Question

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

Answer 1

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

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

And so when `t=2` we have:

`x = 35`

Explanation:

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

`a = (dv)/dt => (dv)/dt = 6t`

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

`v = 3t^2 + A`

Using the initial condition `v=10` when `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`