Particles and distance

Question

My daughter was stumped on this for her homework. We found it posted here but the solution was wrong according to her online assignment. So let’s try again.

A particle that moves along a straight line has velocity

v(t)=t^(2)e^(−4t)

meters per second after t seconds. How many meters will it travel during the first t seconds (from time=0 to time=t)?

Thanks for the help.

Answer 1

Evaluation the same integral of v(t) = t²e^-4t by integrating by parts twice.

∫v(t)dt = ∫t²e^–4tdt

u = t² dv = e^–4tdt

du = 2tdt v = –(1/4)e^–4t

∫v(t)dt = –(1/4)t²e^–4t + (1/2) I where I = ∫te^–4tdt

u = t dv = e^–4tdt

du = dt v = –(1/4)e^–4t

I = –(1/4)te^–4t + 1/4∫e^–4tdt = –(1/4t)te^–4t +1/4(-1/4e^–4t) = –(1/4)te^–4t – (1/16)e^–4t

∫v(t)dt = –(1/4)t²e^–4t + (1/2) I = –(1/4)t²e^–4t + (1/2)[–(1/4)te^–4t – (1/16)e^–4t]

∫v(t)dt = e^–4t (–t²/4 – t/8 – 1/32)

Evaluated from 0 to t this gives the displacement function x(t) below:

x(t) = e^–4t (–t²/4 – t/8 – 1/32) –(–1/32) = e^–4t (–t²/4 – t/8 – 1/32) +1/32

Based on the rapidly decaying exponential in this displacement function, it looks like a maximum displacement of 1/32 meters will be reached within a few seconds.

Answer 2

If I am reading your function properly: v(t) = t²e^-4t _...

I arrive at the following, by performing integration-by-parts twice (initially setting u = t² ):

d(t) = ∫ v(t) = e^-4t ( -t²/4 - t/8 - 1/16 )

...don't have time to check my work, so this could be risky!