Question

Suppose a particle travels along a straight line with velocity v(t) = t 2 e −3t meters per second after t seconds. How far does the particle travel during the first 3 seconds? Round your answer to the nearest hundredth of a meter.

Answer 1

0.222 meters

Step-by-step explanation:

Form the definition of instantly velocity:

`v(t)=(dx(t))/(dt)\\dx(t)=v(t)dt`

`dx=(2te^(-3t))dt`

Integrating:

`\int\limits^(x)_(0) {} \, dx=\int\limits^(t)_(0) {2te^(-3t)} \,dt`

Solve the integral by parts:

`u=2t\\ du=2dt\\dv=e^(-3t)\\v=-(1)/(3)e^(-3t)\\`

`x=-(2)/(3)te^(-3t)+\int\limits^(t)_(0) {(2)/(3) e^(-3t) } \, dt`

`x=-(2)/(3)te^(-3t)-(2)/(9)e^(-3t)+(2)/(9)` (Remember to evaluate in t=0 and t=t)

Evaluating x(t) in t=3:

`x(3)=-2e^(-9) -(2)/(9)e^(-9)+(2)/(9)\\x(3)=0.2219` m; which rounded is 0.222 m.