A particle moves from the point (2.0^i + 40^j)m at t=0 with an initial velocity (5.0^i + 4.0^j)ms⁻¹. It is acted upon by a constant force which produces a constant acceleration (4.0^i + 4.0^j)ms⁻². What is the distance of the particle from the origin at time 2s?

The equation of motion is given by:

→s = →ut + (1/2)→at²

where →s is the displacement, →u is the initial velocity, →a is the acceleration, and t is the time.

Given:
→u = (5.0^i + 4.0^j) ms⁻¹
→a = (4.0^i + 4.0^j) ms⁻²
t = 2s

Substituting the values:
→s = (5^i + 4^j) * 2 + (1/2) * (4^i + 4^j) * 2²
→s = (10^i + 8^j) + (8^i + 8^j)
→s = 18^i + 16^j

The initial position is (2.0^i + 40^j)m. Therefore, the final position vector →r_f is:

→r_f = (2^i + 40^j) + (18^i + 16^j) = 20^i + 56^j

The distance from the origin is the magnitude of the final position vector:

|→r_f| = √(20² + 56²) = √(400 + 3136) = √3536 ≈ 59.46 m

However, there seems to be an error in the provided solution. Let's re-examine the calculation. The initial position was incorrectly added. The correct calculation should be:

→s = (5^i + 4^j)2 + 1/2(4^i + 4^j)2^2 = 10^i + 8^j + 8^i + 8^j = 18^i + 16^j

The final position is the initial position + displacement:
→r_f = (2^i + 40^j) + (18^i + 16^j) = 20^i + 56^j

The distance from the origin is:
|→r_f| = √(20² + 56²) = √(400 + 3136) = √3536 ≈ 59.46 m

The provided solution appears to have a mistake in its calculations. The correct distance from the origin at t=2s is approximately 59.46m. None of the given options are correct.