Question

@Michele_Laino

Answer 1

can ou check my answer, i got -3

Answer 2

3 , not -3

Answer 3

distance s which your particle has travelled, is given by the subsequent integral:
\[s(t)=\int\limits_{0}^{3}(t ^{2}-4)dt\]
since, by definition, we have:
\[\frac{ ds }{ dt }=v(t)\]
so, please solve that integral

Answer 4

oops...s not s(t)

Answer 5

3

Answer 6

sorry, this formula:
\[v(t)=t ^{2}-4\]
is right?

Answer 7

5?

Answer 8

I think that there is an error in your formula
v(t)=t^2-4
is it possible?

Answer 9

yes

Answer 10

because I got s=-3 which is impossible!

Answer 11

i got -3 first also

Answer 12

ok!, I understand, then position of our particle is that:

Answer 13

so our particle has made a distance of 3

Answer 14

it is 3, since distance has to be Always positive, whereas the final point where is our particle, is (-3,0)

Answer 15

thanks!