Question

Find the distance, in feet, a particle travels in its first 2 seconds of travel, if it moves according to the velocity equation v(t)= 6t2 - 18t + 12 (in feet/sec).

Answer 1

Answer:

The particle will travel 6 feet in first 2 seconds.

Step-by-step explanation:

We have been given that a particle moves according to the velocity equation $v(t)= 6t^2-18t+12$. We are asked to find the distance that the particle will travel in its first 2 seconds.

$s(t)=\int |v(t)|dt$

$s(t)=\int\limits^2_0 |6t^2-18t+12|dt$

Now, we will eliminate the absolute value sign as:

$s(t)=\int\limits^1_0 6t^2-18t+12dt+\int\limits^2_1 -6t^2+18t-12dt$

$s(t)=[\frac{6t^3}{3}-\frac{18t^2}{2}+12t]^1_0 +[\frac{-6t^3}{3}+\frac{18t^2}{2}-12t]^2_1$

$s(t)=[2t^3-9t^2+12t]^1_0 +[-2t^3+9t^2-12t]^2_1$

$s(2)=2(1)^3-9(1)^2+12(1)-(2(0)^3-9(0)^2+12(0))-2(2)^3+9(2)^2-12(2)-(-2(1)^3+9(1)^2-12(1))$

$s(2)=2-9+12-(0)-16+36-24-(-2+9-12)$

$s(2)=5-(0)-4-(-5)$

$s(2)=5-4+5$

$s(2)=6$

Therefore, the particle will travel 6 feet in first 2 seconds.