Answer:
Step-by-step explanation:
To find the particle's velocity and the force acting on it, we need to differentiate the given trajectory equation with respect to time (t).
Given trajectory: r = (3t^3 + t^2 + 1)i + (3(t + 4)^(3/2))j + 2ln(t + 1)k
Velocity (v):
The velocity vector (v) is obtained by taking the derivative of the position vector (r) with respect to time (t).
v = dr/dt = (d/dt(3t^3 + t^2 + 1))i + (d/dt(3(t + 4)^(3/2)))j + (d/dt(2ln(t + 1)))k
Differentiating each component:
v = (9t^2 + 2t)i + (3/2)(t + 4)^(1/2)j + (2/(t + 1))k
Therefore, the particle's velocity is v = (9t^2 + 2t)i + (3/2)(t + 4)^(1/2)j + (2/(t + 1))k.
Force (F):
The force acting on the particle is the derivative of momentum with respect to time (F = dp/dt). Since the particle's mass (m) is given as 1, force is equivalent to acceleration (F = ma).
Therefore, we need to find the acceleration vector (a).
Acceleration (a) is obtained by taking the derivative of the velocity vector (v) with respect to time (t).
a = dv/dt = (d/dt(9t^2 + 2t))i + (d/dt((3/2)(t + 4)^(1/2)))j + (d/dt(2/(t + 1)))k
Differentiating each component:
a = (18t + 2)i + (3/4)(t + 4)^(-1/2)j - 2/(t + 1)^2k
Therefore, the force acting on the particle is F = (18t + 2)i + (3/4)(t + 4)^(-1/2)j - 2/(t + 1)^2k.
Scalar Product of Force and Velocity (F ⋅ v):
The scalar product of two vectors is calculated by multiplying their corresponding components and adding them up. For F ⋅ v, we have:
F ⋅ v = (18t + 2)(9t^2 + 2t) + (3/4)(t + 4)^(-1/2)(3/2)(t + 4)^(1/2) + (-2/(t + 1)^2)(2/(t + 1))
Simplifying and combining terms:
F ⋅ v = 162t^3 + 36t^2 + 18t + (9/4) + (-4/(t + 1)^3)
Work (W):
To calculate the work done on the particle between t = 0 and t = 1, we integrate the scalar product F ⋅ v with respect to time (t) over the given interval:
W = ∫[0,1] (162t^3 + 36t^2 + 18t + (9/4) - 4/(t + 1)^3) dt
Integrating each term separately and evaluating the definite integral over [0,1], you can find the numerical value of the work done on the particle between t