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cluponka [151]
3 years ago
15

A small metal particle passes downward through a fluid medium while being subjected to the attraction of a magnetic field such t

hat its position is observed to be s = (15t^3 - 3t) mm, where t is measured in seconds. Determine (a) the particle's displacement from t = 2 s to t = 4 s, and (b) the velocity and acceleration of the particle when t = 5 s.
Engineering
1 answer:
bekas [8.4K]3 years ago
4 0

Answer:

a)Δs = 834 mm

b)V=1122 mm/s

a=450\ mm/s^2

Explanation:

Given that

s = 15t^3 - 3t\ mm

a)

When t= 2 s

s = 15t^3 - 3t\ mm

s = 15\times 2^3 - 3\times 2\ mm

s= 114 mm

At t= 4 s

s = 15t^3 - 3t\ mm

s = 15\times 4^3- 3\times 4\ mm

s= 948 mm

So the displacement between 2 s to 4 s

Δs = 948 - 114 mm

Δs = 834 mm

b)

We know that velocity V

V=\dfrac{ds}{dt}

\dfrac{ds}{dt}=45t^2-3

At t=  5 s

V=45t^2-3

V=45\times 5^2-3

V=1122 mm/s

We know that acceleration a

a=\dfrac{d^2s}{dt^2}

\dfrac{d^2s}{dt^2}=90t

a= 90 t

a = 90 x 5

a=450\ mm/s^2

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