The velocity of a particle is given by v=16t^2i+4t^3j+(5t+2k) m/s, where t is in seconds. if the particle is at the origin when t=0, determine the magnitude of the particle's acceleration when t=2s. also what is the x, y,z coordinate position of the particle at this instant

80.16 m/s^2

at t=2 s

x=42.3 m

y=16 m

z=14 m

Explanation:

solution

The x,y,z components of the velocity are donated by the i,j,k vectors.

acceleration is a derivative of velocity with respect to time.

evaluate acceleration at 2 seconds

the magnitude of the acceleration is the square root of the sum of the square of each component of the acceleration.

position is the integral of velocity with respect to time position at a time can be found by taking by taking the definite intergral of each component.

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can you make it more obvious. i can't see it properly