Newtons Laws and integration

Question

A particle of mass m moves along a straight line (which, without loss of generality we may consider to be the $x$-axis) under the infuence of a constant force $F$.

Suppose that the particle starts at $x = 0$ at $t = 0$ with a velocity $v_0{\bf i}$, ($v_0 > 0$). Find:

(a) the speed,

(b) the velocity as a function of time,

(c) the distance traveled after time $t$,

(d) the speed as a function of position ($x$).

For (a) I assumed it was $v_0$ indepenedent of direction

For (b) I integrated

$$m\frac{d^2 x}{dt^2}{\bf i}=F{\bf i}$$

between 0 and t to get

$$\frac{d}{dt} x(t) {\bf i}= \left(v_0+\frac{Ft}{m}\right){\bf i}$$

Is this integral correct?

For (c) I integrated again to get

$x(t){\bf i}=(v_0 t +\frac{Ft^2}{2m}){\bf i}$

For (d) I believe I need to eliminate time? Should I rearrange b to get

$t=\frac{m}{F} (\frac{dx(t)}{dt}-v_0$)

Then sub back in to c? How do the i components work in this case?

To be a function of position do I need an equation $x = f(x)$?

Answer 1

Since the particle moves only along the x-axis, the displacement/velocity/acceleration have only one component. So the $i$ can be dropped.
I believe part (d) requires an expression of the form $\frac{dx}{dt} = f(x)$, which is the velocity as a function of the position. You have an equation for $t$ from rearranging b (velocity equation). Substituting this into c (the equation for position) will yield a quadratic equation for $\frac{dx}{dt}$ with 2 solutions. Most likely, one of the solutions would yield negative values for velocity which, in this case, is not possible and can be discarded.

Answer 2

It's a simple mechanics problem, so the thing you have to do is to solve de differential equations, which by the way are easy to calculate in this case, so far we have $\mathbf v(\mathbf x=0)=v_0\mathbf i$, so the trick you should use

$$ \frac{d\mathbf v}{dt} =\frac{d\mathbf x}{dt}\frac{d\mathbf v}{d\mathbf x}=\frac{d\mathbf v}{d\mathbf x}\cdot \mathbf v $$

and with that solve for (d).