What is the position of a particle at time $t=2$ if a particle moves along the x axis so that at any time $t>0$ , the velocity is given by $v(t)=4-6t^2$ , and at position $x=7$ at $t=1$ ?

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Answer 1 · 2015-02-05T14:40:09

The answer is: $x=-3$ .

The law that gives the position of the particle at the time $t$ is given by the integral:

$x(t)=int(4-6t^2)dt=4t-6t^3/3+c=4t-2t^3+c$ .

To find $c$ , we can use the intial conditions:

$x(1)=7$ ,

so:

$7=4-2+crArrc=5$ .

So the law becomes:

$x(t)=4t-2t^3+5$ .

Now we can use it to find the positition at the time $t=2$ :

$x(2)=8-16+5=-3$ .

What is the position of a particle at time t=2t=2 if a particle moves along the x axis so that at any time t>0t>0, the velocity is given by v(t)=4-6t^2v(t)=4-6t^2, and at position x=7x=7 at t=1t=1?