The path of a particle can be expressed as: r = 6t i + (3t2-4) j + 7 k, please determine the acceleration of this particle.

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1. The path of a particle can be expressed as: r 6t i (30-4)j 7 k, please determine the acceleration of this particle.
A particle is located at r(t) = 14t i + 6t^2 j. Find its position, velocity and acceleration at t = 2 s.
( Problem 3) If field J is expressed as below at 712, π2, 3T2) , determine the vector component of J that is: J.= r sin θ cos gar-cos 2θ sin φǎθ + tan-In ra, (i) Parallel to az (ii) Normal to surface φ= 372 (iii) Tangential to the spherical surface r 2
The velocity of a particle traveling in a straight line is given by v (6t-3t2) m/s, where t is in seconds. Suppose that s 0 when t0. a. Determine the particle's deceleration when t3.6s b. Determine the particle's position when t 3.6 s C. How far has the particle traveled during the 3.6-s time interval? d. What is the average speed of the particle for the time period given in previous part?
A particle moves along the path r-8tH + (t3 + 5)j) m, where t s in seconds. Determine the magnitudes of the particle's velocity and acceleration whent-3 s.
A particle has a position vs time equation r=(3t^2) i + (4t^2 - t^3) j j = vertical direction with no gravity a) find the time when particle reaches peak of its path b) find the x displacement when the magnitude of the acceleration is a minimum
v(t) = (10 + 6t) i hat − 6t^2 j hat Where t is in seconds and v is in m/s. What is the displacement and acceleration as a function of time?
Use this theorem to find the curvature. r(t) = 6t i + 8 sin(t) j + 8 cos(t) k
The speed of a point is v = 2 i + 3T2 j (ft / s ) . At t = 0 its position is r = -i + 2j (foot) . Finding r t = 2 s
11. The velocity of a particle is v)-(-2t) i+ (6t+2)jm/s What is the magnitude and direction of the acceleration of the particle at t 3 s?