contestada

A particle moves along the x-axis so that at time t≥0t\ge 0t≥0 its position is given by x(t)=−t3+6t2−9t+42.x(t)=-t^3+6t^2-9t+42.x(t)=−t3+6t2−9t+42. Determine the total distance traveled by the particle from 0≤t≤4.0\le t \le 4.0≤t≤4.

Respuesta :

JeanaShupp

Answer: 4 units

Step-by-step explanation:

The given position function: [tex]x(t)=-t^3+6t^2-9t+42.[/tex], where t≥0.

To determine:  Total distance traveled by the particle from 0≤ t ≤ 4.0.

At t=0,

[tex]x(0)=-(0)^3+6(0)^2-9(0)+42=42\\[/tex]

At, t=4,

[tex]x(4)=-(4)^3+6(4)^2-9(4)+42\\\\=-64+6(16)-36+42=38[/tex]

Total distance traveled by the particle from 0≤ t ≤ 4.0 = |42-38| units

= 4 units

Hence, the distance traveled by the particle from 0≤t≤4.0 is 4 units.

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