Given a radioactive particle that encounters a shielding barrier, it is known to

Question

Given a radioactive particle that encounters a shielding barrier, it is known to decelerate and come to rest within the shielding in 6 microseconds (6 s). In this example, presume the distance (d in microns (m)) the particle will travel in the time period after hitting the shielding is given by the following equation.

Given the bounds t0 = 0 s and tmax = 6 s

d= t^3 - 18t^2 + 108t

What is the maximum distance the particle will travel into the shielding?

What is the velocity of the particle at t = 4 s?

What is the acceleration of the particle at t = 3 s?

Explanation / Answer

Substitute t =  6 s

d =  (6 s)^3 -18 ( 6 s)^2 +108*( 6 s)

d = 0.000648 m

velocity = v= differentiation of distNCE TRAVELLED

V = 3*t^2 -36*t +108

v = 108 m/s

Acceleration = -36 m/s2

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