Irodov 1.36: Only the X-Projection Ever Mattered – Amit & Deepali Sinha Foundation

Problem Irodov 1.36

A small particle travels in a fixed sense along a prescribed curve. At every point of the path its tangential acceleration equals the scalar product $w_\tau = \vect a\cdot\vect\tau$ , where $\vect a$ is a constant vector directed along the $x$ axis and $\vect\tau$ is the unit vector tangent to the path, pointing along the particle’s velocity. The particle’s speed is negligible at the point where $x = 0$ . Determine how the particle’s speed depends on its coordinate $x$ .

Given

$w_\tau = \vect a\cdot\vect\tau$ , $\vect a\parallel x$
$v=0$ at $x=0$

Find

v(x)

Solution

Step 1Read the tangential acceleration as a projection

The tangential (speed-changing) acceleration is the dot product of the constant vector $\vect a$ with the unit tangent $\vect\tau$ . Since $\vect\tau = d\vect r/ds$ (the derivative of position with respect to arc length $s$ ) and $\vect a$ points along $x$ with constant magnitude $a$ ,

w_\tau = \vect a\cdot\vect\tau = a\,\vect{i}\cdot\frac{d\vect r}{ds} = a\,\frac{dx}{ds}. \tag{1}

Step 2Rewrite

dv/dt

in terms of

x

and

s

By definition $w_\tau = dv/dt$ . Using the chain rule twice, first through $s$ and then through $x$ ,

\frac{dv}{dt} = \frac{dv}{ds}\frac{ds}{dt} = v\frac{dv}{ds}. \tag{2}

Combining with (1),

v\frac{dv}{ds} = a\frac{dx}{ds}. \tag{3}

Both sides are rates with respect to the same arc length, so $ds$ cancels and the two differentials are left paired.

v\,dv = a\,dx

Step 3Integrate from rest at

x=0

The particle starts at rest at $x=0$ , so integrating the paired differentials from that state to a general one runs each side over its own variable.

\int_0^v v'\,dv' = \int_0^x a\,dx' \tag{4}

Evaluating both integrals,

\frac{v^2}{2} = ax.

Multiplying through by 2,

v^2 = 2ax.

v = \sqrt{2ax}

(5)

Check

Dimensionally, $[a][x] = (\text{m/s}^2)(\text{m}) = \text{m}^2/\text{s}^2$ , and its square root is a speed, as required. The result also depends on the path only through $x$ , never through the path’s shape or length, which matches Step 1: the tangential acceleration was always just $a\,dx/ds$ , so the work-like integral $\int w_\tau\,ds = \int a\,dx$ only ever sees the endpoints’ $x$ coordinates.

Remark

The integral (4), which is $\int w_\tau\,ds=\int a\,dx$ , is the work-energy theorem for a uniform ”field” pointing along $x$ : the work done depends only on the net $x$ -displacement, never on the path’s shape, exactly the way gravity’s work depends only on height dropped and not on the path taken to drop it. The constant vector $\vect a$ plays the role a uniform force per unit mass would play in an energy argument, even though no force was ever introduced.