Question

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

Answer 1

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

$a=450\ mm/s^2$