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).