J asked in Science & MathematicsPhysics · 9 months ago

# suppose the position of a particle is i+3j-2k at t=2pi/3 sec, and suppose the particles velocity at any time t is v(t)=2sin(t)i+2cos(t)j+tk?

a) compute the particles position at any time t.

b) compute the acceleration of the particle at any time t.

c) compute the particles unit tangent vector at any time t.

d) compute the tangential component of the particles acceleration at any time t.

e) compute the curvature of the particles path at any time t.

f) compute the normal component of the particles acceleration at any time t.

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• NCS
Lv 7
9 months ago
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Ooh, this is a little messy.

a) position s(t) = ∫ v(t) dt

s(t) = ∫ [2sin(t) i + 2cos(t) j + t k] dt = -2cos(t) i + 2sin(t) j + ½t² k + C

We're told that s(2π/3) = i + 3 j - 2 k, and so

i + 3 j - 2 k = -2cos(2π/3) i + 2sin(2π/3) j + ½(2π/3)² k + C

i + 3 j - 2 k = i + 1.732 j + 2.193 k + C

1.268 j - 4.193 k = C

which gives us

s(t) = -2cos(t) i + [2sin(t) + 1.268] j + [½t² - 4.193] k ◄

b) a(t) = dv/dt = 2cos(t) i - 2sin(t) j + k ◄

c) unit tangent T(t) = s'(t) / ||s'(t)|| = v(t) / ||v(t)||

||v|| = √[(2sin(t))² i + (2cos(t))² j + t² k]

||v|| = √[4cos²t + 4sin²t + t²] = √[4 + t²]

and so

unit T(t) = [2sin(t) i + 2cos(t) j + t k] / √[4 + t²]

d) the tangential component of the acceleration is

a'(t) = da/dt = -2sin(t) i - 2cos(t) j

e) There are a few ways to calculate the curvature. I'll use

κ = ||T'(t)|| / ||s'(t)|| = ||T'(t) || / ||v(t)||

Use the quotient rule on T(t) to find its derivative T'(t)

and you already know that ||v(t)|| = √(4 + t²)

Too much to type here.

f) the normal vector N(t) = T '(t) / ||T '(t)||

Again, way too much to type in this forum.

The process is correct for a thru d, but CHECK THE MATH!

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