) - t^2 (sinθ(t), cosθ(t))

Substituting the given values of r(t) and θ(t), we get:
v(t) = 2t (cos(t), sin(t)) - t^2 (sin(t), cos(t))

Step 2:
Find the magnitude of the velocity vector.

The magnitude of the velocity vector is given by:
|v(t)| = √[v_x(t)^2 + v_y(t)^2]

Substituting the values of v(t), we get:
|v(t)| = √[(2t cos(t) - t^2 sin(t))^2 + (2t sin(t) + t^2 cos(t))^2]

Simplifying, we get:
|v(t)| = √[4t^2 cos^2(t) - 4t^3 cos(t) sin(t) + t^4 sin^2(t) + 4t^2 sin^2(t) + 4t^3 cos(t) sin(t) + t^4 cos^2(t)]

Simplifying further, we get:
|v(t)| = √[4t^2 (cos^2(t) + sin^2(t)) + t^4 (cos^2(t) + sin^2(t))]

Using the identity cos^2(t) + sin^2(t) = 1, we get:
|v(t)| = √[4t^2 + t^4]

Step 3:
Find the kinetic energy at t = 2.

Substituting t = 2 in the magnitude of the velocity vector, we get:
|v(2)| = √[4(2)^2 + (2)^4]
|v(2)| = √[16 + 16]
|v(2)| = √32

The kinetic energy at t = 2 is given by:
KE = (1/2) m |v(2)|^2

Since the mass of the particle is given as 1 unit, we have:
KE = (1/2) |v(2)|^2

Sub

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