Question

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.

Answer 1

Answer: 4 units

Step-by-step explanation:

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

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

At t=0,

$x(0)=-(0)^3+6(0)^2-9(0)+42=42$

At, t=4,

$x(4)=-(4)^3+6(4)^2-9(4)+42\\\\=-64+6(16)-36+42=38$

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.