Variable Force and Acceleration

This lesson covers problems where the force (and hence the acceleration) is not constant but depends on time, position, or velocity. Calculus is the essential tool.

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1 Three Cases

Case	Equation of motion	Method
$F = f(t)$F=f(t)	$m\frac{\mathrm{d}v}{\mathrm{d}t} = f(t)$mdtdv​=f(t)	Integrate w.r.t. $t$t to find $v(t)$v(t), then integrate again for $x(t)$x(t)
$F = f(x)$F=f(x)	$m v\frac{\mathrm{d}v}{\mathrm{d}x} = f(x)$mvdxdv​=f(x)	Integrate w.r.t. $x$x (work-energy approach)
$F = f(v)$F=f(v)	$m\frac{\mathrm{d}v}{\mathrm{d}t} = f(v)$mdtdv​=f(v)	Separate variables: $\frac{m\,\mathrm{d}v}{f(v)} = \mathrm{d}t$f(v)mdv​=dt

The key identity linking the cases is:

$$a = \frac{\mathrm{d}v}{\mathrm{d}t} = v\frac{\mathrm{d}v}{\mathrm{d}x}$$a=dtdv​=vdxdv​

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2 Case 1: $F = f(t)$F=f(t) — Force as a Function of Time

Method

$m\frac{\mathrm{d}v}{\mathrm{d}t} = f(t)$mdtdv​=f(t)

Integrate: $v = v_0 + \frac{1}{m}\int_0^t f(\tau)\,\mathrm{d}\tau$v=v0​+m1​∫0t​f(τ)dτ

Then: $x = x_0 + \int_0^t v(\tau)\,\mathrm{d}\tau$x=x0​+∫0t​v(τ)dτ

Example 1

A particle of mass 2 kg starts from rest. A force $F = 6t$F=6t N acts on it. Find the velocity and displacement at $t = 3$t=3 s.

Solution

$2\frac{\mathrm{d}v}{\mathrm{d}t} = 6t$2dtdv​=6t

$\frac{\mathrm{d}v}{\mathrm{d}t} = 3t$dtdv​=3t

$v = \frac{3t^2}{2}$v=23t2​ (using $v = 0$v=0 at $t = 0$t=0).

At $t = 3$t=3: $v = 3(9)/2 = 13.5$v=3(9)/2=13.5 m s$^{-1}$−1.

$x = \int_0^3 \frac{3t^2}{2}\,\mathrm{d}t = \left[\frac{t^3}{2}\right]_0^3 = \frac{27}{2} = 13.5$x=∫03​23t2​dt=[2t3​]03​=227​=13.5 m.

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3 Case 2: $F = f(x)$F=f(x) — Force as a Function of Position

Method

Using $a = v\,\mathrm{d}v/\mathrm{d}x$a=vdv/dx:

$mv\frac{\mathrm{d}v}{\mathrm{d}x} = f(x)$mvdxdv​=f(x)

$\int mv\,\mathrm{d}v = \int f(x)\,\mathrm{d}x$∫mvdv=∫f(x)dx

$\frac{1}{2}mv^2 = \int f(x)\,\mathrm{d}x + C$21​mv2=∫f(x)dx+C

This is the work-energy theorem in integral form.

Example 2

A particle of mass 3 kg starts from rest at $x = 0$x=0. A force $F = 12 - 2x$F=12−2x N acts on it. Find the velocity when $x = 4$x=4 m.

Solution

$\frac{1}{2}(3)v^2 = \int_0^4 (12 - 2x)\,\mathrm{d}x = \left[12x - x^2\right]_0^4 = 48 - 16 = 32$21​(3)v2=∫04​(12−2x)dx=[12x−x2]04​=48−16=32

$\frac{3}{2}v^2 = 32$23​v2=32

$v^2 = 64/3$v2=64/3

$v = 8/\sqrt{3} \approx 4.62$v=8/3​≈4.62 m s$^{-1}$−1.

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Example 3 — Inverse Square Law Force

A particle of mass $m$m is attracted towards a fixed point O by a force $F = k/x^2$F=k/x2 (directed towards O). The particle starts from rest at distance $a$a from O. Find the speed at distance $x$x from O.

Solution

Taking the direction towards O as positive:

$mv\frac{\mathrm{d}v}{\mathrm{d}x} = -\frac{k}{x^2}$mvdxdv​=−x2k​ (negative because force is towards O but $x$x is measured outward)

Wait — let us set up carefully. Let $x$x be the distance from O, decreasing. The force towards O is $+k/x^2$+k/x2, and $v = \mathrm{d}x/\mathrm{d}t < 0$v=dx/dt<0 (moving towards O). Use energy:

$\frac{1}{2}mv^2 - 0 = \int_a^x \frac{k}{s^2}\,(-\mathrm{d}s) = k\left[\frac{1}{s}\right]_a^x = k\left(\frac{1}{x} - \frac{1}{a}\right)$21​mv2−0=∫ax​s2k​(−ds)=k[s1​]ax​=k(x1​−a1​)