ACPW help (1 Viewer)

Jkitty · Jul 4, 2015

the velocity of a particle is given by dx/dt=4x cms^-1. if the particle is initially at rest 5cm to the right of the origin
a) find the equation of its displacement in terms of t
d))show that the particle is never at the origin

InteGrand · Jul 4, 2015

Jkitty said:
the velocity of a particle is given by dx/dt=4x cms^-1. if the particle is initially at rest 5cm to the right of the origin
a) find the equation of its displacement in terms of t
d))show that the particle is never at the origin

$(a) $\frac{\mathrm{d}x}{\mathrm{d}t}=4x\Rightarrow \frac{\mathrm{d}t}{\mathrm{d}x}=\frac{1}{4x}$.$

$Integrating with respect to $x$, $t=\frac{1}{4}\ln x + c$, for some constant $c$.$

$Substitute $x=5,t=0$, then $0=\frac{1}{4}\ln 5 + c\Rightarrow c = -\frac{1}{4}\ln 5$.$

$So $t-c=\frac{1}{4}\ln x \Rightarrow 4(t-c)=\ln x \Rightarrow \ln x = 4\left(t+\frac{1}{4}\ln 5\right)=4t + \ln 5$.$

$Hence $x=\mathrm{e}^{4t+\ln 5}=\mathrm{e}^{\ln 5}\mathrm{e}^{4t}$$

$\Rightarrow x(t)=5\mathrm{e}^{4t}.$

$(b) Since $x(t)=5\mathrm{e}^{4t}$, $x$ is always positive, since $\mathrm{e}^{4t}$ is always positive, so $x\neq 0$, as required.$

$(Incidentally, we cannot have the particle be at rest when it is at $x=5$ like the question states at the start, since at that point, $v=\frac{\mathrm{d}x}{\mathrm{d}t}=4\times 5 = 20 \neq 0$.)$

InteGrand · Jul 4, 2015

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ACPW help (1 Viewer)

Jkitty

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InteGrand

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