The acceleration of a particle at time $t$ $t$ seconds is given by $a = 6 t$ $a = 6t$ . Given that $v = 10$ $v=10$ when $t = 0$ $t=0$ and that $x = 7$ $x=7$ when $t = 0$ $t=0$ find $x$ $x$ when $t = 2$ $t=2$ ?

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Answer 1 · 2017-02-18T22:34:34

The position $x$ $x$ at time $t$ $t$ is given by;

$x = t^{3} + 10 t + 7$ $x = t^3 + 10t + 7$

And so when $t = 2$ $t=2$ we have:

$x = 35$ $x = 35$

Acceleration ( $a$ $a$ ) is defined as the rate of change of velocity ( $v$ $v$ ) wrt time ( $t$ $t$ ). thus

$a = \frac{d v}{d t} \Rightarrow \frac{d v}{d t} = 6 t$ $a = (dv)/dt => (dv)/dt = 6t$

This is a First Order separable Differential Equation which we can just integrate to get:

$v = 3 t^{2} + A$ $v = 3t^2 + A$

Using the initial condition $v = 10$ $v=10$ when $t = 0$ $t=0$ we get:

$\ \ \ 10 = 0 + A => A=10$
$:. v = 3t^2 + 10$

Velocity ( $v$ ) is defined as the rate of change of position ( $x$ ) wrt time ( $t$ ). thus

$v = (dx)/dt => (dx)/dt = 3t^2 + 10$

Again this is a First Order separable Differential Equation which we can just integrate to get:

$x = t^3 + 10t + B$

And using the initial condition $x=7$ when $t=0$ we get:

$7 = 0 + 0 + B => B= 7$

Hence the position $x$ at time $t$ is given by;

$x = t^3 + 10t + 7$

And so when $t=2$ we have:

$x = 2^3 + 10*2 + 7 =35$

The acceleration of a particle at time ttt seconds is given by a = 6ta=6t a = 6t. Given that v=10v=10v=10 when t=0t=0t=0 and that x=7x=7x=7 when t=0t=0t=0 find xxx when t=2t=2t=2?