A particle moves in three dimensional space as an equation r= (t^2+1.34t)i^+(3.34t-2)j^+(2t^3-4t^2)k^

where is in unit of meter and in second.
a.Sketch the path of this particle
b.Calculate velocity at t=2 s
c.Calculate acceleration at t=2 s

I can not sketch it for you.

v = (2t+1.34) i + 3.34 j + (6t^2 - 8t) k
so put in t = 2

a = 2 i + (12t-8) k
again use 2 for t

a. To sketch the path of the particle described by the equation r = (t^2 + 1.34t)i + (3.34t - 2)j + (2t^3 - 4t^2)k, we can focus on each component separately to understand how they vary with time.

The x-component of the position vector r is given by (t^2 + 1.34t)i. This component is a quadratic equation in t, so the path of the particle in the x-direction will be a parabolic curve.

The y-component of the position vector r is given by (3.34t - 2)j. This component is a linear equation in t, so the path of the particle in the y-direction will be a straight line.

The z-component of the position vector r is given by (2t^3 - 4t^2)k. This component is also a quadratic equation in t, so the path of the particle in the z-direction will also be a parabolic curve.

Combining these three components, the path of the particle will be a three-dimensional curve that consists of a parabolic curve in the x-z plane and a straight line in the y-direction.

b. To calculate the velocity of the particle at t = 2s, we need to differentiate the position vector r with respect to time (t).

The velocity vector v is given by v = dr/dt, where dr represents the derivative of r with respect to t.

Taking the derivative of each component of r separately, we get:
dx/dt = d(t^2 + 1.34t)/dt = 2t + 1.34
dy/dt = d(3.34t - 2)/dt = 3.34
dz/dt = d(2t^3 - 4t^2)/dt = 6t^2 - 8t

Substituting t = 2s into these expressions, we find:
dx/dt at t=2s = 2(2) + 1.34 = 5.34 m/s
dy/dt at t=2s = 3.34 m/s
dz/dt at t=2s = 6(2)^2 - 8(2) = 16 m/s

Therefore, the velocity of the particle at t = 2s is v = 5.34i + 3.34j + 16k m/s.

c. To calculate the acceleration of the particle at t = 2s, we need to take the derivative of the velocity vector v with respect to time (t).

The acceleration vector a is given by a = dv/dt, where dv represents the derivative of v with respect to t.

Taking the derivative of each component of v, we get:
d^2x/dt^2 = d(2t + 1.34)/dt = 2
d^2y/dt^2 = 0 (since dy/dt is a constant)
d^2z/dt^2 = d(6t^2 - 8t)/dt = 12t - 8

Substituting t = 2s into these expressions, we find:
d^2x/dt^2 at t=2s = 2 m/s^2
d^2y/dt^2 at t=2s = 0 m/s^2
d^2z/dt^2 at t=2s = 12(2) - 8 = 16 m/s^2

Therefore, the acceleration of the particle at t = 2s is a = 2i + 0j + 16k m/s^2.