Find the distance the particle travels

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A particle that moves along a straight line has velocity $$v(t)=t^2e^{-3t}$$ meters per second after t seconds. Find the distance the particle travels during the first t seconds.

________________meters (Your answer should be a function of $$t$$)


shouldnt i just integrate that velocity function? cause if you integrate velocity, you get distance right? well i did, but got the wrong answer. here's my answer:

$$-1/27*e^{-3t}(9t^2+6t+2)$$

i also used math programs to integrate the problem just to make i didnt make any mistakes, but it still won't take my answer.

 

[da_willem]

Both your method and your answer look correct to me.

EDIT: Vincentchan is absolutely correct. Choose C so that x(t=0)=0.

 

[vincentchan]

$$-1/27*e^{-3t}(9t^2+6t+2)$$
wrong

the correct one is:
$$-1/27*e^{-3t}(9t^2+6t+2)+C$$

now you need to determine C by the initial condition... (what is the distance travels when t = 0?)

 

[ProBasket]

$$-1/27*e^{-3t}(9t^2+6t+2)$$
wrong

the correct one is:
$$-1/27*e^{-3t}(9t^2+6t+2)+C$$

now you need to determine C by the initial condition... (what is the distance travels when t = 0?)

so are you telling me to plug in 0 for t? if so, i did and got -0.074074 and it was wrong.

 

[da_willem]

so are you telling me to plug in 0 for t? if so, i did and got -0.074074 and it was wrong.

The thing wrong with your answer: $$-1/27*e^{-3t}(9t^2+6t+2)$$ is that it says that at time t=0 you have traveled -2/27 m. This is because you (and I) forgot the integration constant. You can use this to fix your initial conditions.

Now if at time t=0 you indeed traveled -2/27m you inegration constant can be set zero and your iniial answer is correct. But as it asks for the distance traveled at time t, thi has to be zero at t=0. So choose your integration constant such that x(t=0)=0.