Calculating Velocity & Acceleration of a Particle at r(t)

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SUMMARY

The discussion focuses on calculating the velocity and acceleration vectors of a particle defined by the position function r(t) = (2cos(t))i + (3sin(t))j + 4tk at t = π/2. The velocity vector is derived as v(t) = (-2sin(t))i + (3cos(t))j + 4k. The speed of the particle is determined to be |v(t)| = 2√5, achieved by substituting t = π/2 into the velocity equation and applying the Pythagorean theorem for vector magnitudes.

PREREQUISITES
  • Understanding of vector calculus
  • Familiarity with trigonometric identities, specifically sin²(t) + cos²(t) = 1
  • Knowledge of vector magnitude calculations
  • Basic proficiency in calculus, particularly differentiation and integration
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  • Explore the concept of particle motion in three-dimensional space
  • Investigate the relationship between speed, direction, and vector representation
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Students in physics or mathematics, particularly those studying mechanics and motion, as well as educators looking for examples of particle dynamics in three-dimensional space.

JoeSabs
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Homework Statement


r(t) is the position of a particle in space at time t. Find the particle's velocity and acceleration vectors. Then find the particle's speed and direction of motion at the given value of t. Write the particle's velocity at that time as the product of its speed and direction.

r(t)=(2cost)i+(3sint)j+4tk; t=pi/2


Homework Equations


sin^2+cos^2=1
length of v= sqrt(vi+vj+vk) if vi is the first coefficient of velocity equation, vj 2nd, vk 3rd.


The Attempt at a Solution


v(t)= (-2sint)i+(3cost)j+4k

|v(t)| (length)=2sqrt(5), but I don't know how to get this.

sqrt(((-2sin(t))^2)+((3cos(t))^2)+(4^2))
sqrt((4sin^2(t))+(9cos^2(t))+(16))
?
Are you supposed to somehow remove the coefficients to use cos^2+sin^2? I tried this to no avail. I also tried using the sin^2(x)= (1-cos2x)/2, and the cos one too, but that didn't work either.
 
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It looks to me like you have the correct answer. Just put t=Pi/2 into your last expression. You get sqrt(20) which is the same as 2sqrt(5).
 


Ah, I see! Plug in before trying to simplify. That's very easy, just out of the normal order of operations I'm used to. Thanks a lot!
 

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